aarjaneiro/planetmath huggingface dataset:
inverse of matrix with small-rank adjustment

inverse of matrix with small-rank adjustment

Suppose that an

n

×

n

matrix

B

is obtained by adding a
small-rank adjustment

X

⁢

R

⁢

Y

T

to matrix

A

,

B

=

A

+

X

⁢

R

⁢

Y

T

,

where

X

and

Y

are

n

×

r

matrices, and

R

is an

r

×

r

matrix. Assume that the

inverse

of

A

is known and

r

is much
smaller than

n

. The following formula for

B

-

1

is often
useful,

B

-

1

=

A

-

1

-

A

-

1

⁢

X

⁢

(

R

-

1

+

Y

T

⁢

A

-

1

⁢

X

)

-

1

⁢

Y

T

⁢

A

-

1

provided that all inverses in the formula exist.

In particular, when

r

=

1

and

A

=

I

, we have

(

I

+

x

⁢

y

T

)

-

1

=

I

-

x

⁢

y

T

1

+

y

T

⁢

x

for any

n

×

1

column vectors

x

and

y

such that

1

+

y

T

⁢

x

≠

0

.

Title

inverse of matrix with small-rank adjustment

Canonical name

InverseOfMatrixWithSmallrankAdjustment

Date of creation

2013-03-22 15:46:06

Last modified on

2013-03-22 15:46:06

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

8

Author

kshum (5987)

Entry type

Theorem

Classification

msc 15A09

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

LaTeXML
invertible matrices are dense in set of nxn matrices

invertible matrices are dense in set of nxn matrices

If

A

is any

n

×

n

matrix with real or

complex

entries,
Then there are

invertible matrices

arbitrarily close to

A

,
under any norm for the

n

×

n

matrices.

This is easily proven as follows. Take any invertible matrix

B

(e.g.

B

=

I

), and consider the function
(for

t

∈

ℝ

or

ℂ

)

p

⁢

(

t

)

=

det

⁡

(

(

1

-

t

)

⁢

A

+

t

⁢

B

)

.

Clearly,

p

is a

polynomial function

. It is not identically zero, for

p

⁢

(

1

)

=

det

⁡

B

≠

0

.
But a non-zero

polynomial

has only finitely many zeroes,
So given any single point

t

0

, if

t

is close enough but unequal to

t

0

,

p

⁢

(

t

)

must be non-zero. In particular, applying this for

t

0

=

0

,
we see that the matrix

(

1

-

t

)

⁢

A

+

t

⁢

B

is

invertible

for small

t

≠

0

.
And the distance of this matrix from

A

is

|

t

|

⁢

∥

B

-

A

∥

,
which becomes small as

t

gets small.

Title

invertible matrices are dense in set of nxn matrices

Canonical name

InvertibleMatricesAreDenseInSetOfNxnMatrices

Date of creation

2013-03-22 15:38:51

Last modified on

2013-03-22 15:38:51

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

5

Author

stevecheng (10074)

Entry type

Theorem

Classification

msc 15A09

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

LaTeXML
Moore-Penrose generalized inverse

Moore-Penrose generalized inverse

Let

A

be an

m

×

n

matrix with entries in

ℂ

. The

Moore-Penrose generalized inverse

, denoted by

A

†

, is an

n

×

m

matrix with entries in

ℂ

, such that

1.

A

⁢

A

†

⁢

A

=

A

2.

A

†

⁢

A

⁢

A

†

=

A

†

3.

A

⁢

A

†

and

A

†

⁢

A

are both

Hermitian

Remarks

•

The Moore-Penrose generalized inverse of a given matrix is unique.

•

If

A

†

is the Moore-Penrose generalized inverse of

A

, then

(

A

†

)

T

is the Moore-Penrose generalized inverse of

A

T

.

•

If

A

=

B

⁢

C

such that

(a)

A

∈

ℂ

m

×

n

,

B

∈

ℂ

m

×

r

, and

C

∈

ℂ

r

×

n

,

(b)

r

=

rank

⁡

(

A

)

=

rank

⁡

(

B

)

=

rank

⁡

(

C

)

, then

A

†

=

C

∗

⁢

(

C

⁢

C

∗

)

-

1

⁢

(

B

∗

⁢

B

)

-

1

⁢

B

∗

.

