aarjaneiro/planetmath huggingface dataset:
nullity

nullity

The

nullity

of a

linear mapping

is the

dimension

of the mapping’s kernel.
For a linear mapping

T

:

V

→

W

, the nullity of

T

gives the
number of

linearly independent

solutions to the equation

T

⁢

(

v

)

=

0

,

v

∈

V

.

The nullity is zero if and only if the linear
mapping in question is injective.

Title

nullity

Canonical

name

Nullity

Date of creation

2013-03-22 12:24:06

Last modified on

2013-03-22 12:24:06

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

6

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A03

Related topic

RankLinearMapping

Related topic

RankNullityTheorem

Generated on Fri Feb 9 18:18:59 2018 by

LaTeXML
proof of rank-nullity theorem

proof of rank-nullity theorem

Let

T

:

V

→

W

be a

linear mapping

, with

V

finite-dimensional

. We wish to show that

dim

⁡

V

=

dim

⁡

Ker

T

+

dim

⁡

Img

T

The images of a basis of

V

will span

Img

T

, and hence

Img

T

is
finite-dimensional. Choose then a basis

w

1

,

…

,

w

n

of

Img

T

and choose preimages

v

1

,

…

,

v

n

∈

U

such that

w

i

=

T

⁢

(

v

i

)

,

i

=

1

⁢

…

⁢

n

Choose a basis

u

1

,

…

,

u

k

of

Ker

T

. The result will follow
once we show that

u

1

,

…

,

u

k

,

v

1

,

…

,

v

n

is a basis of

V

.

Let

v

∈

V

be given. Since

T

⁢

(

v

)

∈

Img

T

, by definition, we can
choose scalars

b

1

,

…

,

b

n

such that

T

⁢

(

v

)

=

b

1

⁢

w

1

+

…

⁢

b

n

⁢

w

n

.

Linearity of

T

now implies that

T

⁢

(

b

1

⁢

v

1

+

…

+

b

n

⁢

v

n

-

v

)

=

0

,

and hence we can choose scalars

a

1

,

…

,

a

k

such that

b

1

⁢

v

1

+

…

+

b

n

⁢

v

n

-

v

=

a

1

⁢

u

1

+

…

⁢

a

k

⁢

u

k

.

Therefore

u

1

,

…

,

u

k

,

v

1

,

…

,

v

n

span

V

.

Next, let

a

1

,

…

,

a

k

,

b

1

,

…

,

b

n

be scalars such that

a

1

⁢

u

1

+

…

+

a

k

⁢

u

k

+

b

1

⁢

v

1

+

…

+

b

n

⁢

v

n

=

0

.

By applying

T

to both sides of this equation it follows that

b

1

⁢

w

1

+

…

+

b

n

⁢

w

n

=

0

,

and since

w

1

,

…

,

w

n

are

linearly independent

that

b

1

=

b

2

=

…

=

b

n

=

0

.

Consequently

a

1

⁢

u

1

+

…

+

a

k

⁢

u

k

=

0

as well, and since

u

1

,

…

,

u

k

are also assumed to be linearly
independent we conclude that

a

1

=

a

2

=

…

=

a

k

=

0

also. Therefore

u

1

,

…

,

u

k

,

v

1

,

…

,

v

n

are linearly
independent, and are therefore a basis. Q.E.D.

Title

proof of rank-nullity theorem

Canonical name

ProofOfRanknullityTheorem

Date of creation

2013-03-22 12:25:13

Last modified on

2013-03-22 12:25:13

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

4

Author

rmilson (146)

Entry type

Proof

Classification

msc 15A03

Generated on Fri Feb 9 18:19:06 2018 by

LaTeXML
proof of the dimension theorem for subspaces

proof of the dimension theorem for subspaces

Let

S

and

T

be

subspaces

of a

vector space

.
By the

rank-nullity theorem

and the

second isomorphism theorem

(for modules)
we have

dim

⁡

(

S

+

T

)

=

dim

⁡

S

+

dim

⁡

(

(

S

+

T

)

/

S

)

=

dim

⁡

S

+

dim

⁡

(

T

/

(

S

∩

T

)

)

.

Therefore

dim

⁡

(

S

+

T

)

+

dim

⁡

(

S

∩

T

)

=

dim

⁡

S

+

dim

⁡

(

T

/

(

S

∩

T

)

)

+

dim

⁡

(

S

∩

T

)

=

dim

⁡

S

+

dim

⁡

T

,

by the rank-nullity theorem again.

