aarjaneiro/planetmath huggingface dataset:
strain transformation

strain transformation

Let

E

be a

Euclidean plane

. Fix a line

ℓ

in

E

and a real number

r

≠

0

. Take any point

p

∈

E

. Drop a line

m

p

from

p

perpendicular

to

ℓ

. Denote

d

⁢

(

p

,

ℓ

)

the

distance

from

p

to

ℓ

. Then there is a unique point

p

′

on

m

p

such that

d

⁢

(

p

′

,

ℓ

)

=

r

⋅

d

⁢

(

p

,

ℓ

)

.

The

function

s

r

:

E

→

E

such that

s

r

⁢

(

p

)

=

p

′

is called a

strain transformation

, or simply a

strain

.

One can visualize a strain

stretches

a geometric figure if

|

r

|

>

1

and compresses it if

|

r

|

<

1

. If

r

=

1

, then

s

r

is the

identity function

, the only time when a strain is a

rigid motion

. For example, let

ℓ

be the

x

-axis and

C

be a circle in the upper half plane of the

x

-

y

plane. Then the following diagrams show how a strain transforms

C

:

Again, if

ℓ

is the

x

-axis, then

s

r

is the function that sends

(

x

,

y

)

to

(

x

,

r

⁢

y

)

. Representing the

ordered pairs

as

column vectors

and

s

r

as a matrix , we have

s

r

⁢

(

x

y

)

=

(

1

0

0

r

)

⁢

(

x

y

)

=

(

x

r

⁢

y

)

.

Nevertheless, a strain, as a (

non-singular

)

linear transformation

, takes lines to lines, and

parallel lines

to parallel lines.

In general, given any finite dimensional

vector space

V

over a field

k

, a strain

s

r

is a non-singular

diagonalizable

linear transformation on

V

such that

s

r

leaves a

subspace

W

of

codimension

1

fixed.

0

≠

r

∈

k

is called the

strain coefficient

.

Remark

. By choosing an appropriate base for

V

of

dimension

n

,

s

r

can be represented as a

diagonal matrix

whose

diagonals

are

1

in at least

n

-

1

cells and

r

in at most one cell.

It is easy to see that every non-singular diagonalizable linear transformation on

V

can be written as a

product

of

n

strains, where

n

=

dim

⁡

(

V

)

.

Title

strain transformation

Canonical

name

StrainTransformation

Date of creation

2013-03-22 17:25:45

Last modified on

2013-03-22 17:25:45

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

5

Author

CWoo (3771)

Entry type

Definition

Classification

msc 15A04

Synonym

strain

Defines

strain coefficient

Generated on Fri Feb 9 18:22:25 2018 by

LaTeXML
symmetrizer

symmetrizer

Let

V

be a

vector space

over a field

F

. Let

n

be an integer, where

n

<

char

⁢

(

F

)

if

char

⁢

(

F

)

≠

0

. Let

S

n

be the

symmetric group

on

{

1

,

…

,

n

}

.

The

linear operator

S

:

V

⊗

n

→

V

⊗

n

defined by:

S

=

1

n

!

⁢

∑

σ

∈

S

n

P

⁢

(

σ

)

is called the

symmetrizer

.
Here

P

⁢

(

σ

)

is the

permutation operator

.
It is clear that

P

⁢

(

σ

)

⁢

S

=

S

⁢

P

⁢

(

σ

)

=

S

for all

σ

∈

S

n

.

Let

S

be the symmetrizer for

V

⊗

n

. Then an order-n tensor

A

is

symmetric

(

http://planetmath.org/SymmetricTensor

) if and only

S

⁢

(

A

)

=

A

.

Proof

If

A

is then

S

⁢

(

A

)

=

1

n

!

⁢

∑

σ

∈

S

n

P

⁢

(

σ

)

⁢

A

=

1

n

!

⁢

∑

σ

∈

S

n

A

=

A

.

If

S

⁢

(

A

)

=

A

then

P

⁢

(

σ

)

⁢

A

=

P

⁢

(

σ

)

⁢

S

⁢

(

A

)

=

P

⁢

(

σ

)

⁢

S

⁢

(

A

)

=

S

⁢

(

A

)

=

A

for all

σ

∈

S

n

, so

A

is .