For example, let

A

=

(

1

1

i

0

1

0

)

.

Transform

A

to its

row echelon form

to get a

decomposition

of

A

=

B

⁢

C

, where

B

=

(

1

1

0

1

)

⁢

and

⁢

C

=

(

1

0

i

0

1

0

)

.

It is readily verified that

2

=

rank

⁡

(

A

)

=

rank

⁡

(

B

)

=

rank

⁡

(

C

)

.
So

A

†

=

1

2

⁢

(

1

-

1

0

2

-

i

i

)

.

We check that

A

⁢

A

†

=

I

⁢

and

⁢

A

†

⁢

A

=

1

2

⁢

(

1

0

i

0

2

0

-

i

0

1

)

are both Hermitian. Furthermore,

A

⁢

A

†

⁢

A

=

A

and

A

†

⁢

A

⁢

A

†

=

A

†

. So,

A

†

is the Moore-Penrose generalized inverse of

A

.

Title

Moore-Penrose generalized inverse

Canonical name

MoorePenroseGeneralizedInverse

Date of creation

2013-03-22 14:31:31

Last modified on

2013-03-22 14:31:31

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Definition

Classification

msc 15A09

Classification

msc 60J10

Synonym

Moore-Penrose pseudoinverse

Related topic

DrazinInverse

Related topic

Pseudoinverse

Generated on Fri Feb 9 18:23:25 2018 by

LaTeXML
proof of inverse of matrix with small-rank adjustment

proof of inverse of matrix with small-rank adjustment

We will first prove the formula when

A

=

I

.

Suppose that

R

-

1

+

Y

T

⁢

X

is

invertible

. Thus

(

R

-

1

+

Y

T

⁢

X

)

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

=

I

.

and

R

-

1

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

+

Y

T

⁢

X

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

=

I

.

Multiply by

X

⁢

R

from the left, and multiply by

Y

T

from the
right, we get

X

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

⁢

Y

T

+

X

⁢

R

⁢

Y

T

⁢

X

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

⁢

Y

T

=

X

⁢

R

⁢

Y

T

.

The

right hand side

is equal to

B

-

I

, while the left hand side can
be factorized as

(

I

+

X

⁢

R

⁢

Y

T

)

⁢

X

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

⁢

Y

T

.

So,

B

⋅

(

X

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

⁢

Y

T

)

=

B

-

I

.

After rearranging, we obtain

I

=

B

⁢

(

I

-

X

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

⁢

Y

T

)

.

Therefore

(

I

+

X

⁢

R

⁢

Y

T

)

-

1

=

I

-

X

⁢

(

R

-

1

+

Y

T

⁢

X

)

-

1

⁢

Y

T

(*)

For the general case

B

=

A

+

X

⁢

R

⁢

Y

T

, consider

B

⁢

A

-

1

=

I

+

X

⁢

R

⁢

Y

T

⁢

A

-

1

.

We can apply (*) with

Y

T

replaced by

Y

T

⁢

A

-

1

.

Title

proof of inverse of matrix with small-rank adjustment

Canonical name

ProofOfInverseOfMatrixWithSmallrankAdjustment

Date of creation

2013-03-22 15:46:08

Last modified on

2013-03-22 15:46:08

Owner

kshum (5987)

Last modified by

kshum (5987)

Numerical id

4

Author

kshum (5987)

Entry type

Proof

Classification

msc 15A09

Generated on Fri Feb 9 18:23:27 2018 by

LaTeXML
proof of matrix inverse calculation by Gaussian elimination

proof of matrix inverse calculation by Gaussian elimination

Let

A

be an

invertible matrix

, and

A

-

1

its

inverse

, whose
columns are

A

1

-

1

,

⋯

,

A

n

-

1

.
Then, by

definition

of

matrix inverse

,

A

⁢

A

-

1

=

I

n

.
But this implies

A

⁢

A

1

-

1

=

e

1

,

…

,

A

⁢

A

n

-

1

=

e

n

,
with

e

1

,

⋯

,

e

n

being the first,

…

,

n

-th column of

I

n

respectively.

A

being non

singular

(or

invertible

), for all

k

≤

n

,

A

⁢

A

k

-

1

=

e

k

has a solution for

A

k

-

1

, which can
be found by

Gaussian elimination

of

[

A

∣

e

k

]

.

