Intermediate mathematics/Numbers

## Scalars

See also: Peano axiomsw, Hyperoperation, Algebraic extension
The basis of all of mathematics is the "Next" function. See Graph theoryw.

Next(0)=1
Next(1)=2
Next(2)=3
Next(3)=4
Next(4)=5
We might express this by saying that One differs from nothing as two differs from one. This defines the Natural numbersw (denoted

N

0

{\displaystyle \mathbb{N}\_0}

). Natural numbers are those used for counting.

These have the convenient property of being transitivew. That means that if a<b and b<c then it follows that a<c. In fact they are totally orderedw. See Order theory.
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### Integers

Additionw (See Tutorial:arithmetic) is defined as repeatedly calling the Next function, and its inverse is subtractionw. But this leads to the ability to write equations like

1
−
# 3
x

{\displaystyle 1-3=x}

 for which there is no answer among natural numbers. To provide an answer mathematicians generalize to the set of all integersw (denoted

Z

{\displaystyle \mathbb{Z}}

 because zahlen means count in german) which includes negative integers.

The Additive identityw is zero because x + 0 = x.
The absolute value or modulus of x is defined as

x

=

{

x
,

if
x
≥
0

−
x
,

if
x
<

{\displaystyle |x| = \left\{
 \begin{array}{rl}
 x, & \text{if } x \geq 0 \\
 -x, & \text{if } x < 0.
 \end{array}\right.
 }

Integers form a ring (denoted

O

Q

{\displaystyle \mathcal O\_\mathbb{Q}}

) over the field of rational numbers. Ringw is defined below.
**Z***n* or

Z

/

n

Z

{\displaystyle \mathbb{Z}/n\mathbb{Z}}

 is used to denote the set of integers modulo *n* .
Modular arithmetic is essentially arithmetic in the quotient ringw **Z**/*n***Z** (which has *n* elements).
Consider the ring of integersw **Z** and the ideal of even numbers, denoted by 2**Z**. Then the quotient ring **Z** / 2**Z** has only two elements, zero for the even numbers and one for the odd numbers; applying the definition, [*z*] = *z* + 2**Z** := {*z* + 2*y*: 2*y* ∈ 2**Z**}, where 2**Z** is the ideal of even numbers. It is naturally isomorphic to the finite fieldw with two elements, **F**2. Intuitively: if you think of all the even numbers as 0, then every integer is either 0 (if it is even) or 1 (if it is odd and therefore differs from an even number by 1).
An ideal is a special subset of a ring. Ideals generalize certain subsets of the integers, such as the even numbers or the multiples of 3.
A principal ideal is an ideal

I

{\displaystyle I}

 in a ring

R

{\displaystyle R}

 that is generated by a single element

a

{\displaystyle a}

 of

R

{\displaystyle R}

 through multiplication by every element of

R

{\displaystyle R}

.
A prime ideal is a subset of a ring that shares many important properties of a prime number in the ring of integers. The prime ideals for the integers are the sets that contain all the multiples of a given prime number.
The study of integers is called Number theoryw.

a
∣
b

{\displaystyle a \mid b}

 means a divides b.

a
∤
b

{\displaystyle a \nmid b}

 means a does not divide b.

p
a

∣∣
n

{\displaystyle p^a \mid\mid n}

 means pa exactly divides n (i.e. pa divides n but pa+1 does not).
A prime number is a number that can only be divided by itself and one.
If a, b, c, and d are primes and x=abc and y=c2d then:
xy = lcmw \* gcdw = abc2d \* c
Two integers a and b are said to be relatively prime, mutually prime, or coprime if the only positive integer that divides both of them is 1. Any prime number that divides one does not divide the other. This is equivalent to their greatest common divisor (gcd) being 1.
(See Tutorial:least common multiples)

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### Rational numbers

Multiplicationw (See Tutorial:multiplication) is defined as repeated addition, and its inverse is divisionw. But this leads to equations like

3

/

# 2
x

{\displaystyle 3/2=x}

 for which there is no answer. The solution is to generalize to the set of rational numbersw (denoted

Q

{\displaystyle \mathbb{Q}}

) which include fractions (See Tutorial:fractions). Any number which isnt rational is irrationalw. See also p-adic number

The set of all rational numbers except zero forms a multiplicative group which is a set of invertible elements.
Rational numbers form a division algebra because every non-zero element has an inverse. The ability to find the inverse of every element turns out to be quite useful. A great deal of time and effort has been spent trying to find division algebras.
Rational numbers form a Fieldw. A field is a set on which addition, subtraction, multiplication, and division are defined. See below.
The Multiplicative identityw is one because x \* 1 = x.
Division by zero is undefined and undefinablew. 1/0 exists nowhere on the complex planew. It does, however, exist on the Riemann spherew (often called the extended complex plane) where it is surprisingly well behaved. See also Wheel theory and L'Hôpital's rulew.

1

n
o
t
h
i
n
g

=
e
v
e
r
y
t
h
i
n
g

{\displaystyle \frac{1}{nothing}=everything}

(Addition and multiplication are fast but division is slow even for computers.)

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### Irrational and complex numbers

Exponentiationw (See Tutorial:exponents) is defined as repeated multiplication, and its inverses are rootsw and logarithmsw. But this leads to multiple equations with no solutions:

Equations like

2

=
x
.

{\displaystyle \sqrt{2}=x.}

 The solution is to generalize to the set of algebraic numbersw (denoted

A

{\displaystyle \mathbb{A}}

). (See also algebraic integer and algebraically closedw.) To see a proof that the square root of two is irrational see Square root of 2w.

Equations like

2

2

=
x

{\displaystyle 2^{\sqrt{2}}=x}

 The solution (because x is transcendentalw) is to generalize to the set of Real numbersw (denoted

R

{\displaystyle \R}

).

Equations like

−
1

=
x

{\displaystyle \sqrt{-1}=x}

 and

e
x

=
−

{\displaystyle e^x=-1.}

 The solution is to generalize to the set of complex numbersw (denoted

C

{\displaystyle \mathbb{C}}

) by defining iw = sqrt(-1). A single complex number

# z
a
b
i

{\displaystyle z=a+bi}

 consists of a real part *a* and an imaginary part *bi* (See Tutorial:complex numbers). Imaginary numbersw (denoted

I

{\displaystyle \mathbb{I}}

) often occur in equations involving change with respect to **time**. If friction is resistance to motion then imaginary friction would be resistance to change of motion wrt **time**. (In other words, imaginary friction would be mass.) In fact, in the equation for the Spacetime intervalw, **time** itself is an imaginary quantity.
The Complex conjugatew of the complex number

# z
a
b
i

{\displaystyle z=a+bi}

 is

z
¯

=
a
−
b
i
.

{\displaystyle \overline{z}=a-bi.}

 (Not to be confused with the dualw of a vector.)

1

a
b
i

=

1

a
b
i

⋅

a
−
b
i

a
−
b
i

=

a
−
b
i

a
2

b
2

{\displaystyle \frac{1}{a+bi} = \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} = \frac{a-bi}{a^2+b^2}}

Complex numbers form an Algebra over a field (K-algebra) because complex multiplication is Bilinear.

−
100

∗

−
100

=
10
i
∗
10
# i
−
100
≠

−
100
∗
−
100

{\displaystyle \sqrt{-100} \* \sqrt{-100} = 10i \* 10i = -100 \neq \sqrt{-100 \* -100}}

The complex numbers are not orderedw. However the absolute valuew or modulus of a complex number is:

z

=

a
i
b

=

a
2

b
2

A Gaussian integer *a* + *bi* is a Gaussian prime if and only if either:
* one of *a*, *b* is zero and absolute value of the other is a prime number of the form 4*n* + 3 (with n a nonnegative integer), or
* both are nonzero and *a*2 + *b*2 is a prime number (which will *not* be of the form 4*n* + 3).

There are n solutions of

z
n

{\displaystyle \sqrt[n]{z}}

For n>1 the n solutions add up to zero
0^0w = 1. See Empty productw.

log
b

⁡
(
x
# )

log
a

⁡
(
x
)

log
a

⁡
(
b
)

{\displaystyle \log\_b(x) = \frac{\log\_a(x)}{\log\_a(b)}}

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#### Completeness

The set of rational numbers is not complete because it is possible to create an infinite sequence of smaller and smaller rational numbers whose sum converges to a number that is not a rational number.

The set of real numbers is complete.

### Hypercomplex numbers

Complex numbers can be used to represent and perform rotationsw but only in 2 dimensions. Hypercomplex numbersw like quaternionsw (denoted

H

{\displaystyle \mathbb{H}}

), octonionsw (denoted

O

{\displaystyle \mathbb{O}}

), and sedenions (denoted

S

{\displaystyle \mathbb{S}}

) are one way to generalize complex numbers to some (but not all) higher dimensions.

A quaternion can be thought of as a complex number whose coefficients are themselves complex numbers (hence a hypercomplex number).

(
a
b

ı
^

)
(
c
d

ı
^

)

ȷ
^

=
a
b

ı
^

c

ȷ
^

d

ı
^

ȷ
^

=
a
b

ı
^

c

ȷ
^

d

k
^

{\displaystyle (a + b\boldsymbol{\hat{\imath}}) + (c + d\boldsymbol{\hat{\imath}})\boldsymbol{\hat{\jmath}} = a + b\boldsymbol{\hat{\imath}} + c\boldsymbol{\hat{\jmath}} + d\boldsymbol{\hat{\imath}\hat{\jmath}} = a + b\boldsymbol{\hat{\imath}} + c\boldsymbol{\hat{\jmath}} + d\boldsymbol{\hat{k}}}

Where

ı
^

2

=

ȷ
^

2

=

k
^

2

=

ı
^

ȷ
^

k
^

=
−
1

{\displaystyle \boldsymbol{\hat{\imath}}^2 = \boldsymbol{\hat{\jmath}}^2 = \boldsymbol{\hat{k}}^2 = \boldsymbol{\hat{\imath}} \boldsymbol{\hat{\jmath}} \boldsymbol{\hat{k}} = -1}

and

ı
^

ȷ
^

=

k
^

,

ȷ
^

ı
^

=
−

k
^

,

ȷ
^

k
^

=

ı
^

,

k
^

ȷ
^

=
−

ı
^

,

k
^

ı
^

=

ȷ
^

,

ı
^

k
^

=
−

ȷ
^

.

{\displaystyle \begin{alignat}{2}
\boldsymbol{\hat{\imath}}\boldsymbol{\hat{\jmath}} & = \boldsymbol{\hat{k}}, & \qquad \boldsymbol{\hat{\jmath}}\boldsymbol{\hat{\imath}} & = -\boldsymbol{\hat{k}}, \\
\boldsymbol{\hat{\jmath}}\boldsymbol{\hat{k}} & = \boldsymbol{\hat{\imath}}, & \boldsymbol{\hat{k}}\boldsymbol{\hat{\jmath}} & = -\boldsymbol{\hat{\imath}}, \\
\boldsymbol{\hat{k}}\boldsymbol{\hat{\imath}} & = \boldsymbol{\hat{\jmath}}, & \boldsymbol{\hat{\imath}}\boldsymbol{\hat{k}} & = -\boldsymbol{\hat{\jmath}}.
\end{alignat}}

Any real finite-dimensional division algebra over the reals must be:

* isomorphic to **R** or **C** if unitary and commutativew (equivalently: associativew and commutativew)
* isomorphic to the quaternionsw if noncommutativew but associativew
* isomorphic to the octonionsw if non-associativew but alternativew.

The following is known about the dimension of a finite-dimensional division algebra *A* over a field *K*:

* dim *A* = 1 if *K* is algebraically closedw,
* dim *A* = 1, 2, 4 or 8 if *K* is real closed, and
* If *K* is neither algebraically nor real closed, then there are infinitely many dimensions in which there exist division algebras over *K*.

Split-complex numbers (hyperbolic complex numbers) are similar to complex numbers except that i2 = +1.

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### Tetration

Tetrationw is defined as repeated exponentiation and its inverses are called super-root and super-logarithm.

b

a

=

a

a

.

.

.

a

⏟

=

a
↑↑
# b

a
(
a
(
⋯
a
)
)

⏟

b

 copies of

a

b

 copies of

a

{\displaystyle

\begin{matrix}

 {}^{b}a &

 = &

 \underbrace{a^{a^{{}^{.\,^{.\,^{.\,^a}}}}}} &

 = &

 a\uparrow\uparrow b

 = &

 \underbrace{a\uparrow (a\uparrow(\dots\uparrow a))} &

\\

 & & b\mbox{ copies of }a

 &

 & & b\mbox{ copies of }a

\end{matrix}

}

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### Hyperreal numbers

See also: Non-standard calculus
When a quantity, like the charge of a single electron, becomes so small that it is *insignificant* we, quite justifiably, treat it as though it were zero. A quantity that can be treated as though it were zero, even though it **very definitely is not**, is called infinitesimal. If

q

{\displaystyle q}

 is a finite

(
q
⋅
1
)

{\displaystyle ( q \cdot 1 )}

 amount of charge then using Leibniz's notationw

d
q

{\displaystyle dq}

 would be an infinitesimal

(
q
⋅
1

/

∞
)

{\displaystyle ( q \cdot 1/\infty )}

 amount of charge. See Differentialw

Likewise when a quantity becomes so large that a regular finite quantity becomes insignificant then we call it infinite. We would say that the mass of the ocean is infinite

(
M
⋅
∞
)

{\displaystyle ( M \cdot \infty )}

. But compared to the mass of the Milky Way galaxy our ocean is insignificant. So we would say the mass of the Galaxy is doubly infinite

(
M
⋅

∞
2

)

{\displaystyle ( M \cdot \infty^2 )}

.

Infinity and the infinitesimal are called Hyperreal numbersw (denoted

∗

R

{\displaystyle {}^\*\mathbb{R}}

). Hyperreals behave, in every way, exactly like real numbers. For example,

2
⋅
∞

{\displaystyle 2 \cdot \infty}

 is exactly twice as big as

∞
.

{\displaystyle \infty.}

 In reality, the mass of the ocean **is** a real number so it is hardly surprising that it behaves like one. See Epsilon numbers and Big O notation

In ancient times infinity was called the "all".

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### Numbers dont lie. (But they sure help)

From Wikipedia:Mathematical fallacy:

a

=
b

a
2

=
a
b

a
2

−

b
2

=
a
b
−

b
2

(
a
−
b
)
(
a
b
)

=
b
(
a
−
b
)

a
b

=
b

b
b

=
b

2
b

=
b

2

=
1

{\displaystyle
\begin{align}
a &= b \\
a^2 &= ab \\
a^2 - b^2 &= ab - b^2 \\
(a - b)(a + b) &= b(a - b) \\
a + b &= b \\
b + b &= b \\
2b &= b \\
2 &= 1
\end{align}
}

The fallacy is in line 5: the progression from line 4 to line 5 involves division by *a* − *b*, which is zero since *a* = *b*. Since division by zerow is undefined, the argument is invalid.

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### Intervals

[-2,5[ or [-2,5) denotes the intervalw from -2 to 5, including -2 but excluding 5.
[3..7] denotes all integers from 3 to 7.
The set of all reals is unbounded at both ends.
An open interval does not include its endpoints.
Compactness is a property that generalizes the notion of a subset being closed and bounded.
The unit interval is the closed interval [0,1]. It is often denoted I.
The unit square is a square whose sides have length 1.
Often, "the" unit square refers specifically to the square in the Cartesian planew with corners at the four points (0, 0), (1, 0), (0, 1), and (1, 1).
The unit disk in the complex plane is the set of all complex numbers of absolute value less than one and is often denoted

D

{\displaystyle \mathbb {D}}

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## Vectors

See also: Algebraic geometry, Algebraic variety, Scheme, Algebraic manifold, and Linear algebraw
The one dimensional number line can be generalized to a multidimensional Cartesian coordinate systemw thereby creating multidimensional math (i.e. geometryw). See also Curvilinear coordinates

(Note: It is a mistake to think of scalars as one dimensional vectors. Scalars have magnitude but no direction. This becomes clear when multiplying a scalar with a vector. It will become even more clear when we get to multivectors where we see that trivectors are made of 3 vectors, bivectors are made of 2 vectors, vectors are made of 1 vector, and scalars are made of 0 vectors.)

For sets *A* and *B*, the Cartesian productw *A* × *B* is the set of all ordered pairsw (*a*, *b*) where *a* ∈ *A* and *b* ∈ *B*.

R

3

{\displaystyle \mathbb R^{3}}

 is the Cartesian product

R

×

R

×

R

.

{\displaystyle \mathbb{R} \times \mathbb{R} \times \mathbb{R}.}

R

∞

=

R

N

{\displaystyle \mathbb{R}^\infty = \mathbb{R}^\mathbb{N}}

C

3

{\displaystyle \mathbb{C}^3}

 is the Cartesian productw

C

×

C

×

C

{\displaystyle \mathbb{C} \times \mathbb{C} \times \mathbb{C}}

 (See Complexification)

If we think of

R

{\displaystyle \R}

 as the set of real numbers, then the direct product

R

×

R

{\displaystyle \R \times \R}

 is precisely just the Cartesian product.
If we think of

R

{\displaystyle \R}

 as the group of real numbers under addition, then the direct product

R

×

R

{\displaystyle \R \times \R}

 still consists of the Cartesian product. The difference is that

R

×

R

{\displaystyle \R \times \R}

 is now a group. We have to also say how to add their elements.
To add ordered pairs, we define the sum

(
a
,
b
)
(
c
,
d
)

{\displaystyle (a, b) + (c, d)}

 to be

(
a
c
,
b
d
)

{\displaystyle (a + c, b + d)}

.
However, if we think of

R

{\displaystyle \R}

 as the field of real numbers, then the direct product

R

×

R

{\displaystyle \R \times \R}

 does not exist – naively defining

{
(
x
,
y
)

x
,
y
∈

R

}

{\displaystyle \{ (x,y) | x,y \in \mathbb{R} \}}

 in a similar manner to the above examples would not result in a field since the element

(
1
,
0
)

{\displaystyle (1,0)}

 does not have a multiplicative inverse.
If the arithmetic operation is written as +, as it usually is in abelian groups, then we use the direct sum. If the arithmetic operation is written as × or ⋅ or using juxtaposition (as in the expression

x
y

{\displaystyle xy}

A vector spacew is a coordinate spacew with vector additionw and scalar multiplicationw (multiplication of a vector and a scalarw belonging to a fieldw).

e

1

,

e

2

,

e

3

{\displaystyle {\mathbf e\_1} , {\mathbf e\_2} , {\mathbf e\_3} }

 are orthogonalw unitw basis vectors
If

u

,

v

,

x

{\displaystyle {\mathbf u} , {\mathbf v} , {\mathbf x} }

 are arbitrary vectors
and

u
n

,

v
n

,

x
n

{\displaystyle u\_n , v\_n , x\_n}

 are scalars belonging to a field then we can (and usually do) write:
*u

=

u
1

e
1

u
2

e
2

u
3

e
3

=

[

u
1

u
2

u
3

]

{\displaystyle \mathbf{u} = u\_1 \mathbf{e\_1} + u\_2 \mathbf{e\_2} + u\_3 \mathbf{e\_3} = \begin{bmatrix} u\_1 & u\_2 & u\_3 \end{bmatrix}}*
*v

=

v
1

e
1

v
2

e
2

v
3

e
3

=

[

v
1

v
2

v
3

]

{\displaystyle \mathbf{v} = v\_1 \mathbf{e\_1} + v\_2 \mathbf{e\_2} + v\_3 \mathbf{e\_3} = \begin{bmatrix} v\_1 & v\_2 & v\_3 \end{bmatrix}}*
*x

=

x
1

e
1

x
2

e
2

x
3

e
3

=

[

x
1

x
2

x
3

]

{\displaystyle \mathbf{x} = x\_1 \mathbf{e\_1} + x\_2 \mathbf{e\_2} + x\_3 \mathbf{e\_3} = \begin{bmatrix} x\_1 & x\_2 & x\_3 \end{bmatrix}}*
See also: Linear independencew
A module generalizes a vector space by allowing multiplication of a vector and a scalar belonging to a ringw.
Coordinate systems define the length of vectors parallel to one of the axes but leave all other lengths undefined. This concept of "length" which only works for certain vectors is generalized as the "normw" which works for all vectors. The norm of vector

v

{\displaystyle \mathbf{v} }

 is denoted

‖

v

‖
.

{\displaystyle \|\mathbf{v}\|.}

 The double bars are used to avoid confusion with the absolute value of the function.

Taxicab metricw (called *L**1* norm. See *L**p* space. Sometimes called Lebesgue spaces. See also Lebesgue measurew.) A circle in *L**1* space is shaped like a diamond.

‖

v

# ‖

v
1

v
2

v
3

{\displaystyle \|\mathbf{v}\| = v\_1 + v\_2 + v\_3}

In Euclidean spacew the norm (called *L**2* norm) doesnt depend on the choice of coordinate system. As a result, rigid objects can rotate in Euclidean space. See proof of the Pythagorean theoremw to the right.

‖

v

# ‖

v
1
2

v
2
2

v
3
2

{\displaystyle \|\mathbf{v}\| = \sqrt{v\_1^2 + v\_2^2 + v\_3^2}}

In Minkowski spacew (See Pseudo-Euclidean space) the Spacetime intervalw is

‖
s
# ‖

x
2

y
2

z
2

(
c
t
i

)
2

{\displaystyle \|s\| = \sqrt{x^2 + y^2 + z^2 + (cti)^2}}

 (Where c is the speed of light)
In complex space the most common norm of an n dimensional vector is obtained by treating it as though it were a regular real valued 2n dimensional vector in Euclidean space

‖
z
‖

=

z
1

z
¯

1

⋯

z
n

z
¯

n

{\displaystyle \left\| \boldsymbol{z} \right\| = \sqrt{z\_1 \bar z\_1 + \cdots + z\_n \bar z\_n}}

Infinity norm. (In this space a circle is shaped like a square.)