The theorem says that a is an

eigenvector

of the linear operator

S

corresponding to the

eigenvalue

1. It is easy to verify that

S

2

=

S

, so that

S

is a

projection

onto

S

n

⁢

(

V

)

.

Title

symmetrizer

Canonical name

Symmetrizer

Date of creation

2013-03-22 16:15:44

Last modified on

2013-03-22 16:15:44

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

8

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 15A04

Generated on Fri Feb 9 18:22:27 2018 by

LaTeXML
symplectic complement

symplectic complement

Definition

[

1

,

2

]

Let

(

V

,

ω

)

be a

symplectic vector space

and let

W

be a

vector subspace

of

V

. Then the

symplectic complement

of

W

is

W

ω

=

{

x

∈

V

|

ω

⁢

(

x

,

y

)

=

0

⁢

for all

⁢

y

∈

W

}

.

It is easy to see that

W

ω

is also a vector subspace of

V

.
Depending on the

relation

between

W

and

W

ω

,

W

is given different names.

1.

If

W

⊂

W

ω

, then

W

is an

isotropic subspace

(of

V

).

2.

If

W

ω

⊂

W

, then

W

is an

coisotropic subspace

.

3.

If

W

∩

W

ω

=

{

0

}

, then

W

is an

symplectic subspace

.

4.

If

W

=

W

ω

, then

W

is an

Lagrangian subspace

.

For the symplectic complement, we have the
following

dimension

theorem

.

Theorem

[

1

,

2

]

Let

(

V

,

ω

)

be a symplectic vector
space, and let

W

be a vector subspace of

V

. Then

dim

⁡

V

=

dim

⁡

W

ω

+

dim

⁡

W

.

References

1

D. McDuff, D. Salamon,

Introduction

to Symplectic Topology

,
Clarendon Press, 1997.

2

R. Abraham, J.E. Marsden,

Foundations of Mechanics

,
2nd ed., Perseus Books, 1978.

Title

symplectic complement

Canonical

name

SymplecticComplement

Date of creation

2013-03-22 13:32:25

Last modified on

2013-03-22 13:32:25

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

8

Author

matte (1858)

Entry type

Definition

Classification

msc 15A04

Defines

symplectic complement

Defines

isotropic subspace

Defines

coisotropic subspace

Defines

symplectic subspace

Defines

Lagrangian subspace

Generated on Fri Feb 9 18:22:29 2018 by

LaTeXML
theorem about cyclic subspaces

theorem about cyclic subspaces

Let

k

be field,

V

a

vector space

over

k

,

dim

⁡

V

=

n

, and

T

:

V

→

V

a

linear operator

. Let

W

be a

subspace

of

V

. And let

v

1

,

…

,

v

r

∈

V

such that

W

=

Z

⁢

(

v

1

,

T

)

⁢

⊕

⋯

⁢

⊕

Z

⁢

(

v

r

,

T

)

(see the

cyclic subspace

definition), and

(

m

v

i

,

m

v

j

)

=

1

if

i

≠

j

, where

m

v

denotes the

minimal polynomial

of

v

(or in other words, its

annihilator polynomial

). Then,

Z

⁢

(

v

1

+

⋯

+

v

r

,

T

)

=

Z

⁢

(

v

1

,

T

)

⁢

⊕

⋯

⁢

⊕

Z

⁢

(

v

r

,

T

)

, and

m

v

1

+

⋯

+

v

r

=

m

v

1

⁢

⋯

⁢

m

v

r

.

Title

theorem about cyclic subspaces

Canonical name

TheoremAboutCyclicSubspaces

Date of creation

2013-03-22 14:15:16

Last modified on

2013-03-22 14:15:16

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

12

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 15A04

Generated on Fri Feb 9 18:22:31 2018 by

LaTeXML
there exist additive functions which are not linear

there exist additive functions which are not linear

Example 1

.

There exists a

function

f

:

ℝ

→

ℝ

which is

additive

but not linear.

Proof.

