aarjaneiro/planetmath huggingface dataset:
example of linear least squares

example of linear least squares

The

assumption

of

linear least squares

is that there is a linear relationship between our measurements

z

and the variables to be estimated

x

z

=

M

⁢

x

+

b

(1)

For this example let us assume that our measurements are given in Table 1 and you can see them plotted in Figure 1.

x

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

-1.0

-0.25

0.0

0.25

0.4

0.7

1.0

1.1

1.4

1.8

Table 1:

Example Data

The linear least squares solution to fit the given data is given by the equation

x

f

⁢

i

⁢

t

=

(

A

T

⁢

A

)

-

1

⁢

A

T

⁢

z

(2)

The only not so obvious step before using a tool like Matlab, is to form the

A

matrix, which is a of an

identity

vector and

x

as

column vectors

, such that

A

=

[

1

|

x

]

This is clarified by looking at the example in Matlab,

\PMlinktofile

LinearLeastSquares.mLinearLeastSquares.m. A plot of fitting the measurement data with a line such that it minimizes the the mean square of the error is shown in Figure 1. The equation of the line to fit this data is then

z

=

0.543

⁢

x

+

0.947

Figure 1:

Linear Fit of Example Data

Title

example of linear least squares

Canonical name

ExampleOfLinearLeastSquares

Date of creation

2013-03-22 16:51:19

Last modified on

2013-03-22 16:51:19

Owner

bloftin (6104)

Last modified by

bloftin (6104)

Numerical id

6

Author

bloftin (6104)

Entry type

Example

Classification

msc 15-00

Generated on Fri Feb 9 18:14:17 2018 by

LaTeXML
example of non-diagonalizable matrices

example of non-diagonalizable matrices

Some matrices with real entries which are not

diagonalizable

over

ℝ

are

diagonalizable over the

complex numbers

ℂ

.

For instance,

A

=

(

0

-

1

1

0

)

has

λ

2

+

1

as

characteristic polynomial

.
This

polynomial

doesn’t factor over the reals, but over

ℂ

it does. Its roots are

λ

=

±

i

.

Interpreting the matrix as a

linear transformation

ℂ

2

→

ℂ

2

, it has

eigenvalues

i

and

-

i

and

linearly independent

eigenvectors

(

1

,

-

i

)

,

(

-

i

,

1

)

. So we can

diagonalize

A

:

A

=

(

0

-

1

1

0

)

=

(

1

-

i

-

i

1

)

⁢

(

i

0

0

-

i

)

⁢

(

.5

.5

⁢

i

.5

⁢

i

.5

)

But there exist

real matrices

which aren’t diagonalizable even if

complex

eigenvectors and eigenvalues are allowed.

For example,

B

=

(

0

1

0

0

)

cannot be written as

U

⁢

D

⁢

U

-

1

with

D

diagonal

.

In fact, the characteristic polynomial is

λ

2

and it has only one double root

λ

=

0

.
However the

eigenspace

corresponding to the

0

(kernel) eigenvalue has

dimension

1.

B

⁢

(

v

1

v

2

)

=

(

0

0

)

⇔

v

2

=

0

and thus the eigenspace is

k

⁢

e

⁢

r

⁢

(

B

)

=

s

⁢

p

⁢

a

⁢

n

ℂ

⁢

{

(

1

,

0

)

T

}

, with only one dimension.

There isn’t a

change of basis

where

B

is diagonal.

Title

example of non-diagonalizable matrices

Canonical name

ExampleOfNondiagonalizableMatrices

Date of creation

2013-03-22 14:14:30

Last modified on

2013-03-22 14:14:30

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

14

Author

cvalente (11260)

Entry type

Example

Classification

msc 15-00

Generated on Fri Feb 9 18:14:19 2018 by

LaTeXML
example of rotation matrix

example of rotation matrix

You can use

rotation matrices

to show that if the slope of one line is

m

, then the slope of the line

perpendicular

to it is

-

1

m

:

Let

L

be a line with a slope of

m

passing through

the origin. The rotation matrix

R

π

2

rotates

L

into a line

L

′

perpendicular to

L

:

R

π

/

2

=

(

0

-

1

1

0

)

Every point on

L

can be represented as a multiple of the point

p

→

=

(

1

m

)

.