‖

x

‖

∞

:=
max

(

x
1

,
…
,

x
n

)

.

A Banach space is a normed vector space that is also a complete metric spacew (there are no points missing from it). See #Completeness above.

A manifoldw

M

{\displaystyle \mathbf{M}}

 is a type of topological spacew in which each point has an infinitely small neighbourhoodw that is homeomorphicw to Euclidean spacew. A manifold is locally, but not globally, Euclidean. A Riemannian metric on a manifold allows distances and angles to be measured.

A Tangent space

T

p

M

{\displaystyle \mathbf{T}\_p \mathbf{M}}

 is the set of all vectors tangent to

M

{\displaystyle \mathbf{M}}

 at point p.
Informally, a tangent bundle

T
M

{\displaystyle \mathbf{TM}}

 (red cylinder in image to the right) on a differentiable manifold

M

{\displaystyle \mathbf{M}}

 (blue circle) is obtained by joining all the tangent spaces (red lines) together in a smooth and non-overlapping manner. The tangent bundle always has twice as many dimensions as the original manifold.
A vector bundle is the same thing minus the requirement that it be tangent.
A fiber bundle is the same thing minus the requirement that the fibers be vector spaces.
The cotangent bundle (Dual bundle) of a differentiable manifold is obtained by joining all the cotangent spaces (pseudovector spaces).
The cotangent bundle always has twice as many dimensions as the original manifold.
Sections of that bundle are known as differential one-formsw.

A Lie group is a group that is also a finite-dimensional smooth manifold, in which the group operation is multiplication rather than addition. *n*×*n* invertible matrices (See below) are a Lie group.

A Lie algebra (See Infinitesimal transformation) is a local or linearized version of a Lie group.

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### Multiplication of vectors

In general the multiplication of two vectors is given by:

φ
(
w
,
z
# )
φ

(

∑
i

w
i

e
i

,

∑
j

z
j

e
j

)

=

∑
i

∑
j

w
i

¯

z
j

φ
(

e
i

,

e
j

# )

w

¯

T

Φ

z

.

{\displaystyle \varphi(w,z) = \varphi \left(\sum\_i w\_i e\_i, \sum\_j z\_j e\_j \right) = \sum\_i \sum\_j \overline{w\_i} z\_j \varphi(e\_i, e\_j) = {\overline{\mathbf{w}}}^\mathrm{T} \mathbf{\Phi} \mathbf{z} .}

The components of **Φ** are given by Φ*ij* = *φ*(*e**i*, *e**j*).

The matix **Φ** can take many different forms:

* Bilinear form
* Sesquilinear form
* Symplectic form

#### Dot product

Dot productw (a Scalarw):

u

⋅

v

=
‖

u

‖

‖

v

‖
cos
⁡
(
θ
)

{\displaystyle

\mathbf{u} \cdot \mathbf{v} =

}

u

⋅

v

=

[

u
1

e
1

u
2

e
2

u
3

e
3

]

[

v
1

e
1

v
2

e
2

v
3

e
3

]

=

u
1

v
1

e
1

⋅

e
1

u
1

v
2

e
1

⋅

e
2

u
1

v
3

e
1

⋅

e
3

u
2

v
1

e
2

⋅

e
1

u
2

v
2

e
2

⋅

e
2

u
2

v
3

e
2

⋅

e
3

u
3

v
1

e
3

⋅

e
1

u
3

v
2

e
3

⋅

e
2

u
3

v
3

e
3

⋅

e
3

=

u
1

v
1

(
1
)

u
1

v
2

(
0
)

u
1

v
3

(
0
)

u
2

v
1

(
0
)

u
2

v
2

(
1
)

u
2

v
3

(
0
)

u
3

v
1

(
0
)

u
3

v
2

(
0
)

u
3

v
3

(
1
)

=

u
1

v
1

(
1
)

u
2

v
2

(
1
)

u
3

v
3

(
1
)

=

u
1

v
1

u
2

v
2

u
3

v
3

{\displaystyle

\begin{align}

\mathbf{u}\cdot\mathbf{v} &=

\begin{bmatrix}u\_1

\mathbf{e\_1} \\

u\_2 \mathbf{e\_2} \\

u\_3 \mathbf{e\_3}

\end{bmatrix}

\begin{bmatrix}v\_1 \mathbf{e\_1} & v\_2 \mathbf{e\_2} & v\_3 \mathbf{e\_3}

\end{bmatrix} \\

&= u\_1v\_1\mathbf{e\_1 \cdot e\_1} + u\_1v\_2\mathbf{e\_1 \cdot e\_2} + u\_1v\_3\mathbf{e\_1 \cdot e\_3} + u\_2v\_1\mathbf{e\_2 \cdot e\_1} + u\_2v\_2\mathbf{e\_2 \cdot e\_2} + u\_2v\_3\mathbf{e\_2 \cdot e\_3} + u\_3v\_1\mathbf{e\_3 \cdot e\_1} + u\_3v\_2\mathbf{e\_3 \cdot e\_2} + u\_3 v\_3\mathbf{e\_3 \cdot e\_3} \\

&= u\_1v\_1(1) + u\_1v\_2(0) + u\_1v\_3(0) + u\_2v\_1(0) + u\_2v\_2(1) + u\_2v\_3(0) + u\_3v\_1(0) + u\_3v\_2(0) + u\_3v\_3(1) \\

&= u\_1v\_1(1) + u\_2v\_2(1) + u\_3v\_3(1) \\

&= u\_1v\_1 + u\_2v\_2 + u\_3v\_3

\end{align}}

The dot product between a scalar and a vector is not defined.
The dot product can be generalized to the bilinear formw

β
(

u
,
v

# )

u
T

A
# v
s
c
a
l
a
r

{\displaystyle \beta(\mathbf{u,v}) = u^T Av = scalar}

 where A is an (0,2) tensor. (For the dot product in Euclidean space A is the identity tensor. But in Minkowski space A is the Minkowski metric).
Two vectors are orthogonal if

β
(

u
,
v

# )

{\displaystyle \beta(\mathbf{u,v}) = 0.}

A bilinear form is symmetric if

β
(

u
,
v

# )
β
(

v
,
u

)

{\displaystyle \beta(\mathbf{u,v}) = \beta(\mathbf{v,u})}

Its associated quadratic form is

Q
(

x

# )
β
(

x
,
x

)
.

{\displaystyle Q(\mathbf{x}) = \beta(\mathbf{x,x}).}

In Euclidean space

‖

v

‖
2

=

v

⋅

v

=
Q
(

x

)
.

{\displaystyle \|\mathbf{v}\|^2 = \mathbf{v}\cdot\mathbf{v}= Q(\mathbf{x}).}

A nondegenerate bilinear form is one for which the associated matrix is invertible (its determinate is not zero)

β
(

u
,
v

# )
0

{\displaystyle \beta(\mathbf{u,v})=0 \,}

 for all *v* implies that *u* = 0.
Back to top

#### Inner product

The inner productw is a generalization of the dot product to complex vector space. (Overline represents the conjugate transpose.)

⟨
u
,
v
# ⟩

⟨
v
,
u
⟩

¯

=
u
⋅

v
¯

=
⟨
v
∣
u
⟩

{\displaystyle \langle u,v\rangle=\overline{\langle v,u\rangle}=u\cdot \bar{v}=\langle v \mid u\rangle}

 (See Bra–ket notation.)
The inner product can be generalized to a sesquilinear formw
A complex **Hermitian form** (also called a **symmetric sesquilinear form**), is a sesquilinear form *h* : *V* × *V* → **C** such that

h
(
w
,
z
# )

h
(
z
,
w
)

¯

.

{\displaystyle h(w,z) = \overline{h(z, w)}.}

A is a Hermitian operator iffw

⟨
v
∣
A
u
# ⟩
⟨
A
v
∣
u
⟩
.

{\displaystyle \langle v \mid A u\rangle = \langle A v \mid u\rangle.}

 Often written as

⟨
v
∣
A
∣
u
⟩
.

{\displaystyle \langle v \mid A \mid u\rangle.}

The curl operator,

∇
×

{\displaystyle \nabla\times}

 is Hermitian.
In an inner product space, the norm is determined using the inner product:

‖
x

‖
2

=
⟨
x
,
x
⟩
.

{\displaystyle \|x\|^2=\langle x, x\rangle.\,}

A Hilbert space is an inner product spacew that is also a Complete metric spacew.

From Wikipedia: Parallelogram law:
In an inner product space the parallelogram law is readily established using the properties of the inner product:

‖
x
y

‖
2

=
⟨
x
y
,
x
y
# ⟩
⟨
x
,
x
⟩
⟨
x
,
y
⟩
⟨
y
,
x
⟩
⟨
y
,
y
⟩
,

{\displaystyle \|x+y\|^2=\langle x+y, x+y\rangle= \langle x, x\rangle + \langle x, y\rangle +\langle y, x\rangle +\langle y, y\rangle, \, }

‖
x
−
y

‖
2

=
⟨
x
−
y
,
x
−
y
# ⟩
⟨
x
,
x
⟩
−
⟨
x
,
y
⟩
−
⟨
y
,
x
⟩
⟨
y
,
y
⟩
.

{\displaystyle \|x-y\|^2 =\langle x-y, x-y\rangle= \langle x, x\rangle - \langle x, y\rangle -\langle y, x\rangle +\langle y, y\rangle. \, }

Adding these two expressions:

‖
x
y

‖
2

‖
x
−
y

‖
2

=
2
⟨
x
,
x
⟩
2
⟨
y
,
y
# ⟩
2
‖
x

‖
2

2
‖
y

‖
2

,

which is the parallelogram identity.
If *x* is orthogonal to *y*, then

⟨
x
,

y
# ⟩
0

{\displaystyle \langle x ,\ y\rangle = 0}

 and the above equation for the norm of a sum becomes:

‖
x
y

‖
2

=
⟨
x
,
x
⟩
⟨
x
,
y
⟩
⟨
y
,
x
⟩
⟨
y
,
y
# ⟩
‖
x

‖
2

‖
y

‖
2

,

which is Pythagoras' theorem.

Every Banach space in which the parallelogram identity holds is a Hilbert space. (The parallelogram identity holds in every Hilbert space.)

The inner product is the trace of the outer product.

Inner is horizontal times vertical and shrinks down
Outer is vertical times horizontal and expands out
Back to top

#### Outer product

Outer productw (a tensorw called a dyadicw):

u

⊗

v

.

{\displaystyle \mathbf{u} \otimes \mathbf{v}.}

u

⊗

v

=

[

u
1

e
1

u
2

e
2

u
3

e
3

]

[

v
1

e
1

v
2

e
2

v
3

e
3

]

=

[

u
1

v
1

u
1

v
2

u
1

v
3

u
2

v
1

u
2

v
2

u
2

v
3

u
3

v
1

u
3

v
2

u
3

v
3

]

=

u
1

v
1

⋅

e
1

⊗

e
1

u
1

v
2

⋅

e
1

⊗

e
2

u
1

v
3

⋅

e
1

⊗

e
3

u
2

v
1

⋅

e
2

⊗

e
1

u
2

v
2

⋅

e
2

⊗

e
2

u
2

v
3

⋅

e
2

⊗

e
3

u
3

v
1

⋅

e
3

⊗

e
1

u
3

v
2

⋅

e
3

⊗

e
2

u
3

v
3

⋅

e
3

⊗

e
3

{\displaystyle
\begin{align}
\mathbf{u} \otimes \mathbf{v} &=
\begin{bmatrix}
u\_1 e\_1 \\
u\_2 e\_2 \\
u\_3 e\_3
\end{bmatrix}
\begin{bmatrix}
v\_1 e\_1 & v\_2 e\_2 & v\_3 e\_3
\end{bmatrix} =
\begin{bmatrix}

{\color{red} u\_1 v\_1 } &
{\color{blue} u\_1 v\_2 } &
{\color{blue} u\_1 v\_3 } \\

{\color{blue} u\_2 v\_1 } &
{\color{red} u\_2 v\_2 } &
{\color{blue} u\_2 v\_3 } \\

{\color{blue} u\_3 v\_1 } &
{\color{blue} u\_3 v\_2 } &
{\color{red} u\_3 v\_3 }

\end{bmatrix} \\
&=
{\color{red} u\_1 v\_1 \cdot e\_1 \otimes e\_1 } +
{\color{blue} u\_1 v\_2 \cdot e\_1 \otimes e\_2 } +
{\color{blue} u\_1 v\_3 \cdot e\_1 \otimes e\_3 } +
{\color{blue} u\_2 v\_1 \cdot e\_2 \otimes e\_1 } +
{\color{red} u\_2 v\_2 \cdot e\_2 \otimes e\_2 } +
{\color{blue} u\_2 v\_3 \cdot e\_2 \otimes e\_3 } +
{\color{blue} u\_3 v\_1 \cdot e\_3 \otimes e\_1 } +
{\color{blue} u\_3 v\_2 \cdot e\_3 \otimes e\_2 } +
{\color{red} u\_3 v\_3 \cdot e\_3 \otimes e\_3 }

\end{align}
}

For complexw vectors, it is customary to use the conjugate transposew of **v** (denoted **v**H or **v**\*):

u

⊗

v

=

u

v

H

=

u

v

∗

{\displaystyle \mathbf{u} \otimes \mathbf{v} = \mathbf{u} \mathbf{v}^\mathrm{H} = \mathbf{u} \mathbf{v}^\*\ }

Taking the dot product of **u**⊗**v** and any vector **x** (See Visualization of Tensor multiplicationw) causes the components of **x** not pointing in the direction of **v** to become zero. What remains is then rotated from **v** to **u**. Therefore an outer product rotates one component of a vector and causes all other components to become zero.

e

1

⊗

e

2

⋅

e

2

=

e

1

{\displaystyle \mathbf{e}\_1 \otimes \mathbf{e}\_2 \cdot \mathbf{e}\_2 = \mathbf{e}\_1}

To rotate a vector with 2 components you need the sum of at least 2 outer products (a bivector). But this is still not perfect. Any 3rd component not in the plane of rotation will become zero.
A true 3 dimensional rotation matrix can be constructed by summing three outer products. The first two sum to form a bivector. The third one rotates the axis of rotation zero degrees but is necessary to prevent that dimension from being squashed to nothing.

e

1

⊗

e

2

−

e

2

⊗

e

1

e

3

⊗

e

3

{\displaystyle \mathbf{e}\_1 \otimes \mathbf{e}\_2 - \mathbf{e}\_2 \otimes \mathbf{e}\_1 + \mathbf{e}\_3 \otimes \mathbf{e}\_3}

The Tensor productw is the same as the outer productw but isnt limited to vectors.
Back to top

#### Geometric product

u

v

=

u

⋅

v

u

∧

v

{\displaystyle
\mathbf{u} \mathbf{v} =
\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \wedge \mathbf{v}
}

The geometric productw will be explained in detail below.

#### Wedge product

Wedge productw (a simple bivectorw):

u

∧

v

=

u

⊗

v

−

v

⊗

u

=
[

u

¯

,

v

¯

]

{\displaystyle \mathbf{u} \wedge \mathbf{v} = \mathbf{u} \otimes \mathbf{v} - \mathbf{v} \otimes \mathbf{u} = [\overline{\mathbf{u}}, \overline{\mathbf{v}}]}

The wedge product can be generalized to a symplectic form
The wedge product of 2 vectors is equal to the geometric product minus the inner product.
The wedge product is also called the exterior productw (sometimes mistakenly called the outer product).
The term "exterior" comes from the exterior product of two vectors not being a vector.
Just as a vector has length and direction so a bivector has an area and an orientation.
In three dimensions

u

∧

v

{\displaystyle \mathbf{u} \wedge \mathbf{v}}

 is the dualw of the cross productw which is a pseudovectorw.

u

∧

v

¯

=

u

×

v

{\displaystyle \overline{\mathbf{u} \wedge \mathbf{v}} = \mathbf{u} \times \mathbf{v}}

a
∧
b
∧
# c
a
⊗
b
⊗
c
−
a
⊗
c
⊗
b
c
⊗
a
⊗
b
−
c
⊗
b
⊗
a
b
⊗
c
⊗
a
−
b
⊗
a
⊗
c

{\displaystyle

\mathbf{a \wedge b \wedge c =

a \otimes b \otimes c -

a \otimes c \otimes b +

c \otimes a \otimes b -

c \otimes b \otimes a +

b \otimes c \otimes a -

b \otimes a \otimes c}

}

The triple productw a∧b∧c is a trivector which is a 3rd degree tensor.
In 3 dimensions a trivector is a pseudoscalar so in 3 dimensions every trivector can be represented as a scalar times the unit trivector. See Levi-Civita symbolw

a

⋅
(

b

×

c

)
⋅

e

1

∧

e

2

∧

e

3

{\displaystyle \mathbf{a}\cdot(\mathbf{b}\times \mathbf{c}) \cdot \mathbf{e}\_1 \wedge \mathbf{e}\_2 \wedge \mathbf{e}\_3 }

The dualw of vector **a** is bivector **ā**:

a

¯

=

d
e
f

[

0

−

a
3

a
2

a
3

0

−

a
1

−

a
2

a
1

0

]

{\displaystyle \overline{\mathbf{a}} \quad\stackrel{\rm def}{=} \quad\begin{bmatrix}\,\,0&\!-a\_3&\,\,\,a\_2\\\,\,\,a\_3&0&\!-a\_1\\\!-a\_2&\,\,a\_1&\,\,0\end{bmatrix}}

Back to top

### Covectors

Covectors have direction and magnitude but are not vectors.

The transpose (indicated by T) of a row vector is a column vector and the transpose of a column vector is a row vector.

[

x
1

x
2

…

x
m

]

T

=

[

x
1

x
2

⋮

x
m

]

,

{\displaystyle \begin{bmatrix} x\_1 \; x\_2 \; \dots \; x\_m \end{bmatrix}^{\rm T} = \begin{bmatrix} x\_1 \\ x\_2 \\ \vdots \\ x\_m \end{bmatrix} \,,}

It is customary to represent vectors as row vectors and covectors as column vectors but sometimes this is reversed. Either way it is done purely for convenience and has no deeper meaning.

A vector has units of

e
x

,

e
y

,

e
z

⋯

{\displaystyle e\_x, e\_y, e\_z \cdots}

 while covectors have units of

1

e
x

,

1

e
y

,

1

e
z

,

{\displaystyle \frac{1}{e\_x}, \frac{1}{e\_y}, \frac{1}{e\_z}, }

v

=

[

v
1

e
x

v
2

e
y

v
3

e
z

]

c

=

[

c
1

1

e
x

c
2

1

e
y

c
3

1

e
z

]

=

[

c
1

e
x

c
2

e
y

c
3

e
z

]

{\displaystyle
\begin{align}
\mathbf{v} =
\begin{bmatrix}
 v^1 e\_x & v^2 e\_y & v^3 e\_z
\end{bmatrix} \\

\mathbf{c} =
\begin{bmatrix}
 c\_1 \frac{1}{e\_x} & c\_2 \frac{1}{e\_y} & c\_3 \frac{1}{e\_z}
\end{bmatrix} =
\begin{bmatrix}
 c\_1 e^x & c\_2 e^y & c\_3 e^z
\end{bmatrix}
\end{align}

}

A key feature of the representation of vectors and tensors in terms of indexed components and basis vectors is invariance in the sense that vector components which transform in a covariant manner (or contravariant manner) are paired with basis vectors that transform in a contravariant manner (or covariant manner).

A vector times a covector is a scalar.

[

v
1

v
2

v
3

]

[

c
1

c
2

c
3

]

=

v
1

c
1

v
2

c
2

v
3

c
3

{\displaystyle \begin{bmatrix} v^1 & v^2 & v^3 \end{bmatrix} \begin{bmatrix} c\_1 \\ c\_2 \\ c\_3 \end{bmatrix} = v^1 c\_1 + v^2 c\_2 + v^3 c\_3}

The Mississippi flows at about 3 km per hour. Km per hour has both direction and magnitude and is a vector.

The Mississippi flows downhill about one foot per km. Feet per km has direction and magnitude but is not a vector. Its a covector.

The difference between a vector and a covector becomes apparent when doing a change of units. If we measured in meters instead of km then our unit becomes **smaller** and 3 km per hour become 3000 meters per hour. The numerical value **increases**. Vectors are therefore **contravariant**.

But 1 Foot per km becomes 0.001 feet per meter. The numerical value **decreases**. Covectors are therefore **covariant**.

In Clifford algebra (the algebra of Euclidean space) a covector is sometimes said to be a vector. The terminology here is confusing. 10 miles per hour is a scalar. It becomes a vector when we multiply it by a unit vector in some direction. Its inverse is 0.1 hours per mile multiplied by the inverse of the same unit vector. The inverse of a unit vector is itself. In other words the inverse of a unit vector is a unit vector. So is the inverse of 10 miles per hour times a unit vector a vector or a covector? It is multiplied by a unit vector so can be said to be a vector but it has units of inverse distance so it really is a covector.

Tensors are more complicated. They can be part contravariant and part covariant.

A (1,1) Tensor is the outer product of a vector and a covector. It is one part contravariant and one part covariant and is totally unaffected by a change of units. It is these that we will study in the next section.