Let

V

be the infinite dimensional

vector space

ℝ

over the
field

ℚ

. Since

1

and

2

are two

independent

vectors in

V

, we can extend the set

{

1

,

2

}

to a basis

E

of

V

(notice that here the

axiom of choice

is used).

Now we consider a

linear function

f

:

V

→

ℝ

such that

f

⁢

(

1

)

=

1

while

f

⁢

(

e

)

=

0

for all

e

∈

E

∖

{

1

}

. This function is

ℚ

-linear (i.e. it is additive on

ℝ

) but it is not

ℝ

-linear because

f

⁢

(

2

)

=

0

≠

2

⁢

f

⁢

(

1

)

.
∎

Title

there exist

additive functions

which are not linear

Canonical name

ThereExistAdditiveFunctionsWhichAreNotLinear

Date of creation

2013-03-22 16:17:50

Last modified on

2013-03-22 16:17:50

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

5

Author

paolini (1187)

Entry type

Example

Classification

msc 15A04

Generated on Fri Feb 9 18:22:33 2018 by

LaTeXML
ker⁡L={0} if and only if L is injective

ker

⁡

L

=

{

0

}

if and only if

L

is injective

Theorem

.

A

linear map

between

vector spaces

is injective if and only if its kernel is

{

0

}

.

Proof.

Let

L

:

V

→

W

be a linear map. Suppose

L

is injective, and

L

⁢

(

v

)

=

0

for some vector

v

∈

V

. Also,

L

⁢

(

0

)

=

0

because

L

is linear.
Then

L

⁢

(

v

)

=

L

⁢

(

0

)

, so

v

=

0

. On the other hand, suppose

ker

⁡

L

=

{

0

}

, and

L

⁢

(

v

)

=

L

⁢

(

v

′

)

for vectors

v

,

v

′

∈

V

.
Hence

L

⁢

(

v

-

v

′

)

=

L

⁢

(

v

)

-

L

⁢

(

v

′

)

=

0

because

L

is linear.
Therefore,

v

-

v

′

is in

ker

⁡

L

=

{

0

}

, which means that

v

-

v

′

must be

0

.
∎

Title

ker

⁡

L

=

{

0

}

if and only if

L

is injective

Canonical name

operatornamekerL0IfAndOnlyIfLIsInjective

Date of creation

2013-03-22 14:44:46

Last modified on

2013-03-22 14:44:46

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

11

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 15A04

Generated on Fri Feb 9 18:22:34 2018 by

LaTeXML
example of under-determined polynomial interpolation

example of under-determined polynomial interpolation

Consider the following

interpolation

problem:

Given

x

1

,

y

1

,

x

2

,

y

2

∈

R

with

x

1

≠

x

2

to determine
all cubic polynomials

p

⁢

(

x

)

=

a

⁢

x

3

+

b

⁢

x

2

+

c

⁢

x

+

d

,

x

,

a

,

b

,

c

,

d

∈

ℝ

such that

p

⁢

(

x

1

)

=

y

1

,

p

⁢

(

x

2

)

=

y

2

.

This is a

linear problem

. Let

𝒫

3

denote the

vector space

of
cubic polynomials. The underlying

linear mapping

is the

multi-evaluation mapping

E

:

𝒫

3

→

ℝ

2

,

given by

p

↦

(

p

⁢

(

x

1

)

p

⁢

(

x

2

)

)

,

p

∈

𝒫

3

The interpolation problem in question is represented by the equation

E

⁢

(

p

)

=

(

y

1

y

2

)

where

p

∈

𝒫

3

is the unknown. One can recast the problem into the
traditional form by taking standard bases of

𝒫

3

and

ℝ

2

and
then seeking all possible

a

,

b

,

c

,

d

∈

ℝ

such that

(

(

x

1

)

3

(

x

1

)

2

x

1

1

(

x

2

)

3

(

x

2

)

2

x

2

1

)

⁢

(

a

b

c

d

)

=

(

y

1

y

2

)

However, it is best to treat this problem at an abstract level,
rather than mucking about with

row reduction

. The

Lagrange
interpolation formula

gives us a particular solution, namely the linear
polynomial

p

0

⁢

(

x

)

=

x

-

x

1

x

2

-

x

1

⁢

y

1

+

x

-

x

2

x

1

-

x

2

⁢

y

2

,

x

∈

ℝ

The

general solution

of our interpolation problem is therefore given
as

p

0

+

q

, where

q

∈

𝒫

3

is a solution of the

homogeneous

problem

E

⁢

(

q

)

=

0

.