Notice

p

→

′

=

R

π

2

⁢

p

→

=

(

-

m

1

)

. Since every point on

L

′

can be represented as a multiple of the point

p

→

′

, the slope of

L

′

is

-

1

m

.

Title

example of rotation matrix

Canonical name

ExampleOfRotationMatrix

Date of creation

2013-03-22 15:09:16

Last modified on

2013-03-22 15:09:16

Owner

swiftset (1337)

Last modified by

swiftset (1337)

Numerical id

5

Author

swiftset (1337)

Entry type

Example

Classification

msc 15-00

Related topic

Slope

Related topic

RotationMatrix

Generated on Fri Feb 9 18:14:21 2018 by

LaTeXML
Frobenius inequality

Frobenius inequality

For any matrices

A

,

B

,

C

such that

A

⁢

B

⁢

C

exists, then

rank

⁡

A

⁢

B

+

rank

⁡

B

⁢

C

≤

rank

⁡

A

⁢

B

⁢

C

+

rank

⁡

B

.

References

1

Carl D. Meyer

Matrix Analysis and Applied

Linear Algebra

Title

Frobenius inequality

Canonical name

FrobeniusInequality

Date of creation

2013-03-22 15:30:03

Last modified on

2013-03-22 15:30:03

Owner

georgiosl (7242)

Last modified by

georgiosl (7242)

Numerical id

6

Author

georgiosl (7242)

Entry type

Theorem

Classification

msc 15-00

Generated on Fri Feb 9 18:14:26 2018 by

LaTeXML
graph of equation ERROR(\tmspace)+.1667⁢e⁢m⁢x⁢y= constant

graph of equation

\tmspace

+

.1667

⁢

e

⁢

m

⁢

x

⁢

y

=

constant

Consider the equation

x

⁢

y

=

c

,  i.e.

y

=

c

x

,

(1)

where

c

is a non-zero real constant. Such a dependence between the

real variables

x

and

y

is called an

inverse

proportionality (

http://planetmath.org/

Variation

).

The graph of (1) may be inferred to be a

hyperbola

(

http://planetmath.org/Hyperbola2

), because the curve has two

asymptotes

(see

asymptotes of graph of rational function

) and because the form

x

⁢

y

-

c

=

0

(2)

of the equation is of second degree (

http://planetmath.org/PolynomialRing

) (see conic,

tangent of conic section

).

One can also see the graph of the equation (2) in such a

coordinate system

(

x

′

,

y

′

) where the equation takes a

canonical form

of the hyperbola (

http://planetmath.org/Hyperbola2

). The

symmetry

of (2) with respect to the

variables

x

and

y

suggests to take for the new

coordinate

axes the axis

angle bisectors

y

=

±

x

. Therefore one has to rotate the old coordinate axes

45

∘

, i.e.

{

x

=

x

′

⁢

cos

⁡

45

∘

-

y

′

⁢

sin

⁡

45

∘

=

x

′

-

y

′

2

y

=

x

′

⁢

sin

⁡

45

∘

+

y

′

⁢

cos

⁡

45

∘

=

x

′

+

y

′

2

(3)

(

sin

⁡

45

∘

=

cos

⁡

45

∘

=

1

2

). Substituting (3) into (2) yields

x

′

⁣

2

-

y

′

⁣

2

2

-

c

=

0

,

i.e.

x

′

⁣

2

2

⁢

c

-

y

′

⁣

2

2

⁢

c

=

1

.

(4)

This is recognised to be the equation of a

rectangular hyperbola

with the transversal axis and the conjugate axis (

http://planetmath.org/Hyperbola2

) on the coordinate axes.