Back to top

## Tensors

See also: Matrix norm and Tensor contraction
External links: Review of Linear Algebra and High-Order Tensors

Just as a vector is a sum of unit vectors multiplied by scalars so a tensor is a sum of unit dyads (

e
1

⊗

e
2

{\displaystyle e\_1 \otimes e\_2}

) multiplied by scalars. Each dyad is associated with a certain plane segment having a certain orientation and magnitude. (But a dyad is not the same thing as a bivectorw.)

A simple tensor is a tensor that can be written as a product of tensors of the form

# T
a
⊗
b
⊗
⋯
⊗
d
.

{\displaystyle T=a\otimes b\otimes\cdots\otimes d.}

 (See Outer Product above.) The rank of a tensor *T* is the minimum number of simple tensors that sum to *T*. A bivectorw is a tensor of rank 2.

The order or degree of the tensor is the dimension of the tensor which is the total number of indices required to identify each component uniquely. A vector is a 1st-order tensor.

Complex numbers can be used to represent and perform rotationsw but only in 2 dimensions.

Tensorsw, on the other hand, can be used in any number of dimensions to represent and perform rotations and other linear transformationsw. See the image to the right.

Any affine transformationw is equivalent to a linear transformation followed by a **translationw** of the origin. (The originw is always a fixed point for any linear transformation.) "Translation" is just a fancy word for "move".
Multiplying a tensor and a vector results in a new vector that can not only have a different magnitude but can even point in a completely different direction:

[

a
1

a
2

a
3

b
1

b
2

b
3

c
1

c
2

c
3

]

[

x

y

z

]

=

[

a
1

x

a
2

y

a
3

z

b
1

x

b
2

y

b
3

z

c
1

x

c
2

y

c
3

z

]

{\displaystyle
 \begin{bmatrix}
 a\_1 & a\_2 & a\_3 \\
 b\_1 & b\_2 & b\_3 \\
 c\_1 & c\_2 & c\_3 \\
 \end{bmatrix}
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} =
 \begin{bmatrix}
 a\_1x+a\_2y+a\_3z\\
 b\_1x+b\_2y+b\_3z\\
 c\_1x+c\_2y+c\_3z\\
 \end{bmatrix}
}

Some special cases:

[

a
1

a
2

a
3

b
1

b
2

b
3

c
1

c
2

c
3

]

[

1

0

0

]

=

[

a
1

b
1

c
1

]

{\displaystyle
 \begin{bmatrix}
 {\color{green}a\_1} & a\_2 & a\_3 \\
 {\color{green}b\_1} & b\_2 & b\_3 \\
 {\color{green}c\_1} & c\_2 & c\_3 \\
 \end{bmatrix}
 \begin{bmatrix}
 {\color{red}1}\\
 {\color{red}0}\\
 {\color{red}0}\\
 \end{bmatrix} =
 \begin{bmatrix}
 {\color{green}a\_1}\\
 {\color{green}b\_1}\\
 {\color{green}c\_1}\\
 \end{bmatrix}
}

[

a
1

a
2

a
3

b
1

b
2

b
3

c
1

c
2

c
3

]

[

0

1

0

]

=

[

a
2

b
2

c
2

]

{\displaystyle
 \begin{bmatrix}
 a\_1 & a\_2 & a\_3 \\
 b\_1 & b\_2 & b\_3 \\
 c\_1 & c\_2 & c\_3 \\
 \end{bmatrix}
 \begin{bmatrix}
 0\\
 1\\
 0\\
 \end{bmatrix} =
 \begin{bmatrix}
 a\_2\\
 b\_2\\
 c\_2\\
 \end{bmatrix}
}

[

a
1

a
2

a
3

b
1

b
2

b
3

c
1

c
2

c
3

]

[

0

0

1

]

=

[

a
3

b
3

c
3

]

{\displaystyle
 \begin{bmatrix}
 a\_1 & a\_2 & a\_3 \\
 b\_1 & b\_2 & b\_3 \\
 c\_1 & c\_2 & c\_3 \\
 \end{bmatrix}
 \begin{bmatrix}
 0\\
 0\\
 1\\
 \end{bmatrix} =
 \begin{bmatrix}
 a\_3\\
 b\_3\\
 c\_3\\
 \end{bmatrix}
}

One can also multiply a tensor with another tensor. Each column of the second tensor is transformed exactly as a vector would be.

[

a
1

a
2

a
3

b
1

b
2

b
3

c
1

c
2

c
3

]

[

1

0

0

0

1

0

0

0

1

]

=

[

a
1

a
2

a
3

b
1

b
2

b
3

c
1

c
2

c
3

]

{\displaystyle
 \begin{bmatrix}
 {\color{green}a\_1} & a\_2 & a\_3 \\
 {\color{green}b\_1} & b\_2 & b\_3 \\
 {\color{green}c\_1} & c\_2 & c\_3 \\
 \end{bmatrix}
 \begin{bmatrix}
 {\color{red}1} & 0 & 0 \\
 {\color{red}0} & 1 & 0 \\
 {\color{red}0} & 0 & 1 \\
 \end{bmatrix} =
 \begin{bmatrix}
 {\color{green}a\_1} & a\_2 & a\_3 \\
 {\color{green}b\_1} & b\_2 & b\_3 \\
 {\color{green}c\_1} & c\_2 & c\_3 \\
 \end{bmatrix}
}

And we can also switch things around using a Permutation matrix. (See also Permutation group):

[

1

0

0

0

0

1

0

1

0

]

[

x

y

z

]

=

[

x

z

y

]

{\displaystyle
 \begin{bmatrix}
 1 & 0 & 0 \\
 0 & 0 & 1 \\
 0 & 1 & 0 \\
 \end{bmatrix}
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} =
 \begin{bmatrix}
 x\\
 z\\
 y\\
 \end{bmatrix}
}

[

0

0

1

0

1

0

1

0

0

]

[

x

y

z

]

=

[

z

y

x

]

{\displaystyle
 \begin{bmatrix}
 0 & 0 & 1 \\
 0 & 1 & 0 \\
 1 & 0 & 0 \\
 \end{bmatrix}
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} =
 \begin{bmatrix}
 z\\
 y\\
 x\\
 \end{bmatrix}
}

[

0

1

0

1

0

0

0

0

1

]

[

x

y

z

]

=

[

y

x

z

]

{\displaystyle
 \begin{bmatrix}
 0 & 1 & 0 \\
 1 & 0 & 0 \\
 0 & 0 & 1 \\
 \end{bmatrix}
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} =
 \begin{bmatrix}
 y\\
 x\\
 z\\
 \end{bmatrix}
}

A shift matrix is a special type of permutation matrix:

[

a

b

c

d

a

b

c

d

a

b

c

d

a

b

c

d

]

[

0

0

0

1

1

0

0

0

0

1

0

0

0

0

1

0

]

=

[

b

c

d

a

b

c

d

a

b

c

d

a

b

c

d

a

]

{\displaystyle
 \begin{bmatrix}
 a & b & c & d \\
 a & b & c & d \\
 a & b & c & d \\
 a & b & c & d \\
 \end{bmatrix}
 \begin{bmatrix}
 0 & 0 & 0 & 1 \\
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 \end{bmatrix} =
 \begin{bmatrix}
 b & c & d & a \\
 b & c & d & a \\
 b & c & d & a \\
 b & c & d & a \\
 \end{bmatrix}
}

A projection matrix projects every point onto a line or plane

[

1

0

0

0

1

0

0

0

0

]

[

y

x

z

]

=

[

y

x

0

]

{\displaystyle
\begin{bmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
 y\\
 x\\
 z\\
\end{bmatrix} =
\begin{bmatrix}
 y\\
 x\\
 0\\
\end{bmatrix}
}

Matrices do not in general commute:

(

a

0

0

b

)

(

0

1

1

0

)

=

(

0

a

b

0

)

{\displaystyle
\begin{pmatrix}
 a & 0 \\
 0 & b
\end{pmatrix}
\begin{pmatrix}
 0 & 1 \\
 1 & 0
\end{pmatrix} =
\begin{pmatrix}
 0 & a \\
 b & 0
\end{pmatrix}
}

but

(

0

1

1

0

)

(

a

0

0

b

)

=

(

0

b

a

0

)

{\displaystyle
\begin{pmatrix}
 0 & 1 \\
 1 & 0
\end{pmatrix}
\begin{pmatrix}
 a & 0 \\
 0 & b
\end{pmatrix} =
\begin{pmatrix}
 0 & b \\
 a & 0
\end{pmatrix}
}

The Determinantw of a matrix is the area or volume of the n-dimensional parallelepiped spanned by its column (or row) vectors and is frequently useful.

A

=

a

b

c

d

=
a
d
−
b
c

{\displaystyle |A|=\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc}

For a square

n
×
n

{\displaystyle n \times n}

 matrix *A* whose columns are

a
1

,
…
,

a
n

{\displaystyle \mathbf{a\_1}, \ldots, \mathbf{a\_n} }

, its determinant satisfies

a
1

∧
⋯
∧

a
n

=
det
(
A
)
⋅

e
1

∧
⋯
∧

e
n

{\displaystyle \mathbf{a\_1} \wedge \dots \wedge \mathbf{a\_n} = \det(A) \cdot \mathbf{e\_1} \wedge \dots \wedge \mathbf{e\_n}}

Matrices do have zero divisors:

# A

(

0

1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

)

,

A
2

=

(

0

0

1

0

0

0

0

1

0

0

0

0

0

0

0

0

)

,

(

A
2

)
2

=

(

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

0

)

,

(
(

A
2

)
2

)
2

=
0

{\displaystyle
\begin{align}

A=
\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix}, \quad

A^2=
\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix}, \quad \\

(A^2)^2=
\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}, \quad

((A^2)^2)^2=0

\end{align}

}

# Symmetric tensor

[

x

a

b

a

y

c

b

c

z

]

{\displaystyle \text{Symmetric tensor} =
 \begin{bmatrix}
 x & a & b \\
 a & y & c \\
 b & c & z \\
 \end{bmatrix}
}

# Anti-symmetric tensor

[

0

a

b

−
a

0

c

−
b

−
c

0

]

{\displaystyle \text{Anti-symmetric tensor} =
 \begin{bmatrix}
 0 & a & b \\
 -a & 0 & c \\
 -b & -c & 0 \\
 \end{bmatrix}
}

The Outer product (tensor product) of a vector with itself is a symmetric tensor:

[

x

y

z

]

⊗

[

x

y

z

]

=

[

x
x

x
y

x
z

y
x

y
y

y
z

z
x

z
y

z
z

]

{\displaystyle
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} \otimes
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} =
 \begin{bmatrix}
 xx & xy & xz \\
 yx & yy & yz \\
 zx & zy & zz \\
 \end{bmatrix}
}

The wedge product of 2 vectors is anti-symmetric:

u

∧

v

=

u

⊗

v

−

v

⊗

u

=

[

0

u
1

v
2

−

v
1

u
2

u
1

v
3

−

v
1

u
3

u
2

v
1

−

v
2

u
1

0

u
2

v
3

−

v
2

u
3

u
3

v
1

−

v
3

u
1

u
3

v
2

−

v
3

u
2

0

]

{\displaystyle \mathbf{u} \wedge \mathbf{v} = \mathbf{u} \otimes \mathbf{v} - \mathbf{v} \otimes \mathbf{u} =

\begin{bmatrix}
0 & u\_1 v\_2 - v\_1 u\_2 & u\_1 v\_3 - v\_1 u\_3 \\
u\_2 v\_1 - v\_2 u\_1 & 0 & u\_2 v\_3 - v\_2 u\_3 \\
u\_3 v\_1 - v\_3 u\_1 & u\_3 v\_2 - v\_3 u\_2 & 0
\end{bmatrix}
}

Any

n
×
n

{\displaystyle n \times n}

 matrix *X* with complex entries can be expressed as

# X
A
N

{\displaystyle X = A + N \,}

where
* *A* is diagonalizable
* *N* is nilpotent
* *A* commutes with *N* (i.e. *AN* = *NA*)

This is the Jordan–Chevalley decomposition.

The matrix

P

=

[

1

1

2

2

1

1

2

2

3

3

4

4

3

3

4

4

]

{\displaystyle \mathbf{P} = \begin{bmatrix}
1 & 1 & 2 & 2\\
1 & 1 & 2 & 2\\
3 & 3 & 4 & 4\\
3 & 3 & 4 & 4\end{bmatrix}}

can be partitioned into 4 2×2 blocks

P

11

=

[

1

1

1

1

]

,

P

12

=

[

2

2

2

2

]

,

P

21

=

[

3

3

3

3

]

,

P

22

=

[

4

4

4

4

]

.

{\displaystyle \mathbf{P}\_{11} = \begin{bmatrix}
1 & 1 \\
1 & 1 \end{bmatrix}, \mathbf{P}\_{12} = \begin{bmatrix}
2 & 2\\
2 & 2\end{bmatrix}, \mathbf{P}\_{21} = \begin{bmatrix}
3 & 3 \\
3 & 3 \end{bmatrix}, \mathbf{P}\_{22} = \begin{bmatrix}
4 & 4\\
4 & 4\end{bmatrix}.}

The partitioned matrix can then be written as

P

=

[

P

11

P

12

P

21

P

22

]

.

{\displaystyle \mathbf{P} = \begin{bmatrix}
\mathbf{P}\_{11} & \mathbf{P}\_{12}\\
\mathbf{P}\_{21} & \mathbf{P}\_{22}\end{bmatrix}.}

the matrix product

C

=

A

B

{\displaystyle
\mathbf{C}=\mathbf{A}\mathbf{B}
}

can be formed blockwise, yielding

C

{\displaystyle \mathbf{C}}

 as an

(
m
×
n
)

{\displaystyle (m\times n)}

 matrix with

q

{\displaystyle q}

 row partitions and

r

{\displaystyle r}

 column partitions. The matrices in the resulting matrix

C

{\displaystyle \mathbf{C}}

 are calculated by multiplying:

C

α
β

=

∑

# γ
1

s

A

α
γ

B

γ
β

.

{\displaystyle
\mathbf{C}\_{\alpha \beta} = \sum^s\_{\gamma=1}\mathbf{A}\_{\alpha \gamma}\mathbf{B}\_{\gamma \beta}.
}

Or, using the Einstein notation that implicitly sums over repeated indices:

C

α
β

=

A

α
γ

B

γ
β

.

{\displaystyle
\mathbf{C}\_{\alpha \beta} = \mathbf{A}\_{\alpha \gamma}\mathbf{B}\_{\gamma \beta}.
}

### Change of basis

An *n*-by-*n* square matrixw **A** is **invertible** if there exists an *n*-by-*n* square matrix **A-1** such that

A

A

−
1

=

A

−
1

A

=

I

n

{\displaystyle \mathbf{AA^{-1}} = \mathbf{A^{-1}A} = \mathbf{I}\_n \ }

A matrix is invertible if and only if its determinant is non-zero.

A

A

−
1

=

I

n

A

A

−
1

=
1

A

A

=
1

A

≠
0

{\displaystyle
\begin{align}
\end{align}
}

The standard basis for

R

2

{\displaystyle R^{2}}

 would be:

E
2

=

{

(

1

0

)

,

(

0

1

)

}

{\displaystyle E\_2 = \left\{ \begin{pmatrix} 1\\ 0\end{pmatrix}, \begin{pmatrix} 0\\ 1\end{pmatrix} \right\}}

Given a matrix

M

{\displaystyle M}

 whose columns are the vectors of the new basis of the space (new basis matrix), the new coordinates for a column vector

v

{\displaystyle v}

 are given by the matrix product

M

−
1

v

{\displaystyle M^{-1}v}

.

From Wikipedia:Matrix similarity:

Given a linear transformation:

# y
T
x

{\displaystyle y = Tx}

,
it can be the case that a change of basis can result in a simpler form of the same transformation.

y
′

=
S

x
′

{\displaystyle y' = Sx'}

,
where x' and y' are in the new basis.

x
′

=
P
x

{\displaystyle x' = P x}

y
′

=
P
y

{\displaystyle y' = P y}

and P is the change-of-basis matrix.
To derive T in terms of the simpler matrix, we use:

y
′

=
S

x
′

P
y

=
S
P
x

P

−
1

P
y

=

P

−
1

S
P
x

y

=

(

P

−
1

S
P

)

# x
T
x

{\displaystyle \begin{align}
y' &= Sx' \\
Py &= SPx \\
P^{-1}Py &= P^{-1}SPx \\
y &= \left(P^{-1}SP\right)x = Tx
\end{align}}

Thus, the matrix in the original basis is given by

# T

P

−
1

S
P

{\displaystyle T = P^{-1}SP}

Therefore

# S
P
T

P

−
1

{\displaystyle S = PTP^{-1}}

From Wikipedia:Matrix similarity

Two *n*-by-*n* matrices A and B are called **similar** if

# B

P

−
1

A
P

{\displaystyle B = P^{-1} A P }

for some invertible *n*-by-*n* matrix P.

A transformation *A* ↦ *P*−1*AP* is called a **similarity transformation** or **conjugation** of the matrix A. In the general linear group, similarity is therefore the same as **conjugacy**, and similar matrices are also called **conjugate**.

Back to top

### Diagonalization and the eigendecomposition

Main articles: Eigendecomposition of a matrix
From: Eigenvalues and eigenvectors
An eigenvector of a linear transformation points in a direction in which it is stretched by the transformation and the eigenvalue is the factor by which it is stretched.

Suppose the eigenvectors of *A* form a basis, or equivalently *A* has *n* linearly independent eigenvectors *v*1, *v*2, ..., *v**n* with associated eigenvalues *λ*1, *λ*2, ..., *λ**n*. The eigenvalues need not be distinct. Define a square matrix *Q* whose columns are the *n* linearly independent eigenvectors of *A*,

# Q

[

v
1

v
2

⋯

v
n

]

.

{\displaystyle Q = \begin{bmatrix} v\_1 & v\_2 & \cdots & v\_n \end{bmatrix}.}

Since each column of *Q* is an eigenvector of *A*, right multiplying *A* by *Q* scales each column of *Q* by its associated eigenvalue,

A
# Q

[

λ
1

v
1

λ
2

v
2

⋯

λ
n

v
n

]

.

{\displaystyle AQ = \begin{bmatrix} \lambda\_1 v\_1 & \lambda\_2 v\_2 & \cdots & \lambda\_n v\_n \end{bmatrix}.}

With this in mind, define a diagonal matrix Λ where each diagonal element Λ*ii* is the eigenvalue associated with the *i*th column of *Q*. Then

A
# Q
Q
Λ
.

{\displaystyle AQ = Q\Lambda.}

Because the columns of *Q* are linearly independent, Q is invertible. Right multiplying both sides of the equation by *Q*−1,

# A
Q
Λ

Q

−
1

,

{\displaystyle A = Q\Lambda Q^{-1},}

or by instead left multiplying both sides by *Q*−1,

Q

−
1

A
# Q
Λ
.

{\displaystyle Q^{-1}AQ = \Lambda.}

*A* can therefore be decomposed into a matrix composed of its eigenvectors, a diagonal matrix with its eigenvalues along the diagonal, and the inverse of the matrix of eigenvectors. This is called the eigendecomposition and it is a similarity transformation. Such a matrix *A* is said to be *similar* to the diagonal matrix Λ or *diagonalizable*. The matrix *Q* is the change of basis matrix of the similarity transformation. Essentially, the matrices *A* and Λ represent the same linear transformation expressed in two different bases. The eigenvectors are used as the basis when representing the linear transformation as Λ.

Conversely, suppose a matrix *A* is diagonalizable. Let *P* be a non-singular square matrix such that *P*−1*AP* is some diagonal matrix *D*. Left multiplying both by *P*, *AP* = *PD*. Each column of *P* must therefore be an eigenvector of *A* whose eigenvalue is the corresponding diagonal element of *D*. Since the columns of *P* must be linearly independent for *P* to be invertible, there exist *n* linearly independent eigenvectors of *A*. It then follows that the eigenvectors of *A* form a basis if and only if *A* is diagonalizable.

A matrix that is not diagonalizable is said to be defective. For defective matrices, the notion of eigenvectors generalizes to generalized eigenvectors and the diagonal matrix of eigenvalues generalizes to the Jordan normal form. Over an algebraically closed field, any matrix *A* has a Jordan normal form and therefore admits a basis of generalized eigenvectors and a decomposition into generalized eigenspaces.

From Wikipedia: Eigendecomposition of a matrix

If a matrix **A** can be eigendecomposed and if none of its eigenvalues are zero, then **A** is  nonsingular and its inverse is given by

A

−
1

=

Q

Λ

−
1

Q

−
1

{\displaystyle \mathbf{A}^{-1} = \mathbf{Q}\mathbf{\Lambda}^{-1}\mathbf{Q}^{-1}}

If

A

{\displaystyle \mathbf{A}}

 is a symmetric matrix, since

Q

{\displaystyle \mathbf{Q}}

 is formed from the eigenvectors of

A

{\displaystyle \mathbf{A}}

 it is guaranteed to be an orthogonal matrix, therefore

Q

−
1

=

Q

T

{\displaystyle \mathbf{Q}^{-1} = \mathbf{Q}^\mathrm{T}}

.
Furthermore, because **Λ** is a diagonal matrix, its inverse is easy to calculate:

[

Λ

−
1

]

i
i

=

1

λ
i

{\displaystyle \left[\Lambda^{-1}\right]\_{ii} = \frac{1}{\lambda\_i}}

Back to top

### Normal matrices

A square matrix

A

{\displaystyle A}

 is called diagonalizable or nondefective if it is similar to a diagonal matrix

A diagonal matrixw:

[

6

0

0

0

−
7
i

0

0

0

1
9
i

]

{\displaystyle
 \begin{bmatrix}
6 & 0 & 0 \\
0 & -7i & 0 \\
0 & 0 & 1+9i
 \end{bmatrix}}

The determinate of a diagonal matrix:

a

0

0

0

b

0

0

0

c

=
a
b
c

{\displaystyle
\begin{vmatrix}
a&0&0\\
0&b&0\\
0&0&c
\end{vmatrix}
=abc
}

A superdiagonal entry is one that is directly above and to the right of the main diagonal. A subdiagonal entry is one that is directly below and to the left of the main diagonal. The eigenvalues of diag(*λ*1, ..., *λ**n*) are *λ*1, ..., *λ**n* with associated eigenvectors of *e*1, ..., *e**n*.