A basis of solutions for the latter is, evidently,

q

1

⁢

(

x

)

=

(

x

-

x

1

)

⁢

(

x

-

x

2

)

,

q

2

⁢

(

x

)

=

x

⁢

q

1

⁢

(

x

)

,

x

∈

ℝ

The general solution to our interpolation problem is therefore

p

⁢

(

x

)

=

x

-

x

1

x

2

-

x

1

⁢

y

1

+

x

-

x

2

x

1

-

x

2

⁢

y

2

+

(

a

⁢

x

+

b

)

⁢

(

x

-

x

1

)

⁢

(

x

-

x

2

)

,

x

∈

ℝ

,

with

a

,

b

∈

ℝ

arbitrary. The general

under-determined

interpolation problem is treated in an entirely analogous manner.

Title

example of under-determined polynomial interpolation

Canonical name

ExampleOfUnderdeterminedPolynomialInterpolation

Date of creation

2013-03-22 12:35:22

Last modified on

2013-03-22 12:35:22

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

5

Author

rmilson (146)

Entry type

Example

Classification

msc 15A06

Generated on Fri Feb 9 18:22:39 2018 by

LaTeXML
homogeneous linear problem

homogeneous linear problem

Let

L

:

U

→

V

be a

linear mapping

. A

linear equation

is called

homogeneous

if it has
the form

L

⁢

(

u

)

=

0

,

u

∈

U

.

A homogeneous linear problem always has a

trivial

solution, namely

u

=

0

. The key issue in homogeneous problems
is, therefore, the question of the

existence

of non-trivial solutions,
i.e. whether or not the kernel of

L

is trivial, or equivalently,
whether or not

L

is

injective

.

Title

homogeneous linear problem

Canonical

name

HomogeneousLinearProblem

Date of creation

2013-03-22 12:26:03

Last modified on

2013-03-22 12:26:03

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

5

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A06

Synonym

homogeneous

Related topic

LinearProblem

Generated on Fri Feb 9 18:22:59 2018 by

LaTeXML
linear equation

linear equation

Let

L

:

U

→

V

be a

linear mapping

, and

v

∈

V

an element of
the

codomain

. A

linear equation

is
a

relation

of the form,

L

⁢

(

u

)

=

v

,

where

u

∈

U

is to be considered as the unknown. The
solution set of a linear equation is the set of

u

∈

U

that

satisfy

the
above constraint, or to be more precise, the pre-image

L

-

1

⁢

(

v

)

. The equation is
called

inconsistent

if no solutions exist, that is, if the pre-image is
the

empty set

. Otherwise, the equation is called

consistent

.

The general solution of
a linear equation has the form

u

=

u

p

+

u

h

,

u

p

,

u

h

∈

U

,

where

L

⁢

(

u

p

)

=

v

is a

particular solution

and where

L

⁢

(

u

h

)

=

0

is any solution of the corresponding

homogeneous

problem, i.e. an
element of the kernel of

L

.

Notes.

Elementary

treatments of

linear algebra

focus almost
exclusively on

finite-dimensional linear problems

. They neglect to
mention the underlying

mapping

, preferring to focus instead on
“variables and equations.” However, the scope of the general

concept

is considerably wider, e.g. linear differential equations such as

y

′′

+

y

=

0

.

Title

linear equation

Canonical name

LinearEquation

Date of creation

2013-03-22 12:25:59

Last modified on

2013-03-22 12:25:59

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

8

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A06

Synonym

linear problem

Synonym

linear system

Related topic

HomogeneousLinearProblem

Related topic

FiniteDimensionalLinearProblem

Defines

consistent

Defines

inconsistent

Defines

particular solution

Generated on Fri Feb 9 18:23:01 2018 by

LaTeXML
reduced row echelon form

reduced row echelon form

For a matrix to be in

reduced row echelon form

(or

Hermite normal form

) it has to first satisfy the requirements to be in

row echelon form

and additionally satisfy the following requirements:

1.