The only part that changes between the augmented matrices
constructed is the last column, and these last columns, once the
Gaussian elimination has been performed, correspond to the columns
of

A

-

1

. Because of this, the steps we need to take for the
Gaussian elimination are the same for each augmented matrix.

Therefore, we can solve the

matrix equation

by performing Gaussian
elimination on

[

A

∣

e

1

⋯

e

n

]

, or

[

A

∣

I

n

]

.

Title

proof of matrix inverse calculation by Gaussian elimination

Canonical

name

ProofOfMatrixInverseCalculationByGaussianElimination

Date of creation

2013-03-22 14:15:10

Last modified on

2013-03-22 14:15:10

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 15A09

Generated on Fri Feb 9 18:23:28 2018 by

LaTeXML
Toeplitz matrix

Toeplitz matrix

1

Toeplitz Matrix

A

Toeplitz matrix

is any

n

×

n

matrix with values constant along each (top-left to lower-right)

diagonal

. That is, a Toeplitz matrix has the form

[

a

0

a

1

a

2

⋯

a

n

-

1

a

-

1

a

0

a

1

⋱

⋮

a

-

2

a

-

1

a

0

⋱

a

2

⋮

⋱

⋱

⋱

a

1

a

-

(

n

-

1

)

⋯

a

-

2

a

-

1

a

0

]

Numerical problems involving Toeplitz matrices typically have fast solutions (only

2

⁢

n

-

1

distinct elements need to be solved for, as opposed to

n

2

). For example, the

inverse

of a

symmetric

, positive-definite

n

×

n

Toeplitz matrix can be found in

𝒪

⁢

(

n

2

)

time (

http://planetmath.org/TimeComplexity

).

References

1

Golub and Van Loan,

Matrix Computations

, Johns Hopkins University Press 1993

Title

Toeplitz matrix

Canonical name

ToeplitzMatrix

Date of creation

2013-03-22 13:04:22

Last modified on

2013-03-22 13:04:22

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

8

Author

akrowne (2)

Entry type

Definition

Classification

msc 15A09

Classification

msc 65F35

Classification

msc 15A57

Generated on Fri Feb 9 18:23:30 2018 by

LaTeXML
matrix condition number

matrix condition number

1

Matrix Condition Number

The

condition number

for

matrix inversion

with respect to a

matrix norm

∥

⋅

∥

of a

square matrix

A

is defined by

κ

⁢

(

A

)

=

∥

A

∥

⁢

∥

A

-

1

∥

,

if

A

is non-singular; and

κ

⁢

(

A

)

=

+

∞

if

A

is

singular

.

The condition number is a measure of stability or sensitivity of a matrix (or the

linear system

it represents) to numerical operations. In other words, we may not be able to trust the results of computations on an

ill-conditioned matrix

.

Matrices with condition numbers near 1 are said to be

well-conditioned

. Matrices with condition numbers much greater than one (such as around

10

5

for a

5

×

5

Hilbert matrix

) are said to be

ill-conditioned

.

If

κ

⁢

(

A

)

is the condition number of

A

, then

κ

⁢

(

A

)

measures
a sort of

inverse

distance

from

A

to the set of

singular matrices

,
normalized by

∥

A

∥

.
Precisely, if

A

is

invertible

, and

∥

B

-

A

∥

<

∥

A

-

1

∥

-

1

,
then

B

must also be invertible. On the other hand, in the case of the

2

-norm,
there always exists a singular matrix

B

such that

∥

B

-

A

∥

2

=

∥

A

-

1

∥

2

-

1

(so the distance

estimate

is sharp).

References

1

Golub and Van Loan.

Matrix Computations

, 3rd edition. Johns Hopkins University Press, 1996.