Title

proof of the dimension theorem for subspaces

Canonical name

ProofOfTheDimensionTheoremForSubspaces

Date of creation

2013-03-22 16:35:17

Last modified on

2013-03-22 16:35:17

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

5

Author

yark (2760)

Entry type

Proof

Classification

msc 15A03

Generated on Fri Feb 9 18:19:08 2018 by

LaTeXML
properties of bases

properties of bases

Let

V

be a

vector space

over a field

k

.

1.

V

has a basis.

2.

Every

linearly independent

set in

V

can be expanded into a basis for

V

.

3.

Every

spanning set

of

V

contains a subset that is a basis for

V

.

4.

If

A

,

B

are subsets of

V

such that

A

is linearly independent and

B

spans

V

, then

|

A

|

≤

|

B

|

.

5.

All bases for

V

have the same cardinality (hence it is possible to define the

dimension

of a vector space).

6.

V

and

W

are

isomorphic

iff their bases have the same cardinality.

Remarks

.

•

Property 1 is actually a special case of either property 2 or property 3. If we take

∅

as the given linearly independent set in

V

, and apply property 2, we obtain property 1. Likewise, if we take

V

as the given spanning set of

V

, and apply property 3, we again obtain property 1.

•

The above properties can be generalized to a (left or right) vector space over a

division ring

.

•

However, most of the properties on bases can not be generalized to an arbitrary module over an arbitrary ring. For example, not all modules have bases. But we do have the following: let

M

be a (left) module over a ring

R

. Then

(a)

if

M

has a finite basis, then all bases for

M

are finite.

(b)

if

M

has an infinite basis, then all bases for

M

have the same cardinality.

When a module has a basis, then we call it a

free module

(other characterizations are possible). So free modules behave a bit like vector spaces. However, unlike a vector space, one may not be able to define a dimension on a free module. It is possible that, in a

finitely generated

free module, there are two bases of different cardinalities. For more on this, see the entry on IBN.

Title

properties of bases

Canonical name

PropertiesOfBases

Date of creation

2013-03-22 18:06:00

Last modified on

2013-03-22 18:06:00

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Result

Classification

msc 15A03

Classification

msc 13C05

Classification

msc 16D40

Generated on Fri Feb 9 18:19:19 2018 by

LaTeXML
properties of linear independence

properties of linear independence

Let

V

be a

vector space

over a field

k

. Below are some basic

properties of linear independence

.

1.

S

⊆

V

is never

linearly independent

if

0

∈

S

.

Proof.

Since

1

⋅

0

=

0

. ∎

2.

If

S

is linearly independent, so is any subset of

S

. As a result, if

S

and

T

are linearly independent, so is

S

∩

T

. In

addition

,

∅

is linearly independent, its

spanning set

being the singleton consisting of the

zero vector

0

.

Proof.

If

r

1

⁢

v

1

+

⋯

⁢

r

n

⁢

v

n

=

0

, where

v

i

∈

T

, then

v

i

∈

S

, so

r

i

=

0

for all

i

=

1

,

…

,

n

. ∎

3.

If

S

1

⊆

S

2

⊆

⋯

is a chain of linearly independent subsets of

V

, so is their union.

Proof.

Let

S

be the union. If

r

1

⁢

v

1

+

⋯

⁢

r

n

⁢

v

n

=

0

, then

v

i

∈

S

a

⁢

(

i

)

, for each

i

. Pick the largest

S

a

⁢

(

i

)

so that all

v

i

’s are in it. Since this set is linearly independent,

r

i

=

0

for all

i

.
∎

4.

S

is a basis for

V

iff

S

is a maximal linear independent subset of

V

. Here, maximal means that any

proper superset

of

S

is linearly dependent.

Proof.

If

S

is a basis for

V

, then it is linearly independent and spans

V

. If we take any vector

v

∉

S

, then

v

can be expressed as a

linear combination

of

elements

in

S

, so that

S

∪

{

v

}

is no longer linearly independent, for the

coefficient

in front of

v

is non-zero. Therefore,

S

is maximal.