The first non-zero element in any row must be 1.

2.

The first element of value 1 in any row must be the only non-zero value in its column.

An example of a matrix in reduced row echelon form could be:

(

0

1

2

6

0

1

0

0

4

0

0

0

0

0

1

1

0

0

1

1

0

0

0

0

0

0

1

0

4

1

0

0

0

0

0

0

0

1

2

1

0

0

0

0

0

0

0

0

0

0

)

Title

reduced row echelon form

Canonical name

ReducedRowEchelonForm

Date of creation

2013-03-22 12:14:14

Last modified on

2013-03-22 12:14:14

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

9

Author

mathcam (2727)

Entry type

Definition

Classification

msc 15A06

Related topic

RowEchelonForm

Related topic

GaussianElimination

Related topic

DeterminingRankOfMatrix

Defines

Hermite normal form

Generated on Fri Feb 9 18:23:03 2018 by

LaTeXML
row echelon form

row echelon form

A matrix is said to be in

row echelon form

if each non-zero row has more leading zeros than the previous row. Row-echelon form is the key idea underlying the

Gaussian elimination

algorithm and

LU factorization

.

Let us give the precise definition. Let

(

M

i

⁢

j

)

be an

n

×

m

matrix. For each row

i

=

1

,

…

,

n

define the pivot position

P

i

to
be either the minimum value of

j

=

1

,

…

,

m

for which

M

i

⁢

j

≠

0

, or

∞

if the row consists entirely of zeros. A matrix is in

echelon form

if for all

i

>

1

, either

P

i

=

∞

or

P

i

-

1

<

P

i

.

Examples of matrices in row echelon form include,

(

0

2

1

0

0

1

0

0

0

)

,

(

5

0

1

3

2

0

0

4

1

0

0

0

0

0

7

)

Note that if a matrix is an echelon form, then necessarily rows which
are composed completely of zeros will be grouped at the bottom of the
matrix. Also note that if several rows have the same number of
leading zeros then the matrix is not in row echelon form unless the
rows in question are composed entirely of zeros.

Title

row echelon form

Canonical name

RowEchelonForm

Date of creation

2013-03-22 12:14:11

Last modified on

2013-03-22 12:14:11

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

12

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A06

Related topic

GaussianElimination

Defines

echelon form

Generated on Fri Feb 9 18:23:05 2018 by

LaTeXML
uniqueness of a sparse solution

uniqueness of a sparse solution

Let

𝔽

be a field, and let

m

,

n

∈

ℕ

∖

{

0

}

.

We denote by

∥

x

∥

0

the Hamming weight of a

column vector

x

=

[

x

1

,

…

,

x

n

]

T

∈

𝔽

n

,

i.e.,

∥

x

∥

0

=

#

⁢

{

j

∈

{

1

,

…

,

n

}

:

x

j

≠

0

}

.

Consider a non-negative integer

k

,

and an

m

×

n

matrix

A

whose entries belong to

𝔽

.

Theorem 1

(Donoho and Elad)

The following conditions are

equivalent

:

(1)

for each column vector

y

∈

F

m

there exists at most one

x

∈

F

n

such that

A

⁢

x

=

y

and

∥

x

∥

0

≤

k

,

(2)

k

<

1

2

⁢

spark

⁢

(

A

)

.

The standard proof of the

theorem

(see, for

instance

,

[

2

]

) goes as follows.

Proof.

First, suppose that condition

(1)

is not satisfied. Then there exist column vectors

v

,

w

∈

𝔽

n

such that

v

≠

w

,

A

⁢

v

=

A

⁢

w

,

and

max

⁡

{

∥

v

∥

0

,

∥

w

∥

0

}

≤

k

.

Consequently,

v

-

w

∈

𝔽

n

∖

{

𝟎

}

and

A

⁢

(

v

-

w

)

=

𝟎

.

Moreover, by the definition of the Hamming weight,

∥

v

-

w

∥

0

≤

2

⁢

k

.