Title

matrix condition number

Canonical

name

MatrixConditionNumber

Date of creation

2013-03-22 13:04:17

Last modified on

2013-03-22 13:04:17

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

10

Author

stevecheng (10074)

Entry type

Definition

Classification

msc 15A12

Classification

msc 65F35

Synonym

matrix condition number

Synonym

condition number

Related topic

PropertyOfMatrixConditionNumber

Defines

ill-conditioned

Defines

well-conditioned

Generated on Fri Feb 9 18:23:32 2018 by

LaTeXML
block determinants

block determinants

If

A

and

D

are

square matrices

•

If

A

-

1

exists, then

det

⁡

(

A

B

C

D

)

=

det

⁡

(

A

)

⁢

det

⁡

(

D

-

C

⁢

A

-

1

⁢

B

)

•

If

D

-

1

exists, then

det

⁡

(

A

B

C

D

)

=

det

⁡

(

D

)

⁢

det

⁡

(

A

-

B

⁢

D

-

1

⁢

C

)

The matrices

D

-

C

⁢

A

-

1

⁢

B

and

A

-

B

⁢

D

-

1

⁢

C

are called the

Schur complements

of

A

and

D

, respectively.

Mention that

•

If

A

,

D

are square matrices, then

det

⁡

(

A

B

O

D

)

=

det

⁡

(

A

)

⁢

det

⁡

(

D

)

, where

O

is a

zero matrix

.

•

Also we have that

det

⁡

(

A

O

O

B

)

=

det

⁡

(

A

)

⁢

det

⁡

(

B

)

.

•

Another useful result for

block determinants

is the following.

As

J

=

(

O

I

-

I

O

)

is a

symplectic matrix

,
we have that

det

⁡

J

=

1

. Using now the fact that

det

⁡

M

⁢

N

=

det

⁡

(

M

)

⁢

det

⁡

(

N

)

for any

M

,

N

square matrices, we have that

det

⁡

(

O

A

B

O

)

=

det

⁡

(

O

A

B

O

)

⁢

det

⁡

J

=

-

det

⁡

(

A

)

⁢

det

⁡

(

B

)

This holds for any square matrices

A

,

B

and for the last point

A

,

B

have also the same order. They do not need to be

invertible

.

Title

block determinants

Canonical name

BlockDeterminants

Date of creation

2013-03-22 15:25:57

Last modified on

2013-03-22 15:25:57

Owner

georgiosl (7242)

Last modified by

georgiosl (7242)

Numerical id

20

Author

georgiosl (7242)

Entry type

Theorem

Classification

msc 15A15

Related topic

SchurComplement

Related topic

DeterminantsOfSomeMatricesOfSpecialForm

Generated on Fri Feb 9 18:23:37 2018 by

LaTeXML
Cauchy-Binet formula

Cauchy-Binet formula

Let

A

be an

m

×

n

matrix and

B

an

n

×

m

matrix. Then
the

determinant

of their product

C

=

A

⁢

B

can be written as a sum of products of
minors of

A

and

B

:

|

C

|

=

∑

1

≤

k

1

<

k

2

<

⋯

<

k

m

≤

n

A

⁢

(

1

2

⋯

m

k

1

k

2

⋯

k

m

)

⁢

B

⁢

(

k

1

k

2

⋯

k

m

1

2

⋯

m

)

.

Basically, the sum is over the

maximal

(

m

-th order) minors of

A

and

B

.
See the entry on minors (

http://planetmath.org/MinorOfAMatrix

) for notation.

If

m

>

n

, then neither

A

nor

B

have minors of rank

m

, so

|

C

|

=

0

.
If

m

=

n

, this formula reduces to the usual multiplicativity of determinants

|

C

|

=

|

A

⁢

B

|

=

|

A

|

⁢

|

B

|

.

Proof.

Since

C

=

A

⁢

B

, we can write its elements as

c

i

⁢

j

=

∑

k

=

1

n

a

i

⁢

k

⁢

b

k

⁢

j

. Then its determinant is

|

C

|

=

|

∑

k

1

=

1

n

a

1

⁢

k

1

⁢

b

k

1

⁢

1

⋯

∑

k

m

=

1

n

a

1

⁢

k

m

⁢

b

k

m

⁢

m

⋮

⋱

⋮

∑

k

1

=

1

n

a

m

⁢

k

1

⁢

b

k

1

⁢

1

⋯

∑

k

m

=

1

n

a

m

⁢

k

m

⁢

b

k

m

⁢

m

|

=

∑

k

1

,

…

,

k

m

=

1

n

|

a

1

⁢

k

1

⁢

b

k

1

⁢

1

⋯

a

1

⁢

k

m

⁢

b

k

m

⁢

m

⋮

⋱

⋮

a

m

⁢

k

1

⁢

b

k

1

⁢

1

⋯

a

m

⁢

k

m

⁢

b

k

m

⁢

m

|

=

∑

k

1

,

…

,

k

m

=

1

n

A

⁢

(

1

2

⋯

m

k

1

k

2

⋯

k

m

)

⁢

b

k

1

⁢

1

⁢

b

k

2

⁢

2

⁢

⋯

⁢

b

k

m

⁢

m

.