Title

graph of equation

\tmspace

+

.1667

⁢

e

⁢

m

⁢

x

⁢

y

=

constant

Canonical name

GraphOfEquationxyConstant

Date of creation

2013-03-22 17:30:12

Last modified on

2013-03-22 17:30:12

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Derivation

Classification

msc 15-00

Classification

msc 51N20

Synonym

equation

x

⁢

y

=

constant

Related topic

Variation

Related topic

RuledSurface

Related topic

ExactTrigonometryTables

Related topic

Hyperbola2

Related topic

UncertaintyPrinciple

Related topic

Polytrope

Generated on Fri Feb 9 18:14:37 2018 by

LaTeXML
Hadamard conjecture

Hadamard conjecture

There exists a

Hadamard matrix

of order

n

=

4

⁢

m

, for all

m

∈

ℤ

+

.

A Hadamard matrix of order 428 (m=107) has been recently constructed

[

1

]

.

http://math.ipm.ac.ir/tayfeh-r/papersandpreprints/h428.pdf

See here.

A Hadamard matrix of order 764 has also recently been constructed

[

2

]

.

Also, Paley’s theorem guarantees that there always exists a Hadamard matrix

H

n

when

n

is

divisible

by 4 and of the form

2

e

⁢

(

p

m

+

1

)

, for some

positive integers

e and m, and p an

odd prime

and the matrices can be found using

Paley construction

.

This leaves the order of the lowest unknown Hadamard matrix as 668.
There are 13

integers

m

less than 500 for which no Hadamard matrix of order

4

⁢

m

is known:

167

,

179

,

223

,

251

,

283

,

311

,

347

,

359

,

419

,

443

,

479

,

487

,

491

and all of them are primes

congruent

to 3 mod 4.

References

1

H. Kharaghani, B. Tayfeh-Rezaie,

A Hadamard matrix of order 428

, J. Comb.
Designs

13

, (2005), 435-440.

2

D.Z. Doković,

Hadamard matrices of order 764 exist

,

http://arxiv.org/abs/math/0703312v1

preprint.

Title

Hadamard conjecture

Canonical name

HadamardConjecture

Date of creation

2013-03-22 14:07:07

Last modified on

2013-03-22 14:07:07

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

12

Author

Mathprof (13753)

Entry type

Conjecture

Classification

msc 15-00

Synonym

Hadamard’s conjecture

Related topic

HadamardMatrix

Generated on Fri Feb 9 18:14:39 2018 by

LaTeXML
Hadamard matrix

Hadamard matrix

An

n

×

n

matrix

H

=

(

h

i

⁢

j

)

is a

Hadamard matrix

of order

n

if the entries of

H

are either

+

1

or

-

1

and such that

H

⁢

H

T

=

n

⁢

I

,

where

H

T

is the

transpose

of

H

and

I

is the order

n

identity matrix

.

In other words, an

n

×

n

matrix with only

+

1

and

-

1

as its elements is Hadamard if the

inner product

of two distinct rows is

0

and the inner product of a row with itself is

n

.

A few examples of Hadamard matrices are

[

1

1

1

-

1

]

,

[

-

1

1

1

1

1

-

1

1

1

1

1

-

1

1

1

1

1

-

1

]

,

[

1

1

1

1

1

-

1

1

-

1

1

1

-

1

-

1

1

-

1

-

1

1

]

These matrices were first considered as Hadamard

determinants

, because the determinant of a Hadamard matrix satisfies equality in Hadamard’s

determinant theorem

, which states that if

X

=

(

x

i

⁢

j

)

is a matrix of order

n

where

|

x

i

⁢

j

|

≤

1

for all

i

and

j

,

then

d

⁢

e

⁢

t

⁢

(

X

)

≤

n

n

/

2

Property

1:

The order of a Hadamard matrix is

1

,

2

or

4

⁢

n

,

where

n

is an integer.

Property 2:

If the rows and columns of a Hadamard matrix are permuted, the matrix remains Hadamard.

Property 3:

If any row or column is multiplied by

-

1

,

the Hadamard property is retained.

Hence it is always possible to arrange to have the first row and first column of a Hadamard matrix contain
only

+

1

entries. A Hadamard matrix in this form is said to be

normalized

.