Thus,

spark

⁢

(

A

)

=

inf

⁡

{

∥

x

∥

0

:

x

∈

𝔽

n

∖

{

𝟎

}

,

A

⁢

x

=

𝟎

}

≤

2

⁢

k

,

which means that condition

(2)

is not satisfied.

Next, suppose that

(2)

is not satisfied. Then there exists a column vector

u

∈

𝔽

n

∖

{

𝟎

}

such that

A

⁢

u

=

𝟎

and

∥

u

∥

0

≤

2

⁢

k

.

It is easy to see that

u

=

p

-

q

for some

p

,

q

∈

𝔽

n

with

max

⁡

{

∥

p

∥

0

,

∥

q

∥

0

}

≤

k

.

(If

h

:=

∥

u

∥

0

>

k

,

u

=

[

u

1

,

…

,

u

n

]

T

,

and

{

j

∈

{

1

,

…

,

n

}

:

u

j

≠

0

}

=

{

j

1

,

…

,

j

h

}

,

define

p

=

[

p

1

,

…

,

p

n

]

T

by

p

j

=

{

u

j

,

if

j

∈

{

j

1

,

…

,

j

k

}

,

0

,

otherwise.

If

∥

u

∥

0

≤

k

,

define

p

=

u

.

)

Since

p

≠

q

and

A

⁢

p

=

A

⁢

q

,

condition

(1)

is not satisfied.

□

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).

2

Compressed Sensing: Theory and Applications

, edited by Y. C. Eldar and G. Kutyniok, Cambridge University Press, 2012.

Title

uniqueness of a sparse

solution

Canonical name

UniquenessOfASparseSolution

Date of creation

2013-03-22 19:36:26

Last modified on

2013-03-22 19:36:26

Owner

kammerer (26336)

Last modified by

kammerer (26336)

Numerical id

54

Author

kammerer (26336)

Entry type

Theorem

Classification

msc 15A06

Generated on Fri Feb 9 18:23:10 2018 by

LaTeXML
adjugate

adjugate

The

adjugate

,

adj

⁡

(

A

)

, of an

n

×

n

matrix

A

, is the

n

×

n

matrix

adj

(

A

)

i

⁢

j

=

(

-

1

)

i

+

j

M

j

⁢

i

(

A

)

(1)

where

M

j

⁢

i

⁢

(

A

)

is the indicated minor of

A

(the

determinant

obtained by deleting row

j

and column

i

from

A

). The adjugate
is also known as the

classical adjoint

, to distinguish it from
the usual usage of “

adjoint

” (

http://planetmath.org/AdjointEndomorphism

) which
denotes the

conjugate transpose

operation

.

An

equivalent

characterization

of the adjugate is the following:

adj

⁡

(

A

)

⁢

A

=

det

⁡

(

A

)

⁢

I

.

(2)

The equivalence of (

1

) and (

2

) follows easily
from the

multi-linearity

properties (

http://planetmath.org/DeterminantAsAMultilinearMapping

) of the determinant.
Thus, the adjugate operation is closely related to the

matrix inverse

.
Indeed, if

A

is

invertible

, the adjugate can be defined as

adj

⁡

(

A

)

=

det

⁡

(

A

)

⁢

A

-

1

Yet another definition of the adjugate is the following:

adj

⁡

(

A

)

=

p

n

-

1

⁢

(

A

)

⁢

I

-

p

n

-

2

⁢

(

A

)

⁢

A

+

p

n

-

3

⁢

(

A

)

⁢

A

2

-

⋯

(3)

+

(

-

1

)

n

-

2

⁢

p

1

⁢

(

A

)

⁢

A

n

-

2

+

(

-

1

)

n

-

1

⁢

A

n

-

1

,

where

p

1

⁢

(

A

)

=

tr

⁡

(

A

)

,

p

2

⁢

(

A

)

,

…

,

p

n

⁢

(

A

)

=

det

⁡

(

A

)

are the

elementary

invariant polynomials of

A

. The latter arise as

coefficients

in the

characteristic polynomial

p

⁢

(

t

)

of

A

, namely

p

⁢

(

t

)

=

det

⁡

(

t

⁢

I

-

A

)

=

t

n

-

p

1

⁢

(

A

)

⁢

t

n

-

1

+

⋯

+

(

-

1

)

n

⁢

p

n

⁢

(

A

)

.