Conversely

, suppose

S

is a maximal linearly independent set in

V

. Let

W

be the span of

S

. If

W

≠

V

, pick an element

v

∈

V

-

W

. Suppose

0

=

r

1

⁢

v

1

+

⋯

⁢

r

n

⁢

v

n

+

r

⁢

v

, where

v

i

∈

S

, then

-

r

⁢

v

=

r

1

⁢

v

1

+

⋯

+

r

n

⁢

v

n

. If

r

≠

0

, then

v

would be in the span of

S

, contradicting the

assumption

. So

r

=

0

, and as a result,

r

i

=

0

, since

S

is linearly independent. This shows that

S

∪

{

v

}

is linearly independent, which is impossible since

S

is assumed to be maximal. Therefore,

W

=

V

.
∎

Remark

. All of the properties above can be generalized to modules over rings, except the last one, where the

implication

is only one-sided: basis implying maximal linear independence.

Title

properties of linear independence

Canonical name

PropertiesOfLinearIndependence

Date of creation

2013-03-22 18:05:37

Last modified on

2013-03-22 18:05:37

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

5

Author

CWoo (3771)

Entry type

Result

Classification

msc 15A03

Generated on Fri Feb 9 18:19:32 2018 by

LaTeXML
rank of a linear mapping

rank of a linear mapping

The

rank

of a linear mapping

L

:

U

→

V

is defined to be
the

dim

⁡

L

⁢

(

U

)

, the

dimension

of the

mapping

’s image. Speaking less
formally, the rank gives the number of

independent

linear constraints
on

u

∈

U

imposed by the

equation

L

⁢

(

u

)

=

0

.

Properties

1.

If

V

is finite-dimensional, then

rank

⁡

L

=

dim

⁡

V

if and only
if

L

is

surjective

.

2.

If

U

is finite-dimensional, then

rank

⁡

L

=

dim

⁡

U

if and only
if

L

is

injective

.

3.

Composition

of linear mappings does not increase rank. If

M

:

V

→

W

is another linear mapping, then

rank

⁡

M

⁢

L

≤

rank

⁡

L

and

rank

⁡

M

⁢

L

≤

rank

⁡

M

.

Equality

holds in the first case if
and only if

M

is an

isomorphism

, and in the second case if and
only if

L

is an isomorphism.

Title

rank of a linear mapping

Canonical name

RankOfALinearMapping

Date of creation

2013-03-22 12:24:03

Last modified on

2013-03-22 12:24:03

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

13

Author

yark (2760)

Entry type

Definition

Classification

msc 15A03

Synonym

rank

Related topic

Nullity

Related topic

RankNullityTheorem2

Generated on Fri Feb 9 18:19:58 2018 by

LaTeXML
rank-nullity theorem

rank-nullity theorem

Let

V

and

W

be

vector spaces

over the same field.
If

ϕ

:

V

→

W

is a

linear mapping

, then

dim

⁡

V

=

dim

⁡

(

ker

⁡

ϕ

)

+

dim

⁡

(

im

⁡

ϕ

)

.

In other words, the

dimension

of

V

is equal to the sum (

http://planetmath.org/CardinalArithmetic

)
of the rank (

http://planetmath.org/RankLinearMapping

) and

nullity

of

ϕ

.

Note that if

U

is a

subspace

of

V

, then this
(applied to the canonical mapping

V

→

V

/

U

) says that

dim

⁡

V

=

dim

⁡

U

+

dim

⁡

(

V

/

U

)

,

that is,

dim

⁡

V

=

dim

⁡

U

+

codim

⁡

U

,

where

codim

denotes codimension.

An alternative way of stating the

rank-nullity theorem

is
by saying that if

0

→

U

→

V

→

W

→

0

is a

short exact sequence

of vector spaces, then

dim

⁡

(

V

)

=

dim

⁡

(

U

)

+

dim

⁡

(

W

)

.

In fact, if

0

→

V

1

→

⋯

→

V

n

→

0

is an

exact sequence

of vector spaces, then

∑

i

=

1

⌊

n

/

2

⌋

V

2

⁢

i

=

∑

i

=

1

⌈

n

/

2

⌉

V

2

⁢

i

-

1

,

that is, the sum of the dimensions of even-numbered terms
is the same as the sum of the dimensions of the odd-numbered terms.