A spectral theorem is a result about when a matrix can be diagonalized. This is extremely useful because computations involving a diagonalizable matrix can often be reduced to much simpler computations.

A matrix is normal if and only if it is diagonalizable.

 A Unitary matrixw is a complex square matrix whose rows (or columns) form an orthonormal basis of

C

n

{\displaystyle \C^n}

 with respect to the usual inner product.

[

a

b

−

e

i
φ

b
∗

e

i
φ

a
∗

]

,

a

2

b

2

=
1
,

{\displaystyle
\begin{bmatrix}
a & b \\
-e^{i\varphi} b^\* & e^{i\varphi} a^\* \\

[

1

0

0

0

0

0

0

0

1

0

0

1

0

0

0

0

0

0

0

1

0

0

1

0

0

]

.

{\displaystyle
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0
\end{bmatrix}.
}
 A Hermitian matrixw is a complex square matrix that is equal to its own conjugate transposew.

# A

A
T

¯

=

A
†

=

A
H

{\displaystyle A = \overline {A^\text{T}} = A^\dagger = A^\text{H}}

 The diagonal elements must be real.

[

2

2
i

−
4

2
−
i

0

i

−
4

−
i

−
1

]

{\displaystyle
\begin{bmatrix}
2 & 2+i & -4 \\
2-i & 0 & i \\
-4 & -i & -1 \\
\end{bmatrix}
}

# A

A

T

.

{\displaystyle A = A^\mathrm{T}.}

[

2

2

−
4

2

0

0

−
4

0

−
1

]

{\displaystyle \begin{bmatrix}
2 & 2 & -4\\
2 & 0 & 0\\
-4 & 0 & -1\end{bmatrix}}
 A Skew-Hermitian matrix is a complex square matrix whose conjugate transpose is its negative.

A
†

=
−
A
,

{\displaystyle A^\dagger = -A,\;}

 The diagonal elements must be imaginary.

[

i

2
i

−
4

−
2
i

0

i

4

i

−
i

]

{\displaystyle
\begin{bmatrix}
i & 2+i & -4 \\
-2+i & 0 & i \\
4 & i & -i \\
\end{bmatrix}
}

[

0

2

−
4

−
2

0

0

4

0

0

]

{\displaystyle \begin{bmatrix}
0 & 2 & -4\\
-2 & 0 & 0\\
4 & 0 & 0\end{bmatrix}}

### Matrix representations

See also: Group representation, Presentation of a group, Abstract algebra
#### Real numbers

If a vector is multiplied with the the identity matrix

I

{\displaystyle I}

 then the vector is completely unchanged:

I
⋅
# v

[

1

0

0

0

1

0

0

0

1

]

[

x

y

z

]

=
1

[

x

y

z

]

=

[

x

y

z

]

{\displaystyle I \cdot v =
 \begin{bmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & 1 \\
 \end{bmatrix}
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} = 1
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} =
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix}
}

And if

# A
a
⋅
I

{\displaystyle A=a \cdot I}

 then

A
⋅
# v

[

a

0

0

0

a

0

0

0

a

]

[

x

y

z

]

=
a

[

x

y

z

]

=

[

a
⋅
x

a
⋅
y

a
⋅
z

]

{\displaystyle A \cdot v =
 \begin{bmatrix}
 a & 0 & 0 \\
 0 & a & 0 \\
 0 & 0 & a \\
 \end{bmatrix}
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} = a
 \begin{bmatrix}
 x\\
 y\\
 z\\
 \end{bmatrix} =
 \begin{bmatrix}
 a\cdot x\\
 a \cdot y\\
 a\cdot z\\
 \end{bmatrix}
}

Therefore

# A
a
⋅
I

{\displaystyle A=a \cdot I}

 can be thought of as the matrix representation of the scalar *a*. The scalar matrices are the centerw of the algebra of matrices.

A
# B

[

a

0

0

0

a

0

0

0

a

]

[

b

0

0

0

b

0

0

0

b

]

=

[

a
b

0

0

0

a
b

0

0

0

a
b

]

{\displaystyle A + B =
 \begin{bmatrix}
 a & 0 & 0 \\
 0 & a & 0 \\
 0 & 0 & a \\
 \end{bmatrix} +
 \begin{bmatrix}
 b & 0 & 0 \\
 0 & b & 0 \\
 0 & 0 & b \\
 \end{bmatrix} =
 \begin{bmatrix}
 a + b & 0 & 0 \\
 0 & a + b & 0 \\
 0 & 0 & a + b \\
 \end{bmatrix}
}

A
⋅
# B

[

a

0

0

0

a

0

0

0

a

]

[

b

0

0

0

b

0

0

0

b

]

=

[

a
b

0

0

0

a
b

0

0

0

a
b

]

{\displaystyle A \cdot B =
 \begin{bmatrix}
 a & 0 & 0 \\
 0 & a & 0 \\
 0 & 0 & a \\
 \end{bmatrix}
 \begin{bmatrix}
 b & 0 & 0 \\
 0 & b & 0 \\
 0 & 0 & b \\
 \end{bmatrix} =
 \begin{bmatrix}
 ab & 0 & 0 \\
 0 & ab & 0 \\
 0 & 0 & ab \\
 \end{bmatrix}
}

1

/

# A

[

1

/

a

0

0

0

1

/

a

0

0

0

1

/

a

]

{\displaystyle 1/A =
 \begin{bmatrix}
 1/a & 0 & 0 \\
 0 & 1/a & 0 \\
 0 & 0 & 1/a \\
 \end{bmatrix}
}

A

=

a

0

0

0

a

0

0

0

a

=

a
3

{\displaystyle |A| =
\begin{vmatrix}
 a & 0 & 0 \\
 0 & a & 0 \\
 0 & 0 & a
\end{vmatrix}
= a^3
}

A
B

=

[

a
b

0

0

0

a
b

0

0

0

a
b

]

{\displaystyle A^B =
 \begin{bmatrix}
 a^b & 0 & 0 \\
 0 & a^b & 0 \\
 0 & 0 & a^b \\
 \end{bmatrix}
}

e
A

=

[

e
a

0

0

0

e
a

0

0

0

e
a

]

{\displaystyle e^A=
 \begin{bmatrix}
 e^a & 0 & 0 \\
 0 & e^a & 0 \\
 0 & 0 & e^a \\
 \end{bmatrix} }

.

ln
⁡
# A

[

ln
⁡
a

0

0

0

ln
⁡
a

0

0

0

ln
⁡
a

]

{\displaystyle \ln A=
 \begin{bmatrix}
 \ln a & 0 & 0 \\
 0 & \ln a & 0 \\
 0 & 0 & \ln a \\
 \end{bmatrix} }

.
(Note: Not all matrices have a logarithm and those matrices that do have a logarithm may have more than one logarithm. The study of logarithms of matrices leads to Lie theory since when a matrix has a logarithm then it is in a Lie group and the logarithm is the corresponding element of the vector space of the Lie algebra.)

Back to top

#### Complex numbers

Complex numbers can also be written in matrix formw in such a way that complex multiplication corresponds perfectly to matrix multiplication:

(
a
i
b
)
(
c
i
d
)

=

[

a

b

−
b

a

]

[

c

d

−
d

c

]

=

[

a
c
−
b
d

a
d
b
c

−
(
a
d
b
c
)

a
c
−
b
d

]

{\displaystyle
\begin{align}
(a+ib)(c+id) &=
\begin{bmatrix}
 a & b \\
 -b & a
\end{bmatrix}
\begin{bmatrix}
 c & d \\
 -d & c
\end{bmatrix} \\
&=
\begin{bmatrix}
 ac-bd & ad+bc \\
 -(ad+bc) & ac-bd
\end{bmatrix}
\end{align}
}

(
i
)
(
i
)

=

[

0

1

−
1

0

]

[

0

1

−
1

0

]

=

[

−
1

0

0

−
1

]

=
−
I

{\displaystyle
\begin{align}
(i)(i) &=
\begin{bmatrix}
 0 & 1 \\
 -1 & 0
\end{bmatrix}
\begin{bmatrix}
 0 & 1 \\
 -1 & 0
\end{bmatrix} \\
&=
\begin{bmatrix}
 -1 & 0 \\
 0 & -1
\end{bmatrix} \\
&= -I
\end{align}
}

The absolute value of a complex number is defined by the Euclidean distance of its corresponding point in the complex plane from the origin computed using the Pythagorean theorem.

z

2

=

a

−
b

b

a

=

a
2

b
2

.

{\displaystyle |z|^2 =
\begin{vmatrix}
 a & -b \\
 b & a
\end{vmatrix}
= a^2 + b^2.
}

Back to top

#### Quaternions

Quaternions can be represented as pairs of complex numbers. From this perspective, quaternions are the result of applying the Cayley–Dickson construction to the complex numbers. This is a generalization of the construction of the complex numbers as pairs of real numbers

(
1
j
)
1
(
1
j
)
i

=
1
∗
1
j
∗
1
1
∗
i
j
∗
i

=

[

1

0

0

1

]

[

1

0

0

1

]

[

0

1

−
1

0

]

[

1

0

0

1

]

[

1

0

0

1

]

[

i

0

0

−
i

]

[

i

0

0

−
i

]

[

0

1

−
1

0

]

=

[

1

0

0

1

]

[

0

1

−
1

0

]

[

i

0

0

−
i

]

[

0

i

i

0

]

=
1
j
i
k

{\displaystyle
\begin{align}
(1 + j)1 + (1 + j)i &=
1\*1 + j\*1 + 1\*i + j\*i \\ &=

\begin{bmatrix}
 1 & 0 \\
 0 & 1
\end{bmatrix}
\begin{bmatrix}
 1 & 0 \\
 0 & 1
\end{bmatrix} +
\begin{bmatrix}
 0 & 1 \\
 -1 & 0
\end{bmatrix}
\begin{bmatrix}
 1 & 0 \\
 0 & 1
\end{bmatrix} +
\begin{bmatrix}
 1 & 0 \\
 0 & 1
\end{bmatrix}
\begin{bmatrix}
 i & 0 \\
 0 & -i
\end{bmatrix} +
\begin{bmatrix}
 i & 0 \\
 0 & -i
\end{bmatrix}
\begin{bmatrix}
 0 & 1 \\
 -1 & 0
\end{bmatrix}\\ &=
\begin{bmatrix}
 1 & 0 \\
 0 & 1
\end{bmatrix} +
\begin{bmatrix}
 0 & 1 \\
 -1 & 0
\end{bmatrix} +
\begin{bmatrix}
 i & 0 \\
 0 & -i
\end{bmatrix} +
\begin{bmatrix}
 0 & i \\
 i & 0
\end{bmatrix} \\ &=
1 + j + i + k
\end{align}
}

There are at least two ways of representing quaternions as matricesw in such a way that quaternion addition and multiplication correspond to matrix addition and matrix multiplicationw.

Using 2 × 2 complex matrices, the quaternion *a* + *bi* + *cj* + *dk* can be represented as

[

z
1

z
2

−

z
2

¯

z
1

¯

]

=

[

a
b
i

c
d
i

−
(
c
−
d
i
)

a
−
b
i

]

=
a

[

1

0

0

1

]

b

[

i

0

0

−
i

]

c

[

0

1

−
1

0

]

d

[

0

i

i

0

]

.

{\displaystyle
\begin{bmatrix}
z\_1 & z\_2 \\
-\overline{z\_2} & \overline{z\_1}
\end{bmatrix} =
\begin{bmatrix}
a+bi & c+di \\
-(c-di) & a-bi
\end{bmatrix}= a
\begin{bmatrix}
 1 & 0 \\
 0 & 1 \\
\end{bmatrix}
+ b
\begin{bmatrix}
 i & 0 \\
 0 & -i \\
\end{bmatrix}
+ c
\begin{bmatrix}
 0 & 1 \\
 -1 & 0 \\
\end{bmatrix}
+ d
\begin{bmatrix}
 0 & i \\
 i & 0 \\
\end{bmatrix}.
}

Multiplying any two Pauli matrices always yields a quaternion unit matrix. See Isomorphism to quaternions below.

By replacing each 0, 1, and i with its 2 × 2 matrix representation that same quaternion can be written as a 4 × 4 real (block) matrix:

[

a

b

c

d

−
b

a

−
d

c

−
c

d

a

−
b

−
d

−
c

b

a

]

=
a

[

1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

1

]

b

[

0

1

0

0

−
1

0

0

0

0

0

0

−
1

0

0

1

0

]

c

[

0

0

1

0

0

0

0

1

−
1

0

0

0

0

−
1

0

0

]

d

[

0

0

0

1

0

0

−
1

0

0

1

0

0

−
1

0

0

0

]

.

{\displaystyle \begin{bmatrix}
 a & b & c & d \\
 -b & a & -d & c \\
 -c & d & a & -b \\
 -d & -c & b & a
\end{bmatrix}= a
\begin{bmatrix}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1
\end{bmatrix}
+ b
\begin{bmatrix}
 0 & 1 & 0 & 0 \\
 -1 & 0 & 0 & 0 \\
 0 & 0 & 0 & -1 \\
 0 & 0 & 1 & 0
\end{bmatrix}
+ c
\begin{bmatrix}
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
 -1 & 0 & 0 & 0 \\
 0 & -1 & 0 & 0
\end{bmatrix}
+ d
\begin{bmatrix}
 0 & 0 & 0 & 1 \\
 0 & 0 & -1 & 0 \\
 0 & 1 & 0 & 0 \\
 -1 & 0 & 0 & 0
\end{bmatrix}.
}

Therefore:

b
⋅
i

=
b
⋅
(

e
1

∧

e
2

e
4

∧

e
3

)

c
⋅
j

=
c
⋅
(

e
1

∧

e
3

e
2

∧

e
4

)

d
⋅
k

=
d
⋅
(

e
1

∧

e
4

e
3

∧

e
2

)

{\displaystyle
\begin{align}
b \cdot i &= b \cdot (e\_1 \wedge e\_2 + e\_4 \wedge e\_3)\\
c \cdot j &= c \cdot (e\_1 \wedge e\_3 + e\_2 \wedge e\_4)\\
d \cdot k &= d \cdot (e\_1 \wedge e\_4 + e\_3 \wedge e\_2)\\
\end{align}
}

However, the representation of quaternions in M(4,ℝ) is not unique. In fact, there exist 48 distinct representations of this form. Each 4x4 matrix representation of quaternions corresponds to a multiplication table of unit quaternions. See Wikipedia:Quaternion#Matrix\_representations.

The obvious way of representing quaternions with 3 × 3 real matrices does not work because:

b

[

0

−
1

0

1

0

0

0

0

0

]

⋅
c

[

0

0

−
1

0

0

0

1

0

0

]

≠
d

[

0

0

0

0

0

−
1

0

1

0

]

.

{\displaystyle b
\begin{bmatrix}
 0 & -1 & 0 \\
 1 & 0 & 0 \\
 0 & 0 & 0
\end{bmatrix}
\cdot c
\begin{bmatrix}
 0 & 0 & -1 \\
 0 & 0 & 0 \\
 1 & 0 & 0
\end{bmatrix}
\neq d
\begin{bmatrix}
 0 & 0 & 0 \\
 0 & 0 & -1 \\
 0 & {\color{red}1} & 0
\end{bmatrix}.
}

Back to top

#### Split-complex numbers

From Wikipedia: Split-complex number

A **split complex number** (or **hyperbolic number**, also **perplex number**, **double number**) has two real number components *x* and *y*, and is written *z* = *x* + *y j*, where *j*2 = 1.

One can easily represent split-complex numbers by matrices. The split-complex number

# z
x
j
y

{\displaystyle z = x + jy}

can be represented by the matrix

z
↦

(

x

y

y

x

)

.

{\displaystyle z \mapsto \begin{pmatrix}x & y \\ y & x\end{pmatrix}.}

Note that in the context of 2 × 2 real matrices there are in fact a great number of different representations of split-complex numbers. The above diagonal representation represents the Jordan canonical form of the matrix representation of the split-complex numbers. For a split-complex number *z* = (*x*, *y*) given by the following matrix representation:

# Z

(

x

y

y

x

)

{\displaystyle Z = \begin{pmatrix}x & y \\ y & x\end{pmatrix}}

its Jordan canonical form is given by:

J
z

=

(

x
y

0

0

x
−
y

)

,

{\displaystyle J\_z = \begin{pmatrix}x + y & 0 \\ 0 & x - y\end{pmatrix} ,}

where

# Z
S

J
z

S

−
1

,

{\displaystyle Z = SJ\_z S^{-1}\ ,}

 and

# S

(

1

−
1

1

1

)

.

{\displaystyle S = \begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix} .}

Back to top

#### Vectors

See also: Split-complex numbers
Unfortunately the matrix representation of a vector is not so obvious. First we must decide what properties the matrix should have. To see consider the square (quadratic form) of a single vector:

Q
(
c
# )

c
2

=
⟨
c
,
c
# ⟩
(
a

e
1

b

e
2

)
2

{\displaystyle Q(c) = c^2 = \langle c , c \rangle = (a\mathbf{e\_1} + b\mathbf{e\_2})^2}

Q
(
c
# )

c
2

=
a
a

e
1

e
1

b
b

e
2

e
2

a
b

e
1

e
2

b
a

e
2

e
1

{\displaystyle Q(c) = c^2 = aa\mathbf{e\_1}\mathbf{e\_1} + bb\mathbf{e\_2}\mathbf{e\_2} + ab\mathbf{e\_1}\mathbf{e\_2} + ba\mathbf{e\_2}\mathbf{e\_1}}

Q
(
c
# )

c
2

=

a
2

(

e
1

e
1

)

b
2

(

e
2

e
2

)
a
b
(

e
1

e
2

e
2

e
1

)

{\displaystyle Q(c) = c^2 = a^2(\mathbf{e\_1}\mathbf{e\_1}) + b^2(\mathbf{e\_2}\mathbf{e\_2}) + ab(\mathbf{e\_1}\mathbf{e\_2} + \mathbf{e\_2}\mathbf{e\_1})}

From the Pythagorean theorem we know that:

c
2

=

a
2

(
1
)

b
2

(
1
)
a
b
(
0
# )
S
c
a
l
a
r

{\displaystyle c^2 = a^2(1) + b^2(1) + ab(0) = Scalar}

So we know that

e
1

e
1

=

e
2

e
2

=
1

{\displaystyle e\_1 e\_1 = e\_2 e\_2 = 1}

e
1

e
2

=
−

e
2

e
1

{\displaystyle e\_1 e\_2 = -e\_2 e\_1 }

This particular Clifford algebra is known as Cl2,0. The subscript 2 indicates that the 2 basis vectors are square roots of +1. See Metric signature. If we had used

c
2

=
−

a
2

−

b
2

{\displaystyle c^2 = -a^2 -b^2}

 then the result would have been Cl0,2.

The set of 3 matrices in 3 dimensions that have these properties are called Pauli matrices. The algebra generated by the three Pauli matrices is isomorphic to the Clifford algebra of ℝ3.

From Wikipedia:Pauli matrices
The Pauli matrices are a set of three 2 × 2 complexw matricesw which are Hermitianw and unitaryw.
They are

σ
1

=

σ
x

=

(

0

1

1

0

)

σ
2

=

σ
y

=

(

0

−
i

i

0

)

σ
3

=

σ
z

=

(

1

0

0

−
1

)

.

{\displaystyle
\begin{align}
 \sigma\_1 = \sigma\_x &=
 \begin{pmatrix}
 0&1\\
 1&0
 \end{pmatrix} \\
 \sigma\_2 = \sigma\_y &=
 \begin{pmatrix}
 0&-i\\
 i&0
 \end{pmatrix} \\
 \sigma\_3 = \sigma\_z &=
 \begin{pmatrix}
 1&0\\
 0&-1
 \end{pmatrix} \,.
\end{align}}

Squaring a Pauli matrix results in a "scalar":

σ
1
2

=

σ
2
2

=

σ
3
2

=

(

1

0

0

1

)

=

σ
0

=
I

{\displaystyle \sigma\_1^2 = \sigma\_2^2 = \sigma\_3^2 =
 \begin{pmatrix}
 1&0\\
 0&1
 \end{pmatrix} = \sigma\_0 = I}

Multiplying any 2 Pauli matrices results in a quaternion. Hence the geometric interpretation of the quaternion units

ı
^

,

ȷ
^

,

k
^

{\displaystyle \boldsymbol{\hat{\imath}}, \boldsymbol{\hat{\jmath}}, \boldsymbol{\hat{k}} \quad}

 as bivectors in 3 dimensional (not 4 dimensional) space.