Hadamard matrices are common in signal processing and coding applications.

Title

Hadamard matrix

Canonical

name

HadamardMatrix

Date of creation

2013-03-22 13:09:45

Last modified on

2013-03-22 13:09:45

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

13

Author

Koro (127)

Entry type

Definition

Classification

msc 15-00

Classification

msc 05B20

Synonym

Hadamard

Related topic

HadamardConjecture

Generated on Fri Feb 9 18:14:41 2018 by

LaTeXML
homogeneous function

homogeneous function

Definition 1

.

Suppose

V

,

W

are a

vector spaces

over

R

,
and

f

:

V

→

W

is a mapping.

•

If there exists an

r

∈

ℝ

, such that

f

⁢

(

λ

⁢

v

)

=

λ

r

⁢

f

⁢

(

v

)

for all

λ

∈

ℝ

and

v

∈

V

, then

f

is
a .

•

If there exists an

r

∈

ℝ

, such that

f

⁢

(

λ

⁢

v

)

=

|

λ

|

r

⁢

f

⁢

(

v

)

for all

λ

∈

ℝ

and

v

∈

V

,
then

f

is
.

•

If there exists an

r

∈

ℝ

, such that

f

⁢

(

λ

⁢

v

)

=

λ

r

⁢

f

⁢

(

v

)

for all

λ

≥

0

and

v

∈

V

, then

f

is
a .

Notes

For any

homogeneous function

as above,

f

⁢

(

0

)

=

0

.

When the of homegeneity is clear one simply talks about

r

-homogeneous functions.

Title

homogeneous function

Canonical

name

HomogeneousFunction

Date of creation

2013-03-22 14:44:37

Last modified on

2013-03-22 14:44:37

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

8

Author

matte (1858)

Entry type

Definition

Classification

msc 15-00

Synonym

positively homogeneous function of degree

Synonym

homogeneous function of degree

Synonym

positively homogeneous function

Related topic

HomogeneousPolynomial

Related topic

SubLinear

Generated on Fri Feb 9 18:14:43 2018 by

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If A and B commute so do A and B^{-1}

If

A

and

B

commute so do

A

and

B

-

1

Theorem 1

.

Let

A

and

B

be

commuting matrices

.
If

B

is

invertible

,
then

A

and

B

-

1

commute,
and if

A

and

B

are invertible, then

A

-

1

and

B

-

1

commute.

Proof.

By

assumption

A

⁢

B

=

B

⁢

A

,

multiplying from the left and from the right by

B

-

1

yields

B

-

1

⁢

A

=

A

⁢

B

-

1

.

The second claim follows similarly.
∎

The statement and proof of this result can obviously be extended to elements of any monoid. In particular, in the case of a group, we see that two

elements

commute if and only if their

inverses

do.

Title

If

A

and

B

commute so do

A

and

B

-

1

Canonical name

IfAAndBCommuteSoDoAAndB1

Date of creation

2013-03-22 15:27:14

Last modified on

2013-03-22 15:27:14

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 15-00

Classification

msc 15A27

Generated on Fri Feb 9 18:14:44 2018 by

LaTeXML
If A∈M_n⁢(R) and A is supertriangular then A^n=0

If

A

∈

M

n

⁢

(

R

)

and

A

is supertriangular then

A

n

=

0

theorem:

Let

R

be commutative ring with

identity

.
If an n-square matrix

A

∈

M

⁢

a

⁢

t

n

⁢

(

R

)

is

supertriangular

then

A

n

=

0

.

proof:

Find the

characteristic polynomial

of

A

by computing the

determinant

of

A

-

t

⁢

I

. The

square matrix

A

-

t

⁢

I

is a

triangular matrix

. The determinant of a triangular matrix is the

product

of the

diagonal

element of the matrix. Therefore the characteristic polynomial is

p

⁢

(

t

)

=

t

n

and by the

Cayley-Hamilton theorem

the matrix

A

satisfies the

polynomial

. That is

A

n

=

0

.