Title

rank-nullity theorem

Canonical name

RanknullityTheorem

Date of creation

2013-03-22 16:35:40

Last modified on

2013-03-22 16:35:40

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

7

Author

yark (2760)

Entry type

Theorem

Classification

msc 15A03

Related topic

RankLinearMapping

Related topic

Nullity

Generated on Fri Feb 9 18:20:11 2018 by

LaTeXML
rank-nullity theorem

rank-nullity theorem

The sum of the rank and the

nullity

of a

linear mapping

gives
the

dimension

of the mapping’s domain. More precisely, let

T

:

V

→

W

be a linear mapping. If

V

is a
finite-dimensional, then

dim

⁡

V

=

dim

⁡

Ker

T

+

dim

⁡

Img

T

.

The intuitive content of the

Rank-Nullity theorem

is the principle that

Every

independent

linear constraint takes away one

degree of freedom

.

The rank is just the number of independent linear constraints on

v

∈

V

imposed
by the equation

T

⁢

(

v

)

=

0

.

The dimension of

V

is the number of unconstrained degrees of freedom.
The nullity is the degrees of freedom in the resulting space of
solutions.
To put it yet another way:

The number of

variables

minus

the number of independent linear
constraints

equals

the number of

linearly independent

solutions.

Title

rank-nullity theorem

Canonical

name

RanknullityTheorem

Date of creation

2013-03-22 12:24:09

Last modified on

2013-03-22 12:24:09

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

8

Author

rmilson (146)

Entry type

Theorem

Classification

msc 15A03

Classification

msc 15A06

Related topic

Overdetermined

Related topic

Underdetermined

Related topic

RankLinearMapping

Related topic

Nullity

Related topic

UnderDetermined

Related topic

FiniteDimensionalLinearProblem

Generated on Fri Feb 9 18:20:13 2018 by

LaTeXML
Riesz’ Lemma

Riesz’ Lemma

Let

E

be a normed space,

S

⊂

E

a closed

proper vector subspace

, and

0

<

α

<

1

. Then there is

x

α

∈

E

∖

S

such that

∥

x

α

∥

=

1

and

∥

s

-

x

α

∥

>

α

for every

s

∈

S

.

Title

Riesz’ Lemma

Canonical name

RieszLemma

Date of creation

2013-03-22 14:56:11

Last modified on

2013-03-22 14:56:11

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

6

Author

gumau (3545)

Entry type

Theorem

Classification

msc 15A03

Classification

msc 54E35

Generated on Fri Feb 9 18:20:15 2018 by

LaTeXML
scalar

scalar

A scalar is a quantity that is

invariant

under

coordinate

transformation

, also known as a tensor of rank 0. For example, the number 1 is a scalar, so is any number or variable

n

∈

ℝ

. The point

(

3

,

4

)

is not a scalar because it is variable under

rotation

.
As such, a scalar can be an element of a field over which a

vector space

is defined.

Title

scalar

Canonical name

Scalar

Date of creation

2013-03-22 11:58:55

Last modified on

2013-03-22 11:58:55

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

8

Author

mathcam (2727)

Entry type

Definition

Classification

msc 15A03

Related topic

Vector

Related topic

EigenvalueOfALinearOperator

Related topic

AxialVector3

Generated on Fri Feb 9 18:20:17 2018 by

LaTeXML
spark of a matrix

spark of a matrix

Let

𝔽

be a field, and let

m

,

n

∈

ℕ

∖

{

0

}

.

Consider an

m

×

n

matrix

A

whose entries belong to

𝔽

.

Let

C

1

,

…

,

C

n

∈

𝔽

m

be the columns of

A

.

Definition 1

The spark of the matrix

A

is defined to be the infimum of the

set of all positive integers

k

with the property that

∃

j

1

,

…

,

j

k

∈

{

1

,

…

,

n

}

:

{

j

1

<

…

<

j

k

,

𝐶

𝑗

1

⁢

, …, C

𝑗

𝑘

⁢

a

⁢

r

⁢

e

⁢

l

⁢

i

⁢

n

⁢

e

⁢

a

⁢

r

⁢

l

⁢

y

⁢

d

⁢

e

⁢

p

⁢

e

⁢

n

⁢

d

⁢

e

⁢

n

⁢

t

⁢

(

o

⁢

v

⁢

e

⁢

r

⁢

𝔽

)

.

The notion of spark of a matrix was introduced by Donoho and Elad in

[

1

]

. It is strictly related to Compressed Sensing. The word “spark” comes from a verbal fusion of “sparse” and “rank”.