Quaternions form a division algebra—every non-zero element has an inverse—whereas Pauli matrices do not.
And multiplying a Pauli matrix and a quaternion results in a Pauli matrix

Multiplication is anticommutative:

σ
1

σ
2

=
−

σ
2

σ
1

{\displaystyle \sigma\_1 \sigma\_2 = - \sigma\_2 \sigma\_1}

σ
2

σ
3

=
−

σ
3

σ
2

{\displaystyle \sigma\_2 \sigma\_3 = - \sigma\_3 \sigma\_2}

σ
3

σ
1

=
−

σ
1

σ
3

{\displaystyle \sigma\_3 \sigma\_1 = - \sigma\_1 \sigma\_3}

And

σ
1

σ
2

σ
3

=

(

i

0

0

i

)

=
i

{\displaystyle \sigma\_1 \sigma\_2 \sigma\_3 = \begin{pmatrix} i&0\\0&i\end{pmatrix} = i}

Exponential of a Pauli vector which is analogous to Euler's formula, extended to quaternions:

e

i
a
(

n
^

⋅

σ
→

)

=
I
cos
⁡

a

i
(

n
^

⋅

σ
→

)
sin
⁡

a

{\displaystyle e^{i a(\hat{n} \cdot \vec{\sigma})} = I\cos{a} + i (\hat{n} \cdot \vec{\sigma}) \sin{a} }

commutationw relations:

[

σ
1

,

σ
2

]

=
2
i

σ
3

[

σ
2

,

σ
3

]

=
2
i

σ
1

[

σ
3

,

σ
1

]

=
2
i

σ
2

[

σ
1

,

σ
1

]

=
0

{\displaystyle \begin{align}
 \left[\sigma\_1, \sigma\_2\right] &= 2i\sigma\_3 \, \\
 \left[\sigma\_2, \sigma\_3\right] &= 2i\sigma\_1 \, \\
 \left[\sigma\_3, \sigma\_1\right] &= 2i\sigma\_2 \, \\
 \left[\sigma\_1, \sigma\_1\right] &= 0\, \\
\end{align}}

anticommutation relations:

{

σ
1

,

σ
1

}

=
2
I

{

σ
1

,

σ
2

}

=
0

.

{\displaystyle \begin{align}
 \left\{\sigma\_1, \sigma\_1\right\} &= 2I\, \\
 \left\{\sigma\_1, \sigma\_2\right\} &= 0\,.\\
\end{align}}

Adding the commutator (

a
b
−
b
a

{\displaystyle ab - ba}

) to the anticommutator (

a
b
b
a

{\displaystyle ab + ba}

) gives the general formula for multiplying any 2 arbitrary "vectors" (or rather their matrix representations):

(

a
→

⋅

σ
→

)
(

b
→

⋅

σ
→

# )
(

a
→

⋅

b
→

)

I
i
(

a
→

×

b
→

)
⋅

σ
→

{\displaystyle (\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) = (\vec{a} \cdot \vec{b}) \, I + i ( \vec{a} \times \vec{b} )\cdot \vec{\sigma}}

If

i

{\displaystyle i}

 is identified with the pseudoscalar

σ
x

σ
y

σ
z

{\displaystyle \sigma\_x \sigma\_y \sigma\_z }

 then the right hand side becomes

a
⋅
b
a
∧
b

{\displaystyle a \cdot b + a \wedge b }

 which is also the definition for the geometric productw of two vectors in geometric algebraw (Clifford algebraw). The geometric product of two vectors is a multivectorw.

Further reading: Quan­tum Me­chan­ics for En­gi­neers and How (not) to teach Lorentz covariance of the Dirac equation

Back to top

## Multivectors

See also: Dirac algebra
External links:

A brief introduction to geometric algebra
A brief introduction to Clifford algebra
The Construction of Spinors in Geometric Algebra
Functions of Multivector Variables
Clifford Algebra Representations

Clifford algebra (Euclidean algebra) is a type of algebra characterized by the geometric productw of scalarsw, vectorsw, bivectorsw, trivectorsw...etc.

Just as a vector has length so a bivector has area and a trivector has volume.

Just as a vector has direction so a bivector has orientation. In three dimensions a trivector has only one possible orientation and is therefore a pseudoscalarw.

In three dimensions
 Bivector
 Psuedoscalar

But in four dimensions a trivector becomes a pseudovectorw and the quadvector becomes the
pseudoscalar.

### Multiplication of arbitrary vectors

Instead of the dot product or the wedge productw Clifford algebra uses the geometric productw.

u

v

=
(

u

x

e

x

u

y

e

y

)
(

v

x

e

x

v

y

e

y

)

=

u

x

v

x

e

x

e

x

u

x

v

y

e

x

e

y

u

y

v

x

e

y

e

x

u

y

v

y

e

y

e

y

=

u

x

v

x

(
1
)

u

y

v

y

(
1
)

u

x

v

y

e

x

e

y

−

u

y

v

x

e

x

e

y

=
(

u

x

v

x

u

y

v

y

)

(
1
)

(

u

x

v

y

−

u

y

v

x

)

e

x
y

=

u

⋅

v

u

∧

v

=

scalar

bivector

{\displaystyle
\begin{align}
\mathbf{u} \mathbf{v} &=
(u\_{x} {\color{blue}e\_{x}} + u\_{y} {\color{blue} e\_{y}} )
(v\_{x} {\color{blue}e\_{x}} + v\_{y} {\color{blue} e\_{y}} ) \\ &=
u\_{x} v\_{x} {\color{red}e\_{x} e\_{x}} +
u\_{x} v\_{y} {\color{green}e\_{x} e\_{y}} +
u\_{y} v\_{x} {\color{green}e\_{y} e\_{x}} +
u\_{y} v\_{y} {\color{red}e\_{y} e\_{y}} \\ &=
u\_{x} v\_{x} {\color{red}(1)} +
u\_{y} v\_{y} {\color{red}(1)} +
u\_{x} v\_{y} {\color{green}e\_{x} e\_{y}} -
u\_{y} v\_{x} {\color{green}e\_{x} e\_{y}}
\\ &=
(u\_{x} v\_{x} + u\_{y} v\_{y}){\color{red}(1)} +
(u\_{x} v\_{y} - u\_{y} v\_{x}) {\color{green}e\_{xy}} \\ &=
\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \wedge \mathbf{v} \\ &=
{\color{red}\text{scalar}} +
{\color{green}\text{bivector}}
\end{align}
}

A scalar plus a bivector (or any number of bladesw of different grade) is called a multivectorw. The idea of adding a scalar and a bivector might seem wrong but in the real world it just means that what appears to be a single equation is in fact a set of simultaneous equations.

For example:

u

v

=
5

{\displaystyle
\mathbf{u} \mathbf{v} = 5
}

would just mean that:

(

u

x

v

x

u

y

v

y

)

(
1
)

=
5

and

(

u

x

v

y

−

u

y

v

x

)

e

x
y

=
0

{\displaystyle
\begin{align}
(u\_{x} v\_{x} + u\_{y} v\_{y}){\color{red}(1)} = 5 \\
 \quad \quad \quad \text{and} \\
(u\_{x} v\_{y} - u\_{y} v\_{x}) {\color{green}e\_{xy}} = 0
\end{align}
}

Back to top

### Rules

All the properties of Clifford algebra derive from a few simple rules.

Let

e

x

,

{\displaystyle {\color{blue}e\_{x}},}

e

y

,

{\displaystyle {\color{blue}e\_{y}},}

 and

e

z

{\displaystyle {\color{blue}e\_{z}}}

 be perpendicular unit vectorsw.

Multiplying two perpendicular vectors results in a bivector:

e
x

e
y

=

e

x
y

{\displaystyle e\_x e\_y = {\color{green}e\_{xy}}}

Multiplying three perpendicular vectors results in a trivector:

e
x

e
y

e
z

=

e

x
y
z

{\displaystyle e\_x e\_y e\_z = {\color{orange}e\_{xyz}}}

Multiplying parallel vectors results in a scalar:

e
x

e
x

=

e

x
x

=

1

{\displaystyle e\_x e\_x = {\color{red}e\_{xx}} = {\color{red}1}}

Clifford algebra is associative therefore the fact that multiplying parallel vectors results in a scalar means that:

(

e

x

e

y

)

(

e

y

)

=

e

x

(

e

y

e

y

)

=

e

x

(
1
)

=

e

x

{\displaystyle
\begin{align}
{\color{green}(e\_{x} e\_{y})} {\color{blue} (e\_{y}) } &=
{\color{blue} e\_{x}} {\color{red}(e\_{y} e\_{y})} \\ &=
{\color{blue} e\_{x}} {\color{red}(1)} \\ &=
{\color{blue} e\_{x}}
\end{align}
}

and:

(

e

x

e

y

)

(

e

y

e

z

)

=

e

x
y
y

e

x
y
z

=

e

x

e

x
y
z

{\displaystyle
\begin{align}
{\color{green} (e\_{x} e\_{y})} {\color{blue} (e\_{y} + e\_{z}) } &=
{\color{blue} e\_{xyy}} + {\color{orange}e\_{xyz}} \\ &=
{\color{blue} e\_{x}} + {\color{orange}e\_{xyz}}
\end{align}
}

and:

(

e

x

e

y

)

(

e

z

)

=

e

x
y
z

{\displaystyle
\begin{align}
{\color{green} (e\_{x} e\_{y}) } {\color{blue} (e\_{z}) } &=
{\color{orange} e\_{xyz}}
\end{align}
}

Rotation from x to y is the negative of rotation from y to x:

e

x
y

=
−

e

y
x

{\displaystyle e\_{xy} = -e\_{yx}}

Therefore:

(

e

x

e

y

)

(

e

x

)

=

e

x

(

e

y

e

x

)

=

−

e

x

(

e

x

e

y

)

=
−

(

e

x

e

x

)

e

y

=
−

(
1
)

e

y

=
−

e

y

{\displaystyle
\begin{align}
 {\color{green}(e\_{x} e\_{y})} {\color{blue} (e\_{x}) } &=
 {\color{blue} e\_{x}} {\color{green}(e\_{y} e\_{x})} \\ &=
 {\color{blue} -e\_{x}} {\color{green}(e\_{x} e\_{y})} \\ &=
-{\color{red} (e\_{x} e\_{x})} {\color{blue} e\_{y}} \\ &=
-{\color{red} (1)} {\color{blue} e\_{y}} \\ &=
-{\color{blue} e\_{y}}
\end{align}
}

Back to top

### Multiplication tables

In one dimension:

1

e

x

e

x

1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1 } &
{\color{blue} e\_{x}}

\\\hline
{\color{blue} e\_{x}} &
{\color{red} 1 }

\end{array}
}

In two dimensions:

1

e

x

e

y

e

x
y

e

x

1

e

x
y

e

y

e

y

−

e

x
y

1

−

e

x

e

x
y

−

e

y

e

x

−
1

{\displaystyle
\begin{array}{r|rr|r}

{\color{red} 1 } &
{\color{blue} e\_{x}} &
{\color{blue} e\_{y}} &
{\color{green} e\_{xy}}

\\\hline
{\color{blue} e\_{x}} &
{\color{red} 1 } &
{\color{green} e\_{xy}} &
{\color{blue} {e\_{y}}}

\\
{\color{blue} e\_{y}} &
{\color{green} -e\_{xy}} &
{\color{red} 1 } &
{\color{blue} -e\_{x}}

\\\hline
{\color{green} e\_{xy}} &
{\color{blue} -e\_{y}} &
{\color{blue} {e\_{x}}} &
{\color{red} {-1}}

\end{array}
}

In three dimensions:

1

e

x

e

y

e

z

e

x
y

e

x
z

e

y
z

e

x
y
z

e

x

1

e

x
y

e

x
z

e

y

e

z

e

x
y
z

e

y
z

e

y

−

e

x
y

1

e

y
z

−

e

x

−

e

x
y
z

e

z

−

e

x
z

e

z

−

e

x
z

−

e

y
z

1

e

x
y
z

−

e

x

−

e

y

e

x
y

e

x
y

−

e

y

e

x

e

x
y
z

−
1

−

e

y
z

e

x
z

−

e

z

e

x
z

−

e

z

−

e

x
y
z

e

x

e

y
z

−
1

−

e

x
y

e

y

e

y
z

e

x
y
z

−

e

z

e

y

−

e

x
z

e

x
y

−
1

−

e

x

e

x
y
z

e

y
z

−

e

x
z

e

x
y

−

e

z

e

y

−

e

x

−
1

{\displaystyle
\begin{array}{r|rrr|rrr|r}

{\color{red} 1 } &
{\color{blue} e\_{x}} &
{\color{blue} e\_{y}} &
{\color{blue} e\_{z}} &
{\color{green} e\_{xy}} &
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}} &
{\color{orange} e\_{xyz}}

\\\hline
{\color{blue} e\_{x}} &
{\color{red} 1 } &
{\color{green} e\_{xy}} &
{\color{green} e\_{xz}} &
{\color{blue} e\_{y}} &
{\color{blue} e\_{z}} &
{\color{orange} e\_{xyz}} &
{\color{green} e\_{yz}}

\\
{\color{blue} e\_{y}} &
{\color{green} -e\_{xy}} &
{\color{red} 1 } &
{\color{green} e\_{yz}} &
{\color{blue} -e\_{x}} &
{\color{orange} -e\_{xyz}} &
{\color{blue} e\_{z}} &
{\color{green} -e\_{xz}}
\\
{\color{blue} e\_{z}} &
{\color{green} -e\_{xz}} &
{\color{green} -e\_{yz}} &
{\color{red} 1 } &
{\color{orange} e\_{xyz}}&
{\color{blue} -e\_{x}} &
{\color{blue} -e\_{y}} &
{\color{green} e\_{xy}}

\\\hline
{\color{green} e\_{xy}} &
{\color{blue} -e\_{y}} &
{\color{blue} e\_{x}} &
{\color{orange} e\_{xyz}} &
{\color{red} -1} &
{\color{green} -e\_{yz}} &
{\color{green} e\_{xz}} &
{\color{blue} -e\_{z}}

\\
{\color{green} e\_{xz}} &
{\color{blue} -e\_{z}} &
{\color{orange} -e\_{xyz}} &
{\color{blue} e\_{x}} &
{\color{green} e\_{yz}} &
{\color{red} -1 } &
{\color{green} -e\_{xy}} &
{\color{blue} e\_{y}}

\\
{\color{green} e\_{yz}} &
{\color{orange} e\_{xyz}} &
{\color{blue} -e\_{z}} &
{\color{blue} e\_{y}} &
{\color{green} -e\_{xz}} &
{\color{green} e\_{xy}} &
{\color{red} -1 } &
{\color{blue} -e\_{x}}

\\\hline
{\color{orange} e\_{xyz}} &
{\color{green} e\_{yz}} &
{\color{green} -e\_{xz}} &
{\color{green} e\_{xy}} &
{\color{blue} -e\_{z}} &
{\color{blue} e\_{y}} &
{\color{blue} -e\_{x}} &
{\color{red} -1 }

\end{array}
}

In four dimensions:

1

e

1

e

2

e

3

e

4

e

12

e

13

e

14

e

23

e

24

e

34

e

123

e

124

e

134

e

234

e

1234

e

1

1

e

12

e

13

e

14

e

2

e

3

e

4

e

123

e

124

e

134

e

23

e

24

e

34

e

1234

e

234

e

2

−

e

12

1

e

23

e

24

−

e

1

−

e

123

−

e

124

e

3

e

4

e

234

−

e

13

−

e

14

−

e

1234

e

34

−

e

134

e

3

−

e

13

−

e

23

1

e

34

e

123

−

e

1

−

e

134

−

e

2

−

e

234

e

4

e

12

e

1234

−

e

14

−

e

24

e

124

e

4

−

e

14

−

e

24

−

e

34

1

e

124

e

123

−

e

1

e

234

−

e

2

−

e

3

−

e

1234

e

12

e

13

e

23

−

e

123

e

12

−

e

2

e

1

e

123

e

124

−
1

−

e

23

−

e

24

e

13

e

14

e

1234

−

e

3

−

e

4

−

e

234

e

134

−

e

34

e

13

−

e

3

−

e

123

e

1

e

134

e

23

−
1

−

e

34

−

e

12

−

e

1234

e

14

e

2

e

234

−

e

4

−

e

124

e

24

e

14

−

e

4

−

e

124

−

e

123

e

1

e

24

e

34

−
1

e

1234

−

e

12

−

e

13

−

e

234

e

2

e

3

e

123

−

e

23

e

23

e

123

−

e

3

e

2

e

234

−

e

13

e

12

e

1234

−
1

−

e

34

e

24

−

e

1

−

e

134

e

124

−

e

4

−

e

14

e

24

e

124

−

e

4

−

e

234

e

2

−

e

14

−

e

1234

e

12

e

34

−
1

−

e

23

e

123

−

e

144

−

e

123

e

3

e

13

e

34

e

134

e

234

−

e

4

e

3

e

1234

−

e

14

e

13

−

e

24

e

23

−
1

−

e

124

e

123

−

e

1

−

e

2

−

e

12

e

123

e

23

−

e

13

e

12

e

1234

−

e

3

e

2

e

234

−

e

1

−

e

134

e

124

−
1

−

e

34

e

24

−

e

14

−

e

4

e

124

e

24

−

e

14

−

e

1234

e

12

−

e

4

−

e

234

e

2

e

123

−

e

1

−

e

123

e

34

−
1

−

e

23

e

13

e

3

e

134

e

34

e

1234

−

e

14

e

13

e

234

−

e

4

e

3

−

e

124

e

123

−

e

1

−

e

24

e

23

−
1

−

e

12

−

e

2

e

234

−

e

1234

e

34

−

e

24

e

23

−

e

134

e

124

−

e

123

−

e

4

e

3

−

e

2

e

14

−

e

13

e

12

−
1

e

1

e

1234

−

e

234

e

134

−

e

124

e

123

−

e

34

e

24

−

e

23

−

e

14

e

13

−

e

12

e

4

−

e

3

e

2

−

e

1

1

{\displaystyle

{\color{red} 1 } &
{\color{black} e\_{1}} &
{\color{black} e\_{2}} &
{\color{black} e\_{3}} &
{\color{black} e\_{4}} &
{\color{black} e\_{12}} &
{\color{black} e\_{13}} &
{\color{black} e\_{14}} &
{\color{black} e\_{23}} &
{\color{black} e\_{24}} &
{\color{black} e\_{34}} &
{\color{black} e\_{123}} &
{\color{black} e\_{124}} &
{\color{black} e\_{134}} &
{\color{black} e\_{234}} &
{\color{red} e\_{1234}}

\\\hline

{\color{black} e\_{1}} &
{\color{red} 1 } &
{\color{black} e\_{12}} &
{\color{black} e\_{13}} &
{\color{black} e\_{14}} &
{\color{black} e\_{2}} &
{\color{black} e\_{3}} &
{\color{black} e\_{4}} &
{\color{black} e\_{123}} &
{\color{black} e\_{124}} &
{\color{black} e\_{134}} &
{\color{black} e\_{23}} &
{\color{black} e\_{24}} &
{\color{black} e\_{34}} &
{\color{red} e\_{1234}} &
{\color{black} e\_{234}}

\\

{\color{black} e\_{2}} &
{\color{black} -e\_{12}} &
{\color{red} 1 } &
{\color{black} e\_{23}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{123}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{3}} &
{\color{black} e\_{4}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{13}} &
{\color{black} -e\_{14}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{34}} &
{\color{black} -e\_{134}}

\\

{\color{black} e\_{3}} &
{\color{black} -e\_{13}} &
{\color{black} -e\_{23}} &
{\color{red} 1 } &
{\color{black} e\_{34}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{134}} &
{\color{black} -e\_{2}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{4}} &
{\color{black} e\_{12}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{14}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{124}}

\\

{\color{black} e\_{4}} &
{\color{black} -e\_{14}} &
{\color{black} -e\_{24}} &
{\color{black} -e\_{34}} &
{\color{red} 1 } &
{\color{black} e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{2}} &
{\color{black} -e\_{3}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{12}} &
{\color{black} e\_{13}} &
{\color{black} e\_{23}} &
{\color{black} -e\_{123}}

\\\hline

{\color{black} e\_{12}} &
{\color{black} -e\_{2}} &
{\color{black} e\_{1}} &
{\color{black} e\_{123}} &
{\color{black} e\_{124}} &
{\color{red} -1 } &
{\color{black} -e\_{23}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{13}} &
{\color{black} e\_{14}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{3}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{134}} &
{\color{black} -e\_{34}}

\\

{\color{black} e\_{13}} &
{\color{black} -e\_{3}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{1}} &
{\color{black} e\_{134}} &
{\color{black} e\_{23}} &
{\color{red} -1 } &
{\color{black} -e\_{34}} &
{\color{black} -e\_{12}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{14}} &
{\color{black} e\_{2}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{24}}

\\

{\color{black} e\_{14}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{124}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{1}} &
{\color{black} e\_{24}} &
{\color{black} e\_{34}} &
{\color{red} -1 } &
{\color{red} e\_{1234}} &
{\color{black} -e\_{12}} &
{\color{black} -e\_{13}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{2}} &
{\color{black} e\_{3}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{23}}

\\

{\color{black} e\_{23}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{3}} &
{\color{black} e\_{2}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{13}} &
{\color{black} e\_{12}} &
{\color{red} e\_{1234}} &
{\color{red} -1 } &
{\color{black} -e\_{34}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{134}} &
{\color{black} e\_{124}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{14}}