QED

Title

If

A

∈

M

n

⁢

(

R

)

and

A

is supertriangular then

A

n

=

0

Canonical name

IfAinMnRAndAIsSupertriangularThenAn0

Date of creation

2013-03-22 13:44:39

Last modified on

2013-03-22 13:44:39

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

12

Author

Daume (40)

Entry type

Theorem

Classification

msc 15-00

Generated on Fri Feb 9 18:14:46 2018 by

LaTeXML
in a vector space, λ⁢v=0 if and only if λ=0 or v is the zero vector

in a vector space,

λ

⁢

v

=

0

if and only if

λ

=

0

or

v

is the zero vector

Theorem

Let

V

be a

vector space

over the field

F

.
Further, let

λ

∈

F

and

v

∈

V

.
Then

λ

⁢

v

=

0

if and only if

λ

is zero, or
if

v

is the

zero vector

, or if both

λ

and

v

are zero.

Proof.

Let us denote by

0

F

and by

1

F

the zero and unit
elements in

F

respectively. Similarly, we denote by

0

V

the zero vector in

V

.
Suppose

λ

=

0

F

.
Then, by axiom 8 (

http://planetmath.org/VectorSpace

), we have that

1

F

⁢

v

+

0

F

⁢

v

=

1

F

⁢

v

,

for all

v

∈

V

. By axiom 6 (

http://planetmath.org/VectorSpace

), there is an

element

in

V

that cancels

1

F

⁢

v

. Adding this element to both yields

0

F

⁢

v

=

0

V

.
Next, suppose that

v

=

0

V

. We claim
that

λ

⁢

0

V

=

0

V

for all

λ

∈

F

.
This follows from the previous claim if

λ

=

0

, so
let us assume that

λ

≠

0

F

. Then

λ

-

1

exists, and
axiom 7 (

http://planetmath.org/VectorSpace

) implies that

λ

⁢

λ

-

1

⁢

v

+

λ

⁢

0

V

=

λ

⁢

(

λ

-

1

⁢

v

+

0

V

)

holds for all

v

∈

V

. Then using
axiom 3 (

http://planetmath.org/VectorSpace

), we have that

v

+

λ

⁢

0

V

=

v

for all

v

∈

V

.
Thus

λ

⁢

0

V

satisfies

the axiom for the zero vector, and

λ

⁢

0

V

=

0

V

for all

λ

∈

F

.

For the other

direction

, suppose

λ

⁢

v

=

0

V

and

λ

≠

0

F

.
Then, using axiom 3 (

http://planetmath.org/VectorSpace

), we have that

v

=

1

F

⁢

v

=

λ

-

1

⁢

λ

⁢

v

=

λ

-

1

⁢

0

V

=

0

V

.

On the other hand, suppose

λ

⁢

v

=

0

V

and

v

≠

0

V

.
If

λ

≠

0

, then the above calculation for

v

is again valid whence

0

V

≠

v

=

0

V

,

which is a

contradiction

, so

λ

=

0

.

□

This result with proof can be found in

[

1

]

, page 6.

References

1

W. Greub,

Linear Algebra

,
Springer-Verlag, Fourth edition, 1975.

Title

in a vector space,

λ

⁢

v

=

0

if and only if

λ

=

0

or

v

is the zero vector

Canonical name

InAVectorSpacelambdaV0IfAndOnlyIflambda0OrVIsTheZeroVector

Date of creation

2013-03-22 13:37:34

Last modified on

2013-03-22 13:37:34

Owner

aoh45 (5079)

Last modified by

aoh45 (5079)

Numerical id

10

Author

aoh45 (5079)

Entry type

Theorem

Classification

msc 15-00

Classification

msc 13-00

Classification

msc 16-00

Generated on Fri Feb 9 18:14:48 2018 by

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invariant subspace

invariant subspace

Let

T

:

V

→

V

be a

linear transformation

of a

vector space

V

. A

subspace

U

⊂

V

is
called a

T

-

invariant subspace

if

T

⁢

(

u

)

∈

U

for all

u

∈

U

.