Remark 2

The following properties hold true:

(i)

spark

⁢

(

A

)

∈

{

1

,

…

,

n

}

∪

{

+

∞

}

,

(ii)

spark

⁢

(

A

)

=

+

∞

⇔

rank

⁢

(

A

)

=

n

,

(iii)

spark

⁢

(

A

)

=

1

if and only if the matrix

A

has a zero column,

(iv)

if

spark

⁢

(

A

)

≠

+

∞

,

then

spark

⁢

(

A

)

≤

rank

⁢

(

A

)

+

1

.

Moreover,

spark

⁢

(

A

)

coincides with the infimum of the set

{

∥

x

∥

0

:

x

∈

F

n

∖

{

0

}

,

A

⁢

x

=

0

}

, where

∥

x

∥

0

is the Hamming weight of

x

and

0

stands for the appropriate

zero vectors

.

References

1

D. L. Donoho and M. Elad, Optimally sparse representation in general (nonorthogonal) dictionaries via

ℓ

1

minimization,

Proc. Natl. Acad. Sci. USA

100

, No. 5: 2197 – 2202 (2003).

Title

spark of a matrix

Canonical name

SparkOfAMatrix

Date of creation

2013-03-22 19:36:23

Last modified on

2013-03-22 19:36:23

Owner

kammerer (26336)

Last modified by

kammerer (26336)

Numerical id

21

Author

kammerer (26336)

Entry type

Definition

Classification

msc 15A03

Related topic

RankOfAMatrix

Generated on Fri Feb 9 18:20:21 2018 by

LaTeXML
standard basis

standard basis

If

R

is a

division ring

, then the

direct sum

(

http://planetmath.org/DirectSum

) of

n

copies of

R

,

R

n

=

R

⊕

⋯

⊕

R

⁢

(n times),

is a

vector space

.

The

standard basis

for

R

n

consists of

n

elements

e

1

=

(

1

,

0

,

…

,

0

)

,

e

2

=

(

0

,

1

,

0

,

…

,

0

)

,

…

e

n

=

(

0

,

…

,

0

,

1

)

where each

e

i

has

1

for its

i

th

component

and

0

for every other component.
The

e

i

are called the

standard basis vectors

.

Title

standard basis

Canonical name

StandardBasis

Date of creation

2013-03-22 14:20:07

Last modified on

2013-03-22 14:20:07

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

9

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 15A03

Related topic

BasalUnits

Defines

standard basis vectors

Generated on Fri Feb 9 18:20:23 2018 by

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Sylvester’s law

Sylvester’s law

Let

V

be an

n

-dimensional

real vector space

and let

Q

:

V

×

V

→

ℝ

be a

symmetric quadratic form

. Then there exists a basis of

V

such that

Q

⁢

(

u

,

v

)

=

u

T

⁢

M

⁢

v

where the matrix

M

is

diagonal

.

Furthermore, for every choice of basis such that

Q

⁢

(

u

,

v

)

=

u

T

⁢

M

⁢

v

with

M

diagonal, the number of

positive

diagonal entries, the number of negative diagonal entries, and the number of zeros on the diagonal will be the same. The number of non-zero entries on the diagonal is known as the

rank

of the

quadratic form

.

To account for the fact that not all the entries on the diagonal may be positive, one defines a quantity known as the

signature

. However, there is more than one definition of the signature in use. Some define the signature as the number of positive diagonal entries minus the number of negative ones. Others define the signature as
the number of strictly positive entries on the diagonal.

Among some people as, for

instance

, in

pseudo-Riemannian geometry

and relativity theory, the term signature is used to refer to a symbolic display like

[

+

+

-

0

]

which shows the number of positive, negative, and zero diagonal entries or a pair of numbers

(

n

,

m

)

which means that there are

n

positive entries and

m

negative ones.

The rank and signature (whichever definition; pick your favorite) of a quadratic form are

invariant

under

change of basis

.

This implies that two quadratic forms over the same

finite-dimensional

real vector space are related by a change of basis if and only if they have the same rank and signature.