\\

{\color{black} e\_{24}} &
{\color{black} e\_{124}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{2}} &
{\color{black} -e\_{14}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{12}} &
{\color{black} e\_{34}} &
{\color{red} -1 } &
{\color{black} -e\_{23}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{144}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{3}} &
{\color{black} e\_{13}}

\\

{\color{black} e\_{34}} &
{\color{black} e\_{134}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{4}} &
{\color{black} e\_{3}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{14}} &
{\color{black} e\_{13}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{23}} &
{\color{red} -1 } &
{\color{black} -e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{2}} &
{\color{black} -e\_{12}}

\\\hline

{\color{black} e\_{123}} &
{\color{black} e\_{23}} &
{\color{black} -e\_{13}} &
{\color{black} e\_{12}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{3}} &
{\color{black} e\_{2}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{134}} &
{\color{black} e\_{124}} &
{\color{red} -1 } &
{\color{black} -e\_{34}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{14}} &
{\color{black} -e\_{4}}

\\

{\color{black} e\_{124}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{14}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{12}} &
{\color{black} -e\_{4}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{2}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{123}} &
{\color{black} e\_{34}} &
{\color{red} -1 } &
{\color{black} -e\_{23}} &
{\color{black} e\_{13}} &
{\color{black} e\_{3}}

\\

{\color{black} e\_{134}} &
{\color{black} e\_{34}} &
{\color{red} e\_{1234}} &
{\color{black} -e\_{14}} &
{\color{black} e\_{13}} &
{\color{black} e\_{234}} &
{\color{black} -e\_{4}} &
{\color{black} e\_{3}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{1}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{23}} &
{\color{red} -1 } &
{\color{black} -e\_{12}} &
{\color{black} -e\_{2}}

\\

{\color{black} e\_{234}} &
{\color{red} -e\_{1234}} &
{\color{black} e\_{34}} &
{\color{black} -e\_{24}} &
{\color{black} e\_{23}} &
{\color{black} -e\_{134}} &
{\color{black} e\_{124}} &
{\color{black} -e\_{123}} &
{\color{black} -e\_{4}} &
{\color{black} e\_{3}} &
{\color{black} -e\_{2}} &
{\color{black} e\_{14}} &
{\color{black} -e\_{13}} &
{\color{black} e\_{12}} &
{\color{red} -1 } &
{\color{black} e\_{1}}

\\\hline

{\color{red} e\_{1234}} &
{\color{black} -e\_{234}} &
{\color{black} e\_{134}} &
{\color{black} -e\_{124}} &
{\color{black} e\_{123}} &
{\color{black} -e\_{34}} &
{\color{black} e\_{24}} &
{\color{black} -e\_{23}} &
{\color{black} -e\_{14}} &
{\color{black} e\_{13}} &
{\color{black} -e\_{12}} &
{\color{black} e\_{4}} &
{\color{black} -e\_{3}} &
{\color{black} e\_{2}} &
{\color{black} -e\_{1}} &
{\color{red} 1 }

\end{array}
}

Back to top

### Basis

Every multivector of the Clifford algebra can be expressed as a linear combination of the canonical basis elements. The basis elements of the Clifford algebra Cℓ3 are

{

1

,

e
x

,

e
y

,

e
z

,

e

x
y

,

e

x
z

,

e

y
z

,

e

x
y
z

}

{\displaystyle \{{\color{red}1}, \, {\color{blue}e\_x}, \, {\color{blue}e\_y}, \, {\color{blue}e\_z}, \, {\color{green}e\_{xy}}, \, {\color{green}e\_{xz}}, \, {\color{green}e\_{yz}}, \, {\color{orange}e\_{xyz}} \}}

 and the general element of Cℓ3 is given by

# A

a
0

(
1
)

a
1

e
x

a
2

e
y

a
3

e
z

a
4

e

x
y

a
5

e

x
z

a
6

e

y
z

a
7

e

x
y
z

.

{\displaystyle A = a\_0{\color{red}(1)} + a\_1 {\color{blue}e\_x} + a\_2 {\color{blue}e\_y} + a\_3 {\color{blue}e\_z} + a\_4 {\color{green}e\_{xy}} + a\_5 {\color{green}e\_{xz}} + a\_6 {\color{green}e\_{yz}} + a\_7 {\color{orange}e\_{xyz}}.}

If

a
0

,

a
1

,

a
2

,

a
3

,

a
4

,

a
5

,

a
6

,

a
7

{\displaystyle a\_0, a\_1, a\_2, a\_3, a\_4, a\_5, a\_6, a\_7 }

 are all real then the Clifford algebra is Cℓ3(**R**). If the coefficients are allowed to be complexw then the Clifford algebra is Cℓ3(**C**).

A multivectorw can be separated into components of different grades:

⟨

A

⟩
0

=

a
0

(
1
)

⟨

A

⟩
1

=

a
1

e
x

a
2

e
y

a
3

e
z

⟨

A

⟩
2

=

a
4

e

x
y

a
5

e

x
z

a
6

e

y
z

⟨

A

⟩
3

=

a
7

e

x
y
z

{\displaystyle
\begin{align}

\langle \mathbf{A} \rangle\_0 = a\_0{\color{red}(1)}\\

\langle \mathbf{A} \rangle\_1 = a\_1 {\color{blue}e\_x} + a\_2 {\color{blue}e\_y} + a\_3 {\color{blue}e\_z}\\

\langle \mathbf{A} \rangle\_2 = a\_4 {\color{green}e\_{xy}} + a\_5 {\color{green}e\_{xz}} + a\_6 {\color{green}e\_{yz}}\\

\langle \mathbf{A} \rangle\_3 = a\_7 {\color{orange}e\_{xyz}}

\end{align}

}

The elements of even grade form a subalgebra because the sum or product of even grade elements always results in an element of even grade. The elements of odd grade do not form a subalgebra.

Back to top

### Hodge dual

Multiplying (using geometric product) any multivector by the pseudoscalar converts the multivector to its dual.

The **Hodge star operator** is a linear operator on the exterior algebra of *V*, mapping *k*-vectors to (*n* – *k*)-vectors, for

0
≤
k
≤
n

{\displaystyle 0\leq k \leq n}

. It has the following property, which defines it completely:

α
∧
(

⋆

β
# )
⟨
α
,
β
⟩

ω

{\displaystyle \alpha \wedge ({\star} \beta) = \langle \alpha,\beta \rangle \,\omega }

 for every pair of *k*-vectors

α
,
β
∈

⋀

k

V
.

{\displaystyle \alpha,\beta\in {\textstyle\bigwedge}^{\!k}V .}

The Hodge star relates the exterior and cross product in three dimensions:

⋆

(

u

∧

v

# )

u

×

v

⋆

(

u

×

v

# )

u

∧

v

.

{\displaystyle {\star} (\mathbf{u} \wedge \mathbf{v}) = \mathbf{u} \times \mathbf{v} \qquad {\star} (\mathbf{u} \times \mathbf {v}) = \mathbf{u} \wedge \mathbf{v} .}

And

⋆
(

w
1

∧
⋯
∧

w
k

)

=

u
1

∧
⋯
∧

u

n
−
k

,

{\displaystyle \star(w\_1\wedge\cdots\wedge w\_k) \,=\, u\_1\wedge\cdots\wedge u\_{n-k},}

Back to top

### Relation to other algebras

C

ℓ
0

(

R

)

{\displaystyle C\ell\_0 (\mathbf{R})}

: Real numbersw (scalarsw). A scalar can (and should) be thought of as zero vectors multiplied together. See Empty productw.

C

ℓ
0

(

C

)

{\displaystyle C\ell\_0 (\mathbf{C})}

: Complex numbersw

C

ℓ
1

(

R

)

{\displaystyle C\ell\_1 (\mathbf{R})}

: Split-complex numbersw

1

e

x

e

x

1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1 } &
{\color{blue} e\_{x}}

\\\hline
{\color{blue} e\_{x}} &
{\color{red} 1 }

\end{array}
}

C

ℓ
1

(

C

)

{\displaystyle C\ell\_1 (\mathbf{C})}

: Bicomplex numbersw

C

ℓ
2
0

(

R

)

{\displaystyle C\ell\_2^0 (\mathbf{R})}

: Complex numbersw (The superscript 0 indicates the even subalgebra)

1

e

x
y

e

x
y

−
1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1} &
{\color{green} e\_{xy}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} {-1}}

\end{array}
}

C

ℓ
3
0

(

R

)

{\displaystyle C\ell\_3^0 (\mathbf{R})}

: Quaternionsw

1

e

x
y

e

x
z

e

y
z

e

x
y

−
1

−

e

y
z

e

x
z

e

x
z

e

y
z

−
1

−

e

x
y

e

y
z

−

e

x
z

e

x
y

−
1

{\displaystyle
\begin{array}{r|rrr}

{\color{red} 1} &
{\color{green} e\_{xy}} &
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} -1 } &
{\color{green} -e\_{yz}} &
{\color{green} e\_{xz}}

\\
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}} &
{\color{red} -1 } &
{\color{green} -e\_{xy}}

\\
{\color{green} e\_{yz}} &
{\color{green} -e\_{xz}} &
{\color{green} e\_{xy}} &
{\color{red} -1 }

\end{array}
}

C

ℓ
3
0

(

C

)

{\displaystyle C\ell\_3^0 (\mathbf{C})}

: Biquaternionsw

Back to top

### Multivector multiplication using tensors

To find the product

A
# B
(

a
0

1

a
1

e

x

a
2

e

y

a
3

e

x
y

)
(

b
0

1

b
1

e

x

b
2

e

y

b
3

e

x
y

)

{\displaystyle
AB =
(a\_0 {\color{red} 1 } + a\_1 {\color{blue} e\_{x}} + a\_2 {\color{blue} e\_{y}} + a\_3 {\color{green} e\_{xy}})
(b\_0 {\color{red} 1 } + b\_1 {\color{blue} e\_{x}} + b\_2 {\color{blue} e\_{y}} + b\_3 {\color{green} e\_{xy}})
}

we have to multiply every component of the first multivector with every component of the second multivector.

A
# B

(

a
0

b
0

1

1

a
0

b
1

1

e

x

a
0

b
2

1

e

y

a
0

b
3

1

e

x
y

)

(

a
1

b
0

e

x

1

a
1

b
1

e

x

e

x

a
1

b
2

e

x

e

y

a
1

b
3

e

x

e

x
y

)

(

a
2

b
0

e

y

1

a
2

b
1

e

y

e

x

a
2

b
2

e

y

e

y

a
2

b
3

e

y

e

x
y

)

(

a
3

b
0

e

x
y

1

a
3

b
1

e

x
y

e

x

a
3

b
2

e

x
y

e

y

a
3

b
3

e

x
y

e

x
y

)

{\displaystyle
\begin{align}
AB = &

(a\_0 b\_0 {\color{red}1}{\color{red}1} + a\_0 b\_1 {\color{red}1}{\color{blue} e\_{x}} + a\_0 b\_2 {\color{red}1}{\color{blue} e\_{y}} + a\_0 b\_3 {\color{red}1}{\color{green} e\_{xy}}) \\ &+

(a\_1 b\_0 {\color{blue} e\_{x}}{\color{red}1} + a\_1 b\_1 {\color{blue} e\_{x}}{\color{blue} e\_{x}} + a\_1 b\_2 {\color{blue} e\_{x}}{\color{blue} e\_{y}} + a\_1 b\_3 {\color{blue} e\_{x}}{\color{green} e\_{xy}}) \\ &+

(a\_2 b\_0 {\color{blue} e\_{y}}{\color{red}1} + a\_2 b\_1 {\color{blue} e\_{y}}{\color{blue} e\_{x}} + a\_2 b\_2 {\color{blue} e\_{y}}{\color{blue} e\_{y}} + a\_2 b\_3 {\color{blue} e\_{y}}{\color{green} e\_{xy}}) \\ &+

(a\_3 b\_0 {\color{green} e\_{xy}}{\color{red}1} + a\_3 b\_1 {\color{green} e\_{xy}}{\color{blue} e\_{x}} + a\_3 b\_2 {\color{green} e\_{xy}}{\color{blue} e\_{y}} + a\_3 b\_3 {\color{green} e\_{xy}}{\color{green} e\_{xy}})

\end{align}
}

Then we reduce each of the 16 resulting terms to its standard form.

A
# B

(

a
0

b
0

1

a
0

b
1

e

x

a
0

b
2

e

y

a
0

b
3

e

x
y

)

(

a
1

b
0

e

x

a
1

b
1

1

a
1

b
2

e

x
y

a
1

b
3

e

y

)

(

a
2

b
0

e

y

−

a
2

b
1

e

x
y

a
2

b
2

1

−

a
2

b
3

e

x

)

(

a
3

b
0

e

x
y

−

a
3

b
1

e

y

a
3

b
2

e

x

−

a
3

b
3

1

)

{\displaystyle
\begin{align}
AB = &

(a\_0 b\_0 {\color{red}1} + a\_0 b\_1 {\color{blue} e\_{x}} + a\_0 b\_2 {\color{blue} e\_{y}} + a\_0 b\_3 {\color{green} e\_{xy}}) \\ &+

(a\_1 b\_0 {\color{blue} e\_{x}} + a\_1 b\_1 {\color{red}1} + a\_1 b\_2 {\color{green} e\_{xy}} + a\_1 b\_3 {\color{blue} e\_{y}}) \\ &+

(a\_2 b\_0 {\color{blue} e\_{y}} - a\_2 b\_1 {\color{green} e\_{xy}} + a\_2 b\_2 {\color{red}1} - a\_2 b\_3 {\color{blue} e\_{x}}) \\ &+

(a\_3 b\_0 {\color{green} e\_{xy}} - a\_3 b\_1 {\color{blue} e\_{y}} + a\_3 b\_2 {\color{blue} e\_{x}} - a\_3 b\_3 {\color{red}1})

\end{align}
}

Finally we collect like products into the four components of the final multivector.

A
# B

(

a
0

b
0

a
1

b
1

a
2

b
2

−

a
3

b
3

)

1

(

a
1

b
0

a
0

b
1

a
3

b
2

−

a
2

b
3

)

e

x

(

a
2

b
0

−

a
3

b
1

a
0

b
2

a
1

b
3

)

e

y

(

a
3

b
0

−

a
2

b
1

a
1

b
2

a
0

b
3

)

e

x
y

{\displaystyle
\begin{align}
AB = &
( a\_0 b\_0 + a\_1 b\_1 + a\_2 b\_2 - a\_3 b\_3 ) {\color{red}1} \\ &+
( a\_1 b\_0 + a\_0 b\_1 + a\_3 b\_2 - a\_2 b\_3 ) {\color{blue} e\_{x} } \\ &+
( a\_2 b\_0 - a\_3 b\_1 + a\_0 b\_2 + a\_1 b\_3 ) {\color{blue} e\_{y} } \\ &+
( a\_3 b\_0 - a\_2 b\_1 + a\_1 b\_2 + a\_0 b\_3 ) {\color{green} e\_{xy} }
\end{align}
}

This is all very tedious and error-prone. It would be nice if there was some way to cut straight to the end. Tensor notation allows us to do just that.

To find the tensor that we need we first need to know which terms end up as scalars, which terms end up as vectors...etc. There is an easy way to do this and it involves the multiplication table.

First lets start with an easy one.

Back to top

#### Complex numbers

The multiplication table for

C

ℓ
2
0

(

R

)

{\displaystyle C\ell\_2^0 (\mathbf{R})}

 (Which is isomorphic to complex numbers)

1

e

x
y

e

x
y

−
1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1} &
{\color{green} e\_{xy}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} -1 }

\end{array}
}

We can see then that:

(

1

e

x
y

e

x
y

1

)

(

1

e

x
y

)

=

(

1

⋅

1

e

x
y

e

x
y

e

x
y

1

1

e

x
y

)

=

(

1

−

1

e

x
y

e

x
y

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

{\color{red} 1} &
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xy}} &
{\color{red} 1 }

\end{array}
\right)

\left(
\begin{array}{rrrr}

{\color{red} 1} \\
{\color{green} e\_{xy}}

\end{array}
\right) &=

\left(
\begin{array}{lcl}

{\color{red} 1} \cdot
{\color{red} 1} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xy}}
{\color{red} 1} &
+ &
{\color{red} 1 }
{\color{green} e\_{xy}}

\end{array}
\right) \\ &=

\left(
\begin{array}{lcl}

{\color{red} 1} &
- &
{\color{red} 1 }

\\
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xy}}

\end{array}
\right)

\end{align}
}

It worked! All the terms in the first row are scalars and all the terms in the second row are bivectors. This is exactly what we are looking for.

(

b
0

⋅

1

b
1

⋅

e

x
y

b
1

⋅

e

x
y

b
0

⋅

1

)

(

a
0

⋅

1

a
1

⋅

e

x
y

)

=

(

b
0

a
0

⋅

1

⋅

1

b
1

a
1

⋅

e

x
y

e

x
y

b
1

a
0

⋅

e

x
y

1

b
0

a
1

⋅

1

e

x
y

)

=

(

b
0

a
0

⋅

1

−

b
1

a
1

⋅

1

b
1

a
0

⋅

e

x
y

b
0

a
1

⋅

e

x
y

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

b\_0 & \cdot &
{\color{red} 1} \quad &
b\_1 & \cdot &
{\color{green} e\_{xy}}

\\
b\_1 & \cdot &
{\color{green} e\_{xy}} \quad &
b\_0 & \cdot &
{\color{red} 1 }

\end{array}
\right)

\left(
\begin{array}{rrrr}

a\_0 & \cdot &
{\color{red} 1} \\
a\_1 & \cdot &
{\color{green} e\_{xy}}

\end{array}
\right) &=

\left(
\begin{array}{lcl}

b\_0 a\_0 & \cdot &
{\color{red} 1} \cdot
{\color{red} 1} &
+ &
b\_1 a\_1 & \cdot &
{\color{green} e\_{xy}}
{\color{green} e\_{xy}}

\\
b\_1 a\_0 & \cdot &
{\color{green} e\_{xy}}
{\color{red} 1} &
+ &
b\_0 a\_1 & \cdot &
{\color{red} 1 }
{\color{green} e\_{xy}}

\end{array}
\right) \\ &=

\left(
\begin{array}{lcl}

b\_0 a\_0 & \cdot &
{\color{red} 1} &
- &
b\_1 a\_1 & \cdot &
{\color{red} 1 }

\\
b\_1 a\_0 & \cdot &
{\color{green} e\_{xy}} &
+ &
b\_0 a\_1 & \cdot &
{\color{green} e\_{xy}}

\end{array}
\right)

\end{align}
}

Pay special attention to the signs in the final matrix above.

Therefore to find the product

(

a
0

1

a
1

e

x
y

)
(

b
0

1

b
1

e

x
y

)

{\displaystyle
(a\_0 {\color{red} 1 } + a\_1 {\color{green} e\_{xy}} )
(b\_0 {\color{red} 1 } + b\_1 {\color{green} e\_{xy}} )
}

We would multiply:

(

b
0

−

b
1

b
1

b
0

)

(

a
0

a
1

)

=

(

b
0

a
0

−

b
1

a
1

b
1

a
0

b
0

a
1

)

{\displaystyle

\left(
\begin{array}{rr}

{\color{red} b\_0 } &
{\color{green} -b\_1 }

\\
{\color{green} b\_1 } &
{\color{red} b\_0 }

\end{array}
\right)

\left(
\begin{array}{r}

{\color{red} a\_0 } \\
{\color{green} a\_1 }

\end{array}
\right) =

\left(
\begin{array}{rcr}

{\color{red} b\_0 }
{\color{red} a\_0 }&
- &
{\color{red} b\_1 }
{\color{red} a\_1 }

\\
{\color{green} b\_1 }
{\color{green} a\_0 }&
+ &
{\color{green} b\_0 }
{\color{green} a\_1 }

\end{array}
\right)
}

Each row of the final matrix has exactly the right terms with exactly the right signs.

The vector above represents a complex number. You should think of the first column of the matrix above as representing another complex number. All the other terms in the matrix are just there to make our lives a little bit easier. See Advanced\_mathematics#Column\_vectors.

It works. It works so well that complex numbers can be represented as matricesw as:

[

a

−
b

b

a

]

{\displaystyle
\begin{bmatrix}
 a & -b \\
 b & a
\end{bmatrix}
}

Which corresponds perfectly to a multiplication table for complex numbers:

1

−
i

i

1

{\displaystyle
\begin{array}{r|r}

{\color{red} 1} &
{\color{green} -i}

\\\hline
{\color{green} i} &
{\color{red} 1 }

\end{array}
}

1

−

e

x
y

e

x
y

1

{\displaystyle \quad \quad
\begin{array}{r|r}

{\color{red} 1} &
{\color{green} -e\_{xy}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} 1 }

\end{array}
}

Back to top

#### Quaternions

The multiplication table for

C

ℓ
3
0

(

R

)

{\displaystyle C\ell\_3^0 (\mathbf{R})}

 (Which is isomorphic to quaternions) is:

1

e

x
y

e

x
z

e

y
z

e

x
y

−
1

−

e

y
z

e

x
z

e

x
z

e

y
z

−
1

−

e

x
y

e

y
z

−

e

x
z

e

x
y

−
1

# i

e

y
z

# j

e

x
z

# k

e

x
y

{\displaystyle
\begin{array}{r|rrr}

{\color{red} 1} &
{\color{green} e\_{xy}} &
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} -1 } &
{\color{green} -e\_{yz}} &
{\color{green} e\_{xz}}

\\
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}} &
{\color{red} -1 } &
{\color{green} -e\_{xy}}

\\
{\color{green} e\_{yz}} &
{\color{green} -e\_{xz}} &
{\color{green} e\_{xy}} &
{\color{red} -1 }

\end{array} \quad \quad \quad
\begin{array}{l}
i = e\_{yz} \\
j = e\_{xz} \\
k = e\_{xy}
\end{array}
}

The entire 2nd row of the multiplication table is just

e

x
y

{\displaystyle {\color{green} e\_{xy}}}

 multiplied by the entire first row.