If

U

is an invariant subspace, then the

restriction

of

T

to

U

gives a well defined linear transformation of

U

. Furthermore,
suppose that

V

is

n

-dimensional and that

v

1

,

…

,

v

n

is a
basis of

V

with the first

m

vectors giving a basis of

U

. Then,
the representing matrix of the

transformation

T

relative to this
basis takes the form

(

A

B

0

C

)

where

A

is an

m

×

m

matrix representing the restriction
transformation

T

|

U

:

U

→

U

relative to the basis

v

1

,

…

,

v

m

.

Title

invariant subspace

Canonical

name

InvariantSubspace

Date of creation

2013-03-22 12:19:55

Last modified on

2013-03-22 12:19:55

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

9

Author

rmilson (146)

Entry type

Definition

Classification

msc 15-00

Related topic

LinearTransformation

Related topic

Invariant

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Jacobian and chain rule

Jacobian and chain rule

Let

u

,

v

be

differentiable functions

of

x

,

y

and

x

,

y

be differentiable functions of

s

,

t

.  Then the connection

∂

⁡

(

u

,

v

)

∂

⁡

(

s

,

t

)

=

∂

⁡

(

u

,

v

)

∂

⁡

(

x

,

y

)

⋅

∂

⁡

(

x

,

y

)

∂

⁡

(

s

,

t

)

(1)

between the Jacobian determinants is in .

Proof.

Starting from the right hand side of (1), where one can multiply the

determinants

(

http://planetmath.org/Determinant2

) similarly as the corresponding matrices (

http://planetmath.org/MatrixMultiplication

), we have

|

u

x

u

y

v

x

v

y

|

⋅

|

x

s

x

t

y

s

y

t

|

=

|

u

x

⁢

x

s

+

u

y

⁢

y

s

u

x

⁢

x

t

+

u

y

⁢

y

t

v

x

⁢

x

s

+

v

y

⁢

y

s

v

x

⁢

x

t

+

v

y

⁢

y

t

|

=

|

u

s

u

t

v

s

v

t

|

.

Here, the last stage has been written according to the general

chain rule

(

http://planetmath.org/ChainRuleSeveralVariables

).  But thus we have arrived at the left hand side of the equation (1), which hereby has been proved.

Remark.

The rule (1) is only a visualisation of the more general one concerning the case of functions of

n

variables.

Title

Jacobian and chain rule

Canonical name

JacobianAndChainRule

Date of creation

2013-03-22 18:59:45

Last modified on

2013-03-22 18:59:45

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

4

Author

pahio (2872)

Entry type

Example

Classification

msc 15-00

Classification

msc 26B05

Classification

msc 26B10

Generated on Fri Feb 9 18:14:52 2018 by

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Jacobi’s theorem

Jacobi’s theorem

Jacobi’s Theorem

Any

skew-symmetric matrix

of odd order has

determinant

equal to

0

.

Proof.

Suppose

A

is an

n

×

n

square matrix

.
For the determinant, we then have

det

⁡

A

=

det

⁡

A

T

, and

det

⁡

(

-

A

)

=

(

-

1

)

n

⁢

det

⁡

A

. Thus, since

n

is odd, and

A

T

=

-

A

, we have

det

⁡

A

=

-

det

⁡

A

, and the theorem follows.

□

0.0.1

Remarks

1.

According to

[

1

]

, this theorem was given by

Carl Gustav Jacob Jacobi

(1804-1851)

[

2

]

in 1827.

2.

The

2

×

2

matrix

(

0

1

-

1

0

)

shows that Jacobi’s theorem does not hold for

2

×

2

matrices. The determinant of the

2

⁢

n

×

2

⁢

n

block matrix

with
these

2

×

2

matrices on the

diagonal

equals

(

-

1

)

n

. Thus Jacobi’s theorem
does not hold for matrices of even order.

3.