Title

Sylvester’s law

Canonical name

SylvestersLaw

Date of creation

2013-03-22 12:58:58

Last modified on

2013-03-22 12:58:58

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 15A03

Synonym

Sylvester’s theorem

Related topic

PseudoRiemannianManifold

Defines

rank

Defines

signature

Generated on Fri Feb 9 18:20:25 2018 by

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symmetric tensor

symmetric tensor

Let

V

be a

vector space

over a field. Let

S

n

be the

symmetric group

on

{

1

,

…

,

n

}

. An order-n tensor (

http://planetmath.org/TensorProduct

)

A

∈

V

⊗

n

is said to
be

if

P

⁢

(

σ

)

⁢

A

=

A

for all

σ

∈

S

n

,
where

P

⁢

(

σ

)

is the

permutation operator

associated to

σ

.
The set of

symmetric tensors

in

V

⊗

n

is denoted by

S

n

⁢

(

V

)

.

Title

symmetric tensor

Canonical

name

SymmetricTensor

Date of creation

2013-03-22 16:15:41

Last modified on

2013-03-22 16:15:41

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

5

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 15A03

Generated on Fri Feb 9 18:20:27 2018 by

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theorem for the direct sum of finite dimensional vector spaces

theorem for the direct sum of finite dimensional vector spaces

Theorem

Let

S

and

T

be

subspaces

of a

finite dimensional

vector space

V

. Then

V

is the

direct sum

of

S

and

T

, i.e.,

V

=

S

⊕

T

,
if and only if

dim

⁡

V

=

dim

⁡

S

+

dim

⁡

T

and

S

∩

T

=

{

0

}

.

Proof.

Suppose that

V

=

S

⊕

T

. Then, by definition,

V

=

S

+

T

and

S

∩

T

=

{

0

}

.
The dimension theorem for subspaces states that

dim

⁡

(

S

+

T

)

+

dim

⁡

S

∩

T

=

dim

⁡

S

+

dim

⁡

T

.

Since the

dimension

of the

zero vector space

{

0

}

is zero, we have that

dim

⁡

V

=

dim

⁡

S

+

dim

⁡

T

,

and the first

direction

of the claim follows.

For the other direction, suppose

dim

⁡

V

=

dim

⁡

S

+

dim

⁡

T

and

S

∩

T

=

{

0

}

. Then the
dimension theorem theorem for subspaces implies that

dim

⁡

(

S

+

T

)

=

dim

⁡

V

.

Now

S

+

T

is a subspace of

V

with the same dimension
as

V

so,
by Theorem 1 on this page (

http://planetmath.org/VectorSubspace

),

V

=

S

+

T

. This proves the second direction.

□

Title

theorem for the direct sum of finite dimensional vector spaces

Canonical name

TheoremForTheDirectSumOfFiniteDimensionalVectorSpaces

Date of creation

2013-03-22 13:36:17

Last modified on

2013-03-22 13:36:17

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

8

Author

matte (1858)

Entry type

Theorem

Classification

msc 15A03

Generated on Fri Feb 9 18:20:28 2018 by

LaTeXML
unit vector

unit vector

A

unit vector

is a unit-length element of

Euclidean space

.
Equivalently, one may say that the norm of a unit vector is equal
to

1

, and write

∥

𝐮

∥

=

1

, where

𝐮

is the vector in
question.

Let

𝐯

be a non-zero vector. To

normalize

𝐯

is to find
the unique unit vector with the same

direction

as

𝐯

. This is done
by multiplying

𝐯

by the

reciprocal

of its length; the
corresponding unit vector is given by

𝐮

=

𝐯

∥

𝐯

∥

.

Note:

The concept of a unit vector and normalization makes
sense in any

vector space

equipped with a real or

complex norm

.
Thus, in quantum mechanics one represents states as unit vectors
belonging to a (possibly)

infinite-dimensional

Hilbert space. To
obtain an expression for such states one normalizes
the results of a calculation.

Example:

Consider

ℝ

3

and the vector

𝐯

=

(

1

,

2

,

3

)

. The norm (length) is

14

. Normalizing, we obtain
the unit vector

𝐮

pointing in the same direction, namely

𝐮

=

(

1

14

,

2

14

,

3

14

)

.

Title

unit vector

Canonical name

UnitVector

Date of creation

2013-03-22 11:58:50

Last modified on

2013-03-22 11:58:50

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

16

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A03

Related topic

VectorNorm

Related topic

NormedVectorSpace

Defines

normalize

Generated on Fri Feb 9 18:20:30 2018 by

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