The entire 3rd row of the multiplication table is just

e

x
z

{\displaystyle {\color{green} e\_{xz}}}

 multiplied by the entire first row.

The entire 4th row of the multiplication table is just

e

y
z

{\displaystyle {\color{green} e\_{yz}}}

 multiplied by the entire first row.

We can see then that if we multiply each row by the first row again then we get:

(

1

e

x
y

e

x
z

e

y
z

e

x
y

1

e

y
z

e

x
z

e

x
z

e

y
z

1

e

x
y

e

y
z

e

x
z

e

x
y

1

)

(

1

e

x
y

e

x
z

e

y
z

)

=

(

1

⋅

1

e

x
y

e

x
y

e

x
z

e

x
z

e

y
z

e

y
z

e

x
y

1

1

e

x
y

e

y
z

e

x
z

e

x
z

e

y
z

e

x
z

1

e

y
z

e

x
y

1

e

x
z

e

x
y

e

y
z

e

y
z

1

e

x
z

e

x
y

e

x
y

e

x
z

1

e

y
z

)

=

(

1

−

1

−

1

−

1

e

x
y

e

x
y

e

x
y

−

e

x
y

e

x
z

−

e

x
z

e

x
z

e

x
z

e

y
z

e

y
z

−

e

y
z

e

y
z

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

{\color{red} 1} &
{\color{green} e\_{xy}} &
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}}

\\
{\color{green} e\_{xy}} &
{\color{red} 1 } &
{\color{green} e\_{yz}} &
{\color{green} e\_{xz}}

\\
{\color{green} e\_{xz}} &
{\color{green} e\_{yz}} &
{\color{red} 1 } &
{\color{green} e\_{xy}}

\\
{\color{green} e\_{yz}} &
{\color{green} e\_{xz}} &
{\color{green} e\_{xy}} &
{\color{red} 1 }

\end{array}
\right)

\left(
\begin{array}{rrrr}

{\color{red} 1} \\
{\color{green} e\_{xy}} \\
{\color{green} e\_{xz}} \\
{\color{green} e\_{yz}}

\end{array}
\right) &=

\left(
\begin{array}{rrrr}

{\color{red} 1} \cdot
{\color{red} 1} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xz}}
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{yz}}
{\color{green} e\_{yz}}

\\
{\color{green} e\_{xy}}
{\color{red} 1} &
+ &
{\color{red} 1 }
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{yz}}
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xz}}
{\color{green} e\_{yz}}

\\
{\color{green} e\_{xz}}
{\color{red} 1} &
+ &
{\color{green} e\_{yz}}
{\color{green} e\_{xy}} &
+ &
{\color{red} 1 }
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{yz}}

\\
{\color{green} e\_{yz}}
{\color{red} 1} &
+ &
{\color{green} e\_{xz}}
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xy}}
{\color{green} e\_{xz}} &
+ &
{\color{red} 1 }
{\color{green} e\_{yz}}

\end{array}
\right) \\ &=

\left(
\begin{array}{llll}

{\color{red} 1} &
- &
{\color{red} 1 } &
- &
{\color{red} 1 } &
- &
{\color{red} 1 }

\\
{\color{green} e\_{xy}} &
+ &
{\color{green} e\_{xy}}&
+ &
{\color{green} e\_{xy}} &
- &
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xz}} &
- &
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xz}} &
+ &
{\color{green} e\_{xz}}

\\
{\color{green} e\_{yz}} &
+ &
{\color{green} e\_{yz}} &
- &
{\color{green} e\_{yz}} &
+ &
{\color{green} e\_{yz}}

\end{array}
\right)

\end{align}
}

This works because we have in effect multiplied each term by a second term twice. In other words we have multiplied every term by the square of another term and the square of every term is either 1 or -1.

Therefore to find the product

(

a
0

1

a
1

e

x
y

a
2

e

x
z

a
3

e

y
z

)
(

b
0

1

b
1

e

x
y

b
2

e

x
z

b
3

e

y
z

)

{\displaystyle
(a\_0 {\color{red} 1 } + a\_1 {\color{green} e\_{xy}} + a\_2 {\color{green} e\_{xz}} + a\_3 {\color{green} e\_{yz}})
(b\_0 {\color{red} 1 } + b\_1 {\color{green} e\_{xy}} + b\_2 {\color{green} e\_{xz}} + b\_3 {\color{green} e\_{yz}})
}

We would multiply:

(

b
0

−

b
1

−

b
2

−

b
3

b
1

b
0

b
3

−

b
2

b
2

−

b
3

b
0

b
1

b
3

b
2

−

b
1

b
0

)

(

a
0

a
1

a
2

a
3

)

=

(

b
0

a
0

−

b
1

a
1

−

b
2

a
2

−

b
3

a
3

b
1

a
0

b
0

a
1

b
3

a
2

−

b
2

a
3

b
2

a
0

−

b
3

a
1

b
0

a
2

b
1

a
3

b
3

a
0

b
2

a
1

−

b
1

a
2

b
0

a
3

)

{\displaystyle

\left(
\begin{array}{rrrr}

{\color{red} b\_0 } &
{\color{green} -b\_1 } &
{\color{green} -b\_2 } &
{\color{green} -b\_3 }

\\
{\color{green} b\_1 } &
{\color{red} b\_0 } &
{\color{green} b\_3 } &
{\color{green} {-b\_2 }}

\\
{\color{green} b\_2 } &
{\color{green} -b\_3 } &
{\color{red} b\_0 } &
{\color{green} b\_1 }

\\
{\color{green} b\_3 } &
{\color{green} b\_2 } &
{\color{green} {-b\_1 }} &
{\color{red} b\_0 }

\end{array}
\right)

\left(
\begin{array}{rrrr}

{\color{red} a\_0 } \\
{\color{green} a\_1 } \\
{\color{green} a\_2 } \\
{\color{green} a\_3 }

\end{array}
\right) =

\left(
\begin{array}{rrrr}

{\color{red} b\_0 }
{\color{red} a\_0 } &
- &
{\color{red} b\_1 }
{\color{red} a\_1 } &
- &
{\color{red} b\_2 }
{\color{red} a\_2 } &
- &
{\color{red} b\_3 }
{\color{red} a\_3 }

\\
{\color{green} b\_1 }
{\color{green} a\_0 } &
+ &
{\color{green} b\_0 }
{\color{green} a\_1 } &
+ &
{\color{green} b\_3 }
{\color{green} a\_2 } &
- &
{\color{green} {b\_2 }}
{\color{green} a\_3 }

\\
{\color{green} b\_2 }
{\color{green} a\_0 } &
- &
{\color{green} b\_3 }
{\color{green} a\_1 } &
+ &
{\color{green} b\_0 }
{\color{green} a\_2 } &
+ &
{\color{green} b\_1 }
{\color{green} a\_3 }

\\
{\color{green} b\_3 }
{\color{green} a\_0 } &
+ &
{\color{green} b\_2 }
{\color{green} a\_1 } &
- &
{\color{green} {b\_1 }}
{\color{green} a\_2 } &
+ &
{\color{green} b\_0 }
{\color{green} a\_3 }

\end{array}
\right)
}

Just as complex numbers can be represented as matricesw, so a quaternion can be represented as:

(

a

−
b

−
c

−
d

b

a

d

−
c

c

−
d

a

b

d

c

−
b

a

)

=

(

(

a

−
b

b

a

)

−

(

c

d

−
d

c

)

(

c

−
d

d

c

)

(

a

b

−
b

a

)

)

{\displaystyle
\left(
\begin{array}{rrrr}
 a & -b & -c & -d \\
 b & a & d &-c \\
 c & -d & a & b \\
 d & c & -b & a
\end{array}
\right)=

\left(
\begin{array}{rr}

\left(
\begin{array}{rr}
 a & -b \\
 b & a
\end{array}
\right) &

\left(
\begin{array}{rr}
c & d \\
-d &c
\end{array}
\right) \\

\left(
\begin{array}{rr}
 c & -d \\
 d & c
\end{array}
\right) &

\left(
\begin{array}{rr}
a & b \\
-b & a
\end{array}
\right)

\end{array}
\right)
}

Which corresponds to a multiplication table for quaternions:

1

−

e

x
y

−

e

x
z

−

e

y
z

e

x
y

1

e

y
z

−

e

x
z

e

x
z

−

e

y
z

1

e

x
y

e

y
z

e

x
z

−

e

x
y

1

{\displaystyle
\begin{array}{r|rrr}

{\color{red} 1} &
{\color{green} -e\_{xy}} &
{\color{green} -e\_{xz}} &
{\color{green} -e\_{yz}}

\\\hline
{\color{green} e\_{xy}} &
{\color{red} 1 } &
{\color{green} e\_{yz}} &
{\color{green} -e\_{xz}}

\\
{\color{green} e\_{xz}} &
{\color{green} -e\_{yz}} &
{\color{red} 1 } &
{\color{green} e\_{xy}}

\\
{\color{green} e\_{yz}} &
{\color{green} e\_{xz}} &
{\color{green} -e\_{xy}} &
{\color{red} 1 }

\end{array}
}

Back to top

#### CL2

The multiplication table for

C

ℓ

2

(

R

)

{\displaystyle C\ell\_{2} (\mathbf{R})}

 is:

1

e

x

e

y

e

x
y

e

x

1

e

x
y

e

y

e

y

−

e

x
y

1

−

e

x

e

x
y

−

e

y

e

x

−
1

{\displaystyle
\begin{array}{r|rr|r}

{\color{red} 1 } &
{\color{blue} e\_{x}} &
{\color{blue} e\_{y}} &
{\color{green} e\_{xy}}

\\\hline
{\color{blue} e\_{x}} &
{\color{red} 1 } &
{\color{green} e\_{xy}} &
{\color{blue} {e\_{y}}}

\\
{\color{blue} e\_{y}} &
{\color{green} -e\_{xy}} &
{\color{red} 1 } &
{\color{blue} -e\_{x}}

\\\hline
{\color{green} e\_{xy}} &
{\color{blue} -e\_{y}} &
{\color{blue} {e\_{x}}} &
{\color{red} {-1}}

\end{array}
}

We can see then that:

(

1

e

x

e

y

e

x
y

e

x

1

e

x
y

e

y

e

y

e

x
y

1

e

x

e

x
y

e

y

e

x

1

)

(

1

e

x

e

y

e

x
y

)

=

(

1

⋅

1

e

x

e

x

e

y

e

y

e

x
y

e

x
y

e

x

1

1

e

x

e

x
y

e

y

e

y

e

x
y

e

y

1

e

x
y

e

x

1

e

y

e

x

e

x
y

e

x
y

1

e

y

e

x

e

x

e

y

1

e

x
y

)

=

(

1

1

1

−

1

e

x

e

x

e

x

−

e

x

e

y

−

e

y

e

y

e

y

e

x
y

−

e

x
y

e

x
y

e

x
y

)

{\displaystyle
\begin{align}

\left(
\begin{array}{rrrr}

{\color{red} 1 } &
{\color{blue} e\_{x}} &
{\color{blue} e\_{y}} &
{\color{green} e\_{xy}}

\\
{\color{blue} e\_{x}} &
{\color{red} 1 } &
{\color{green} e\_{xy}} &
{\color{blue} {e\_{y}}}

\\
{\color{blue} e\_{y}} &
{\color{green} e\_{xy}} &
{\color{red} 1 } &
{\color{blue} e\_{x}}

\\
{\color{green} e\_{xy}} &
{\color{blue} e\_{y}} &
{\color{blue} {e\_{x}}} &
{\color{red} {1}}

\end{array}
\right)

\left(
\begin{array}{rrrr}

{\color{red} 1 } \\
{\color{blue} e\_{x}} \\
{\color{blue} e\_{y}} \\
{\color{green} e\_{xy}}

\end{array}
\right) &=

\left(
\begin{array}{rrrr}

{\color{red} 1 } \cdot
{\color{red} 1 } &
+ &

{\color{blue} e\_{x}}
{\color{blue} e\_{x}} &
+ &
{\color{blue} e\_{y}}
{\color{blue} e\_{y}} &
+ &

{\color{green} e\_{xy}}
{\color{green} e\_{xy}}

\\
{\color{blue} e\_{x}}
{\color{red} 1 } &
+ &

{\color{red} 1 }
{\color{blue} e\_{x}} &
+ &
{\color{green} e\_{xy}}
{\color{blue} e\_{y}} &
+ &

{\color{blue} {e\_{y}}}
{\color{green} e\_{xy}}

\\
{\color{blue} e\_{y}}
{\color{red} 1 } &
+ &
{\color{green} e\_{xy}}
{\color{blue} e\_{x}} &
+ &
{\color{red} 1 }
{\color{blue} e\_{y}} &
+ &
{\color{blue} e\_{x}}
{\color{green} e\_{xy}}

\\
{\color{green} e\_{xy}}
{\color{red} 1 } &
+ &
{\color{blue} e\_{y}}
{\color{blue} e\_{x}} &
+ &
{\color{blue} {e\_{x}}}
{\color{blue} e\_{y}} &
+ &
{\color{red} {1}}
{\color{green} e\_{xy}}

\end{array}
\right) \\ &=

\left(
\begin{array}{llll}

 {\color{red}1} & + & {\color{red}1} & + & {\color{red}1} & - & {\color{red}1} \\
 {\color{blue}e\_{x}} & + & {\color{blue}e\_{x}} & + & {\color{blue}e\_{x}} & - & {\color{blue}e\_{x}} \\
 {\color{blue}e\_{y}} & - & {\color{blue}e\_{y}} & + & {\color{blue}e\_{y}} & + & {\color{blue}e\_{y}} \\
 {\color{green}e\_{xy}} & - & {\color{green}e\_{xy}} & + & {\color{green}e\_{xy}} & + & {\color{green}e\_{xy}}

\end{array}
\right)

\end{align}
}

Therefore to find the product

(

a
0

1

a
1

e

x

a
2

e

y

a
3

e

x
y

)
(

b
0

1

b
1

e

x

b
2

e

y

b
3

e

x
y

)

{\displaystyle
(a\_0 {\color{red} 1 } + a\_1 {\color{blue} e\_{x}} + a\_2 {\color{blue} e\_{y}} + a\_3 {\color{green} e\_{xy}})
(b\_0 {\color{red} 1 } + b\_1 {\color{blue} e\_{x}} + b\_2 {\color{blue} e\_{y}} + b\_3 {\color{green} e\_{xy}})
}

We would multiply:

(

b
0

b
1

b
2

−

b
3

b
1

b
0

b
3

−

b
2

b
2

−

b
3

b
0

b
1

b
3

−

b
2

b
1

b
0

)

(

a
0

a
1

a
2

a
3

)

=

(

b
0

a
0

b
1

a
1

b
2

a
2

−

b
3

a
3

b
1

a
0

b
0

a
1

b
3

a
2

−

b
2

a
3

b
2

a
0

−

b
3

a
1

b
0

a
2

b
1

a
3

b
3

a
0

−

b
2

a
1

b
1

a
2

b
0

a
3

)

{\displaystyle

\left(
\begin{array}{rrrr}

{\color{red} b\_0 } &
{\color{blue} b\_1 } &
{\color{blue} b\_2 } &
{\color{green} -b\_3 }

\\
{\color{blue} b\_1 } &
{\color{red} b\_0 } &
{\color{green} b\_3 } &
{\color{blue} {-b\_2 }}

\\
{\color{blue} b\_2 } &
{\color{green} -b\_3 } &
{\color{red} b\_0 } &
{\color{blue} b\_1 }

\\
{\color{green} b\_3 } &
{\color{blue} -b\_2 } &
{\color{blue} {b\_1 }} &
{\color{red} b\_0 }

\end{array}
\right)

\left(
\begin{array}{rrrr}

{\color{red} a\_0 } \\
{\color{blue} a\_1 } \\
{\color{blue} a\_2 } \\
{\color{green} a\_3 }

\end{array}
\right) =

\left(
\begin{array}{rrrr}

{\color{red} b\_0 }
{\color{red} a\_0 } &
+ &
{\color{red} b\_1 }
{\color{red} a\_1 } &
+ &
{\color{red} b\_2 }
{\color{red} a\_2 } &
- &
{\color{red} b\_3 }
{\color{red} a\_3 }

\\
{\color{blue} b\_1 }
{\color{blue} a\_0 } &
+ &
{\color{blue} b\_0 }
{\color{blue} a\_1 } &
+ &
{\color{blue} b\_3 }
{\color{blue} a\_2 } &
- &
{\color{blue} {b\_2 }}
{\color{blue} a\_3 }

\\
{\color{blue} b\_2 }
{\color{blue} a\_0 } &
- &
{\color{blue} b\_3 }
{\color{blue} a\_1 } &
+ &
{\color{blue} b\_0 }
{\color{blue} a\_2 } &
+ &
{\color{blue} b\_1 }
{\color{blue} a\_3 }

\\
{\color{green} b\_3 }
{\color{green} a\_0 } &
- &
{\color{green} b\_2 }
{\color{green} a\_1 } &
+ &
{\color{green} {b\_1 }}
{\color{green} a\_2 } &
+ &
{\color{green} b\_0 }
{\color{green} a\_3 }

\end{array}
\right)
}

Back to top

### Squares of pseudoscalars are either 1 or -1

In 0 dimensions:

(
1

)
2

=
1

{\displaystyle (1)^2 = 1}

In 1 dimension:

(

e

1

)
2

=

e

11

=
1

{\displaystyle (e\_{1})^2 = e\_{11} = 1}

In 2 dimensions:

(

e

12

)
2

=

e

1212

=
−

e

11

22

=
−
1

{\displaystyle (e\_{12})^2 = e\_{1212} = -e\_{11 \color{red}{22}} = -1}

In 3 dimensions:

(

e

123

)
2

=

e

123123

=

e

1212

33

=
−
1

{\displaystyle (e\_{123})^2 = e\_{123123} = e\_{1212 \color{red}{33}} = -1}

In 4 dimensions:

(

e

1234

)
2

=

e

12341234

=
−

e

123123

44

=
1

{\displaystyle (e\_{1234})^2 = e\_{12341234} = -e\_{123123 \color{red}{44}} = 1}

In 5 dimensions:

(

e

12345

)
2

=

e

1234512345

=

e

12341234

55

=
1

{\displaystyle (e\_{12345})^2 = e\_{1234512345} = e\_{12341234 \color{red}{55}} = 1}

In 6 dimensions:

(

e

123456

)
2

=

e

123456123456

=
−

e

1234512345

66

=
−
1

{\displaystyle (e\_{123456})^2 = e\_{123456123456} = -e\_{1234512345 \color{red}{66}} = -1}

In 7 dimensions:

(

e

1234567

)
2

=

e

12345671234567

=

e

123456123456

77

=
−
1

{\displaystyle (e\_{1234567})^2 = e\_{12345671234567} = e\_{123456123456 \color{red}{77}} = -1}

In 8 dimensions:

(

e

12345678

)
2

=

e

1234567812345678

=
−

e

12345671234567

88

=
1

{\displaystyle (e\_{12345678})^2 = e\_{1234567812345678} = -e\_{12345671234567 \color{red}{88}} = 1}

In 9 dimensions:

(

e

123456789

)
2

=

e

123456789123456789

=

e

1234567812345678

99

=
1

{\displaystyle (e\_{123456789})^2 = e\_{123456789123456789} = e\_{1234567812345678 \color{red}{99}} = 1}

Back to top

### Bivectors in higher dimensions

Multiplying a complex number by an imaginary number rotates the complex number 90° in the complex plane and shrinks or enlarges it.

To perform a general rotation without enlarging it requires

e

θ
i

=
cos
⁡

θ

i
sin
⁡

θ

,

{\displaystyle e^{\theta i} = \cos{\theta} + i \sin{\theta},}

Multipying a vector by a simple bivector rotates the vector 90° in the plane of the bivector and shrinks or enlarges it.

To perform a general rotation without enlarging it requires

e

θ
i

=

e

θ

e

12

=
cos
⁡

θ

e

12

sin
⁡

θ

,

{\displaystyle e^{\theta i} = e^{\theta\mathbf{e}\_{12}} = \cos{\theta} + \mathbf{e}\_{12}\sin{\theta},}

which when multiplied by any vector rotates it through an angle *θ* about the origin:

(

x
′

e

1

y
′

e

2

# )
(
x

e

1

y

e

2

)

e

i
θ

.

{\displaystyle (x'\mathbf{e}\_1 + y'\mathbf{e}\_2) = (x\mathbf{e}\_1 + y\mathbf{e}\_2)e^{i\theta}.}

In four dimensions a rigid object can rotate in two different ways simultaneously. Such a rotation can only be represented as the sum of two simple bivectors.

In six dimensions a rigid object can rotate in three different ways simultaneously. Such a rotation can only be represented as the sum of three simple bivectors.