For

n

=

3

, any

antisymmetric matrix

A

can be written
as

A

=

(

0

-

v

3

v

2

v

3

0

-

v

1

-

v

2

v

1

0

)

for some real

v

1

,

v

2

,

v

3

, which can be written as a
vector

v

=

(

v

1

,

v

2

,

v

3

)

. Then

A

is the matrix representing the
mapping

u

↦

v

×

u

, that is, the

cross product

with
respect to

v

. Since

A

⁢

v

=

v

×

v

=

0

, we have

det

⁡

A

=

0

.

References

1

H. Eves,

Elementary Matrix

Theory

,
Dover publications, 1980.

2

The MacTutor History of Mathematics archive,

http://www-gap.dcs.st-and.ac.uk/ history/Mathematicians/Jacobi.html

Carl Gustav Jacob Jacobi

Title

Jacobi’s theorem

Canonical name

JacobisTheorem

Date of creation

2013-03-22 13:33:06

Last modified on

2013-03-22 13:33:06

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

13

Author

Koro (127)

Entry type

Theorem

Classification

msc 15-00

Generated on Fri Feb 9 18:14:54 2018 by

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Kronecker product

Kronecker product

Definition.

Let

A

=

(

a

i

⁢

j

)

be a

n

×

n

matrix and
let

B

be a

m

×

m

matrix. Then the

Kronecker product

of

A

and

B

is
the

m

⁢

n

×

m

⁢

n

block matrix

A

⊗

B

=

(

a

11

⁢

B

⋯

a

1

⁢

n

⁢

B

⋮

⋱

⋮

a

n

⁢

1

⁢

B

⋯

a

n

⁢

n

⁢

B

)

.

The Kronecker product is also known as the

direct product

or the

tensor product

[

1

]

.

Fundamental properties

[

1

,

2

]

1.

The product is

bilinear

. If

k

is a scalar, and

A

,

B

and

C

are

square matrices

, such that

B

and

C

are of the same order, then

A

⊗

(

B

+

C

)

=

A

⊗

B

+

A

⊗

C

,

(

B

+

C

)

⊗

A

=

B

⊗

A

+

C

⊗

A

,

k

⁢

(

A

⊗

B

)

=

(

k

⁢

A

)

⊗

B

=

A

⊗

(

k

⁢

B

)

.

2.

If

A

,

B

,

C

,

D

are square matrices such that the products

A

⁢

C

and

B

⁢

D

exist, then

(

A

⊗

B

)

⁢

(

C

⊗

D

)

exists and

(

A

⊗

B

)

⁢

(

C

⊗

D

)

=

A

⁢

C

⊗

B

⁢

D

.

If

A

and

B

are

invertible matrices

, then

(

A

⊗

B

)

-

1

=

A

-

1

⊗

B

-

1

.

3.

If

A

and

B

are square matrices, then for the

transpose

(

A

T

) we have

(

A

⊗

B

)

T

=

A

T

⊗

B

T

.

4.

Let

A

and

B

be square matrices of orders

n

and

m

, respectively.
If

{

λ

i

|

i

=

1

,

…

,

n

}

are the

eigenvalues

of

A

and

{

μ

j

|

j

=

1

,

…

,

m

}

are the eigenvalues of

B

, then

{

λ

i

⁢

μ

j

|

i

=

1

,

…

,

n

,

j

=

1

,

…

,

m

}

are the eigenvalues of

A

⊗

B

. Also,

det

⁡

(

A

⊗

B

)

=

(

det

⁡

A

)

m

⁢

(

det

⁡

B

)

n

,

rank

(

A

⊗

B

)

=

rank

A

⁢

rank

B

,

trace

(

A

⊗

B

)

=

trace

A

⁢

trace

B

,

References

1

H. Eves,

Elementary Matrix

Theory

,
Dover publications, 1980.

2

T. Kailath, A.H. Sayed, B. Hassibi,

Linear estimation

,
Prentice Hall, 2000

Title

Kronecker product

Canonical name

KroneckerProduct

Date of creation

2013-03-22 13:33:31

Last modified on

2013-03-22 13:33:31

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 15-00

Synonym

tensor product (for matrices)

Synonym

direct product

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