The equivalence of (

2

) and (

3

) follows from
the

Cayley-Hamilton theorem

. The latter states that

p

⁢

(

A

)

=

0

, which
in turn implies that

A

⁢

(

A

n

-

1

-

p

1

⁢

(

A

)

⁢

A

n

-

2

+

⋯

+

(

-

1

)

n

-

1

⁢

p

n

-

1

⁢

(

A

)

)

=

(

-

1

)

n

-

1

⁢

det

⁡

(

A

)

⁢

I

The adjugate operation enjoys a number of notable
properties:

adj

⁡

(

A

⁢

B

)

=

adj

⁡

(

B

)

⁢

adj

⁡

(

A

)

,

(4)

adj

(

A

t

)

=

adj

(

A

)

t

,

(5)

det

⁡

(

adj

⁡

(

A

)

)

=

det

⁡

(

A

)

n

-

1

.

(6)

Title

adjugate

Canonical name

Adjugate

Date of creation

2013-03-22 13:09:42

Last modified on

2013-03-22 13:09:42

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

17

Author

rmilson (146)

Entry type

Definition

Classification

msc 15A09

Synonym

classical adjoint

Generated on Fri Feb 9 18:23:13 2018 by

LaTeXML
generalized inverse

generalized inverse

Let

A

be an

m

×

n

matrix with entries in

ℂ

. A

generalized inverse

, denoted by

A

-

, is an

n

×

m

matrix with entries in

ℂ

, such that

A

⁢

A

-

⁢

A

=

A

.

Examples

1.

Let

A

=

(

2

3

0

1

2

0

0

0

0

)

.

Then any matrix of the form

A

-

=

(

2

-

3

a

-

1

2

b

c

d

e

)

,

where

a

,

b

,

c

,

d

and

e

∈

ℂ

, is a generalized inverse.

2.

Using the same example from above, if

a

=

b

=

c

=

d

=

e

=

0

, then we have an example of the

Moore-Penrose generalized inverse

, which is a unique matrix.

3.

Again, using the example from above, if

a

=

b

=

c

=

d

=

0

and

e

is any complex number, we have an example of a

Drazin inverse

.

Remark

Generalized inverse of a matrix has found many applications in

statistics

. For example, in

general linear model

, one solves the set of

normal equations

𝐗

T

⁢

𝐗

⁢

𝜷

=

𝐗

T

⁢

𝐘

,

to get the MLE

𝜷

^

of the

parameter

vector

𝜷

. If the design matrix

X

is not of full rank (this occurs often when the model is either an ANOVA or ANCOVA type) and hence

𝐗

T

⁢

𝐗

is

singular

. Then the MLE can be given by

𝜷

^

=

(

𝐗

T

⁢

𝐗

)

-

⁢

𝐗

T

⁢

𝐘

.

Title

generalized inverse

Canonical

name

GeneralizedInverse

Date of creation

2013-03-22 14:31:26

Last modified on

2013-03-22 14:31:26

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

5

Author

CWoo (3771)

Entry type

Definition

Classification

msc 15A09

Classification

msc 62J10

Classification

msc 62J12

Generated on Fri Feb 9 18:23:15 2018 by

LaTeXML
group inverse

group inverse

Let

A

be an

n

×

n

matrix over

ℝ

.
A

group inverse

for

A

is an

n

×

n

matrix

X

such that

A

⁢

X

⁢

A

=

A

(1)

X

⁢

A

⁢

X

=

X

(2)

A

⁢

X

=

X

⁢

A

.

(3)

Such a matrix, when it exists, is unique and is denoted by

A

#

.
A group inverse is a special case of a Drazin inverse.

Title

group inverse

Canonical name

GroupInverse

Date of creation

2013-03-22 17:01:17

Last modified on

2013-03-22 17:01:17

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

5

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 15A09

Generated on Fri Feb 9 18:23:16 2018 by

LaTeXML