From Wikipedia:Bivector

The wedge product of two vectors is a bivector, but not all bivectors are wedge products of two vectors. For example, in four dimensions the bivector

B

=

e

1

∧

e

2

e

3

∧

e

4

=

e

1

e

2

e

3

e

4

=

e

12

e

34

{\displaystyle \mathbf{B} = \mathbf{e}\_1 \wedge \mathbf{e}\_2 + \mathbf{e}\_3 \wedge \mathbf{e}\_4 = \mathbf{e}\_1 \mathbf{e}\_2 + \mathbf{e}\_3\mathbf{e}\_4 = \mathbf{e}\_{12} + \mathbf{e}\_{34}}

cannot be written as the wedge product of two vectors. A bivector that can be written as the wedge product of two vectors is simple. In two and three dimensions all bivectors are simple, but not in four or more dimensions;

A bivector has a real square if and only if it is simple.

(

e

12

)
2

=

e

1212

=
−
1

{\displaystyle
\begin{align}
(\mathbf{e}\_{12})^2 &=
e\_{1212} \\ &=
-1
\end{align}
}

But:

(

e

12

e

34

)
2

=

e

1212

e

1234

e

3412

e

3434

=
−

e

1122

e

1234

e

1234

−

e

3344

=
−
1
2

e

1234

−
1

=
2

e

1234

−
2

{\displaystyle
\begin{align}
(\mathbf{e}\_{12} + \mathbf{e}\_{34})^2 &=
e\_{1212} + e\_{1234} + e\_{3412} + e\_{3434} \\ &=
-e\_{1122} + e\_{1234} + e\_{1234} - e\_{3344} \\ &=
-1 + 2 e\_{1234} -1 \\ &=
2 e\_{1234} - 2
\end{align}
}

Back to top

### Other quadratic forms

The square of a vector is:

v

v

=
(

v

x

e

x

v

y

e

y

)
(

v

x

e

x

v

y

e

y

)

=

v

x

v

x

e

x

e

x

v

x

v

y

e

x

e

y

v

y

v

x

e

y

e

x

v

y

v

y

e

y

e

y

=

v

x

v

x

(
1
)

v

y

v

y

(
1
)

v

x

v

y

e

x

e

y

−

v

y

v

x

e

x

e

y

=
(

v

x

v

x

v

y

v

y

)

(
1
)

(

v

x

v

y

−

v

y

v

x

)

e

x
y

=
(

v

x

2

v

y

2

)

(
1
)

(
0
)

e

x
y

=
(

v

x

2

v

y

2

)

(
1
)

=

scalar

{\displaystyle
\begin{align}
\mathbf{v} \mathbf{v} &=
(v\_{x} {\color{blue}e\_{x}} + v\_{y} {\color{blue}e\_{y}})
(v\_{x} {\color{blue}e\_{x}} + v\_{y} {\color{blue}e\_{y}}) \\ &=
v\_{x} v\_{x} {\color{red}e\_{x} e\_{x}} +
v\_{x} v\_{y} {\color{green}e\_{x} e\_{y}} +
v\_{y} v\_{x} {\color{green}e\_{y} e\_{x}} +
v\_{y} v\_{y} {\color{red}e\_{y} e\_{y}} \\ &=
v\_{x} v\_{x} {\color{red}(1)} +
v\_{y} v\_{y} {\color{red}(1)} +
v\_{x} v\_{y} {\color{green}e\_{x} e\_{y}} -
v\_{y} v\_{x} {\color{green}e\_{x} e\_{y}}
\\ &=
(v\_{x} v\_{x} + v\_{y} v\_{y}){\color{red}(1)} +
(v\_{x} v\_{y} - v\_{y} v\_{x}) {\color{green}e\_{xy}} \\ &=
(v\_{x}^2 + v\_{y}^2){\color{red}(1)} +
(0) {\color{green}e\_{xy}} \\ &=
(v\_{x}^2 + v\_{y}^2){\color{red}(1)} \\ &=
{\color{red}\text{scalar}}
\end{align}
}

(

v

x

2

v

y

2

{\displaystyle v\_{x}^2 + v\_{y}^2}

) is called the quadratic formw. In this case both terms are positive but some Clifford algebras have quadratic forms with negative terms. Some have both positive and negative terms.
From Wikipedia:Clifford algebra:

Every nondegenerate quadratic form on a finite-dimensional real vector space is equivalent to the standard diagonal form:

Q
(
v
# )

v
2

=

v
1
2

⋯

v
p
2

−

v

p
1

2

−
⋯
−

v

p
q

2

,

{\displaystyle Q(v) = v^2 = v\_1^2 + \cdots + v\_p^2 - v\_{p+1}^2 - \cdots - v\_{p+q}^2 ,}

where *n* = *p* + *q* is the dimension of the vector space. The pair of integers (*p*, *q*) is called the signature of the quadratic form. The real vector space with this quadratic form is often denoted **R***p*,*q*. The Clifford algebra on **R***p*,*q* is denoted Cℓ*p*,*q*(**R**). The symbol Cℓ*n*(**R**) means either Cℓ*n*,0(**R**) or Cℓ0,*n*(**R**) depending on whether the author prefers positive-definite or negative-definite spaces.

A standard basisw {*e**i*} for **R***p*,*q* consists of *n* = *p* + *q* mutually orthogonal vectors, *p* of which square to +1 and *q* of which square to −1. The algebra Cℓ*p*,*q*(**R**) will therefore have *p* basis vectors that square to +1 and *q* basis vectors that square to −1.

From Wikipedia:Spacetime algebra:

Spacetime algebra (STA) is a name for the Clifford algebraw Cl1,3(**R**), or equivalently the geometric algebraw G(M4), which can be particularly closely associated with the geometry of special relativityw and relativistic spacetimew. See also Algebra of physical space.

The spacetime algebra may be built up from an orthogonal basis of one time-like vector

γ
0

{\displaystyle \gamma\_0}

 and three space-like vectors,

{

γ
1

,

γ
2

,

γ
3

}

{\displaystyle \{\gamma\_1, \gamma\_2, \gamma\_3\}}

, with the multiplication rule

γ
μ

γ
ν

γ
ν

γ
μ

=
2

η

μ
ν

{\displaystyle \gamma\_\mu \gamma\_\nu + \gamma\_\nu \gamma\_\mu = 2 \eta\_{\mu \nu} }

where

η

μ
ν

{\displaystyle \eta\_{\mu \nu}}

 is the Minkowski metricw with signature ( + − − − ).

Thus:

γ
0
2

=

1

{\displaystyle \gamma\_0^2 = {1}}

γ
1
2

=

γ
2
2

=

γ
3
2

=

−
1

{\displaystyle \gamma\_1^2 = \gamma\_2^2 = \gamma\_3^2 = {-1}}

γ
μ

γ
ν

=
−

γ
ν

γ
μ

{\displaystyle \gamma\_\mu \gamma\_\nu = - \gamma\_\nu \gamma\_\mu}

The basis vectors

γ
k

{\displaystyle \gamma\_k}

 share these properties with the Gamma matrices, but no explicit matrix representation need be used in STA.

C

ℓ

3
,
0

(

R

)

{\displaystyle C\ell\_{3,0} (\mathbf{R})}

:Algebra of physical spacew (Time = scalar)

C

ℓ

3
,
1

(

R

)

{\displaystyle C\ell\_{3,1} (\mathbf{R})}

:Spacetime algebraw (Time = vector)

Back to top

### Regressive product

The **regressive product** (usually referred to as the "meet") is the dual of the exterior product (or "join" in this context)

The dual specification of elements permits, for blades

A

{\displaystyle A}

 and

B

{\displaystyle B}

, the intersection (or meet) where the duality is to be taken relative to the smallest grade blade containing both

A

{\displaystyle A}

 and

B

{\displaystyle B}

 (the join).

C
∨
D
:=
(
(
C

I

−
1

)
∧
(
D

I

−
1

)
)
I

{\displaystyle
C \vee D := ((CI^{-1}) \wedge (DI^{-1}))I
}

Which in 2 or 3 dimensions is

C
∨
D
:=
(
(
−
C
I
)
∧
(
−
D
I
)
)
I

{\displaystyle
C \vee D := ((-CI) \wedge (-DI))I
}

with

I

{\displaystyle I}

 the unit pseudoscalar of the algebra. The regressive product, like the exterior product, is associative.

For example:

(
(

e
x

∧

e
y

)
∨
(

e
y

∧

e
z

)
)
I

=
(
(
−

e
x

∧

e
y

∧
I
)
∧
(
−

e
y

∧

e
z

∧
I
)
)
I

=
(
(
−

e
x

∧

e
y

∧

e
x

∧

e
y

∧

e
z

)
∧
(
−

e
y

∧

e
z

∧

e
x

∧

e
y

∧

e
z

)
)
I

=
(
(

e
z

)
∧
(

e
x

)
)
I

=

e
z

∧

e
x

∧
(

e
x

∧

e
y

∧

e
z

)

=
−

e
y

{\displaystyle
\begin{align}
((e\_x \wedge e\_y) \vee (e\_y \wedge e\_z)) I
&= ((-e\_x \wedge e\_y \wedge I) \wedge (-e\_y \wedge e\_z \wedge I)) I \\
&= ((-e\_x \wedge e\_y \wedge e\_x \wedge e\_y \wedge e\_z) \wedge (-e\_y \wedge e\_z \wedge e\_x \wedge e\_y \wedge e\_z)) I \\
&= ((e\_z) \wedge (e\_x))I \\
&= e\_z \wedge e\_x \wedge (e\_x \wedge e\_y \wedge e\_z) \\
&= -e\_y
\end{align}
}

Back to top

### Rotors

See also: Rotor (mathematics)
From Wikipedia:Geometric algebra
The inverse of a vector (wrt the geometric product) is:

v

−
1

=

1
v

=

v

v
v

=

v

v
⋅
v
v
∧
v

=

v

v
⋅
v

{\displaystyle v^{-1} = \frac{1}{v} = \frac{v}{vv} = \frac{v}{v \cdot v + v \wedge v} = \frac{v}{v \cdot v}}

The **projection** of

v

{\displaystyle v}

 onto

a

{\displaystyle a}

 (or the parallel part) is

v

‖
a

=
(
v
⋅
a
)

a

−
1

{\displaystyle v\_{\| a} = (v \cdot a)a^{-1} }

and the **rejection** of

v

{\displaystyle v}

 from

a

{\displaystyle a}

 (or the orthogonal part) is

v

⊥
a

=
v
−

v

‖
a

=
(
v
∧
a
)

a

−
1

.

{\displaystyle v\_{\perp a} = v - v\_{\| a} = (v\wedge a)a^{-1} .}

The reflection

v
′

{\displaystyle v'}

 of a vector

v

{\displaystyle v}

 along a vector

a

{\displaystyle a}

, or equivalently across the hyperplane orthogonal to

a

{\displaystyle a}

, is the same as negating the component of a vector parallel to

a

{\displaystyle a}

. The result of the reflection will be

v
′

=

−

v

‖
a

v

⊥
a

=

−
(
v
⋅
a
)

a

−
1

(
v
∧
a
)

a

−
1

=

(
−
a
⋅
v
−
a
∧
v
)

a

−
1

=
−
a
v

a

−
1

{\displaystyle v' = {-v\_{\| a} + v\_{\perp a}} = {-(v \cdot a)a^{-1} + (v \wedge a)a^{-1}}
= {(-a \cdot v - a \wedge v)a^{-1}}
= -ava^{-1} }

If *a* is a unit vector then

a

−
1

=

a
1

=
a

{\displaystyle a^{-1}=\frac{a}{1} = a}

 and therefore

v
′

=
−
a
v
a

{\displaystyle v' = -ava }

−
a
v
a

{\displaystyle -ava}

 is called the **sandwich product** which is called a **double-sided** product.

If we have a product of vectors

# R

a
1

a
2

⋯

a
r

{\displaystyle R = a\_1a\_2 \cdots a\_r}

 then we denote the reverse as

R
†

=
(

a
1

a
2

⋯

a
r

)
†

=

a
r

⋯

a
2

a
1

.

{\displaystyle R^\dagger = (a\_1a\_2\cdots a\_r)^\dagger = a\_r\cdots a\_2 a\_1.}

Any rotation is equivalent to 2 reflections.

v
″

=
b

v
′

# b
b
a
v
a
# b
R
v

R
†

{\displaystyle v'' = bv'b = bavab = RvR^\dagger}

R is called a Rotor

# a
b
⋅
a
b
∧
# a
S
c
a
l
a
r
B
i
v
e
c
t
o
# r
M
u
l
t
i
v
e
c
t
o
r

{\displaystyle R = ba = b \cdot a + b \wedge a = Scalar + Bivector = Multivector}

If a and b are unit vectors then the rotor is automatically normalised:

R

R
†

=

R
†

# R
1
.

{\displaystyle RR^\dagger = R^\dagger R=1 .}

2 rotations becomes:

R
2

R
1

M

R
1
†

R
2
†

{\displaystyle R\_2R\_1MR\_1^\dagger R\_2^\dagger}

R2R1 represents Rotor R1 rotated by Rotor R2. This would be called a single-sided transformation. (R2R1R2 would be double-sided.) Therefore rotors do not transform double-sided the same way that other objects do. They transform single-sided.

Back to top

### Quaternions

From Wikipedia:Quaternion#Quaternions\_as\_the\_even\_part\_of\_Cℓ3,0(R):
the *reflection* of a vector *r* in a plane perpendicular to a unit vector *w* can be written:

r

′

=
−
w

r

w
.

{\displaystyle r^{\prime} = - w\, r\, w.}

Two reflections make a rotation by an angle twice the angle between the two reflection planes, so

v

′
′

=

σ
2

σ
1

v

σ
1

σ
2

{\displaystyle v^{\prime\prime} = \sigma\_2 \sigma\_1 \, v \, \sigma\_1 \sigma\_2 }

corresponds to a rotation of 180° in the plane containing σ1 and σ2.

This is very similar to the corresponding quaternion formula,

v

′
′

=
−

k

v

k

.

{\displaystyle v^{\prime\prime} = -\mathbf{k}\, v\, \mathbf{k}. }

In fact, the two are identical, if we make the identification

k

=

σ
2

σ
1

,

i

=

σ
3

σ
2

,

j

=

σ
1

σ
3

{\displaystyle \mathbf{k} = \sigma\_2 \sigma\_1, \mathbf{i} = \sigma\_3 \sigma\_2, \mathbf{j} = \sigma\_1 \sigma\_3}

and it is straightforward to confirm that this preserves the Hamilton relations

i

2

=

j

2

=

k

2

=

i

j

k

=
−

{\displaystyle \mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{i} \mathbf{j} \mathbf{k} = -1.}

In this picture, quaternions correspond not to vectors but to bivectorsw – quantities with magnitude and orientations associated with particular 2D *planes* rather than 1D *directions*. The relation to complex numbersw becomes clearer, too: in 2D, with two vector directions σ1 and σ2, there is only one bivector basis element σ1σ2, so only one imaginary. But in 3D, with three vector directions, there are three bivector basis elements σ1σ2, σ2σ3, σ3σ1, so three imaginaries.

The usefulness of quaternions for geometrical computations can be generalised to other dimensions, by identifying the quaternions as the even part Cℓ+3,0(**R**) of the Clifford algebraw Cℓ3,0(**R**).

The square root of the product of a quaternion with its conjugate is called its **norm**:

‖
q
# ‖

q

q
∗

=

q
∗

q

=

a
2

b
2

c
2

d
2

{\displaystyle \lVert q \rVert = \sqrt{qq^\*} = \sqrt{q^\*q} = \sqrt{a^2 + b^2 + c^2 + d^2}}

From Wikipedia:Quaternions and spatial rotation
A unit quaternion is a quaternion of norm one.
Unit quaternions, also known as versors, provide a convenient mathematical notation for representing orientations and rotations of objects in three dimensions.

Every nonzero quaternion has a multiplicative inverse

(
a
b
i
c
j
d
k

)

−
1

=

1

a
2

b
2

c
2

d
2

(
a
−
b
i
−
c
j
−
d
k
)
.

{\displaystyle (a+bi+cj+dk)^{-1} = \frac{1}{a^2+b^2+c^2+d^2}\,(a-bi-cj-dk).}

Thus quaternions form a division algebra.

The inverse of a unit quaternion is obtained simply by changing the sign of its imaginary components.

A 3-D Euclidean vector such as (2, 3, 4) or (*ax*, *ay*, *az*) can be rewritten as 0 + 2 **i** + 3 **j** + 4 **k** or 0 + *ax* **i** + *ay* **j** + *az* **k**, where **i**, **j**, **k** are unit vectors representing the three Cartesian axes. A rotation through an angle of θ around the axis defined by a unit vector

u
→

=
(

u
x

,

u
y

,

u
z

# )
0

u
x

i

u
y

j

u
z

k

{\displaystyle \vec{u} = (u\_x, u\_y, u\_z) = 0 + u\_x\mathbf{i} + u\_y\mathbf{j} + u\_z\mathbf{k}}

can be represented by a quaternion. This can be done using an extension of Euler's formulaw:

q

=

e

θ
2

(
0

u
x

i

u
y

j

u
z

k

)

=
cos
⁡

θ
2

(
0

u
x

i

u
y

j

u
z

k

)
sin
⁡

θ
2

{\displaystyle \mathbf{q} = e^{\frac{\theta}{2}{(0 + u\_x\mathbf{i} + u\_y\mathbf{j} + u\_z\mathbf{k})}} = \cos \frac{\theta}{2} + (0 + u\_x\mathbf{i} + u\_y\mathbf{j} + u\_z\mathbf{k}) \sin \frac{\theta}{2}}

It can be shown that the desired rotation can be applied to an ordinary vector

p

=
(

p
x

,

p
y

,

p
z

# )
0

p
x

i

p
y

j

p
z

k

{\displaystyle \mathbf{p} = (p\_x, p\_y, p\_z) = 0 + p\_x\mathbf{i} + p\_y\mathbf{j} + p\_z\mathbf{k}}

 in 3-dimensional space, considered as a quaternion with a real coordinate equal to zero, by evaluating the conjugation of **p** by **q**:

p
′

=

q

p

q

−
1

{\displaystyle \mathbf{p'} = \mathbf{q} \mathbf{p} \mathbf{q}^{-1}}

using the Hamilton product

The conjugate of a product of two quaternions is the product of the conjugates in the reverse order.

Conjugation by the product of two quaternions is the composition of conjugations by these quaternions: If **p** and **q** are unit quaternions, then rotation (conjugation) by **pq** is

p
q

v
→

(

p
q

)

−
1

=

p
q

v
→

q

−
1

p

−
1

=

p

(

q

v
→

q

−
1

)

p

−
1

{\displaystyle \mathbf{p q} \vec{v} (\mathbf{p q})^{-1} = \mathbf{p q} \vec{v} \mathbf{q}^{-1} \mathbf{p}^{-1} = \mathbf{p} (\mathbf{q} \vec{v} \mathbf{q}^{-1}) \mathbf{p}^{-1}}

,
which is the same as rotating (conjugating) by **q** and then by **p**. The scalar component of the result is necessarily zero.

The imaginary part

b

i

c

j

d

k

{\displaystyle b\mathbf{i} + c\mathbf{j} + d\mathbf{k}}

 of a quaternion behaves like a vector

v
→

=
(
b
,
c
,
d
)

{\displaystyle \vec{v} = (b,c,d)}

 in three dimension vector space, and the real part a behaves like a scalar in **R**. When quaternions are used in geometry, it is more convenient to define them as a scalar plus a vector:

a
b

i

c

j

d

k

=
a

v
→

.

{\displaystyle a + b\mathbf{i} + c\mathbf{j} + d\mathbf{k} = a + \vec{v}.}

When multiplying the vector/imaginary parts, in place of the rules **i**2 = **j**2 = **k**2 = **ijk** = −1 we have the quaternion multiplication rule:

v
→

w
→

=

v
→

×

w
→

−

v
→

⋅

w
→

,

{\displaystyle \vec{v} \vec{w} = \vec{v} \times \vec{w} - \vec{v} \cdot \vec{w},}

From these rules it follows immediately that (see details):

(
s

v
→

)
(
t

w
→

# )
(
s
t
−

v
→

⋅

w
→

)
(
s

w
→

t

v
→

v
→

×

w
→

)
.

{\displaystyle (s + \vec{v}) (t + \vec{w}) = (s t - \vec{v} \cdot \vec{w}) + (s \vec{w} + t \vec{v} + \vec{v} \times \vec{w}).}

It is important to note, however, that the vector part of a quaternion is, in truth, an "axial" vector or "pseudovector", not an ordinary or "polar" vector.

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Next section: Intermediate mathematics/Functions

# See also

* Introductory mathematics
* Advanced mathematics
* Impostor syndrome
* List of free mathematics books

# External links

* MIT open courseware
* Cheat sheets
* http://mathinsight.org
* https://math.stackexchange.com
* https://www.eng.famu.fsu.edu/~dommelen/quantum/style\_a/IV.\_Supplementary\_Informati.html
* http://www.sosmath.com
* https://webhome.phy.duke.edu/~rgb/Class/intro\_math\_review/intro\_math\_review/node1.html
* Wikiversity:Mathematics
* w:c:4chan-science:Mathematics

# References

1. ↑ Wikipedia:Division algebra
2. ↑ Wikipedia:Cartesian product
3. ↑ Wikipedia:Tangent bundle
4. ↑ Wikipedia:Lie group
5. ↑ Wikipedia:Sesquilinear form
6. ↑ Wikipedia:Outer product
7. ↑ Wikipedia:Tensor (intrinsic definition)
8. ↑ Wikipedia:Tensor
9. ↑ "Pauli matrices". Planetmath website. 28 March 2008. http://planetmath.org/PauliMatrices. Retrieved 28 May 2013.

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