In both steps above, we have used the property that the determinant is

multilinear

in the colums of a matrix.

Note that the terms in the last sum with any two

k

’s the same will
make the minor of

A

vanish. And, for

{

k

1

,

⋯

,

k

m

}

’s
that differ only by a

permutation

, the minor of

A

will simply change
sign according to the parity of the permutation. Hence the determinant of

C

can be rewritten as

|

C

|

=

∑

1

≤

k

1

<

⋯

<

k

m

≤

n

A

⁢

(

1

2

⋯

m

k

1

k

2

⋯

k

m

)

⁢

∑

σ

∈

S

m

sgn

⁡

(

σ

)

⁢

b

k

σ

⁢

(

1

)

⁢

1

⁢

b

k

σ

⁢

(

2

)

⁢

2

⁢

⋯

⁢

b

k

σ

⁢

(

m

)

⁢

m

,

where

S

m

is the

permutation group

on

m

elements.
But the last sum is none other than the determinant

B

⁢

(

k

1

k

2

⋯

k

m

1

2

⋯

m

)

.
Hence we write

|

C

|

=

∑

1

≤

k

1

<

⋯

<

k

m

≤

n

A

⁢

(

1

2

⋯

m

k

1

k

2

⋯

k

m

)

⁢

B

⁢

(

k

1

k

2

⋯

k

m

1

2

⋯

m

)

,

which is the

Cauchy-Binet formula

.
∎

Title

Cauchy-Binet formula

Canonical name

CauchyBinetFormula

Date of creation

2013-03-22 14:07:04

Last modified on

2013-03-22 14:07:04

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

11

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 15A15

Synonym

Binet-Cauchy formula

Related topic

MinorOfAMatrix

Generated on Fri Feb 9 18:23:41 2018 by

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Cayley-Hamilton theorem

Cayley-Hamilton theorem

Let

T

be a

linear operator

on a

finite-dimensional

vector space

V

, and let

c

⁢

(

t

)

be the

characteristic polynomial

of

T

. Then

c

⁢

(

T

)

=

T

0

, where

T

0

is the zero

transformation

. In other words,

T

satisfies its own

characteristic equation

.

Title

Cayley-Hamilton theorem

Canonical name

CayleyHamiltonTheorem

Date of creation

2013-03-22 12:17:43

Last modified on

2013-03-22 12:17:43

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

4

Author

akrowne (2)

Entry type

Theorem

Classification

msc 15A15

Classification

msc 15A18

Generated on Fri Feb 9 18:23:43 2018 by

LaTeXML
Cramer’s rule

Cramer’s rule

Let

A

⁢

x

=

b

be the matrix form of a system of

n

linear equations

in

n

unknowns, with

x

and

b

as

n

×

1

column vectors

and

A

an

n

×

n

matrix. If

det

⁡

(

A

)

≠

0

, then this system has a unique solution, and for each

i

(

1

≤

i

≤

n

) ,

x

i

=

det

⁡

(

M

i

)

det

⁡

(

A

)

where

M

i

is

A

with column

i

replaced by

b

.

Title

Cramer’s rule

Canonical name

CramersRule

Date of creation

2013-03-22 11:55:27

Last modified on

2013-03-22 11:55:27

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

10

Author

akrowne (2)

Entry type

Theorem

Classification

msc 15A15

Generated on Fri Feb 9 18:23:48 2018 by

LaTeXML
determinant as a multilinear mapping

determinant as a multilinear mapping

Let

𝐌

=

(

M

i

⁢

j

)

be an

n

×

n

matrix with entries in a field

K

. The matrix

𝐌

is really the same thing as a list of

n

column
vectors

of size

n

. Consequently, the

determinant

operation

may be
regarded as a

mapping

det

:

K

n

×

…

×

K

n

⏞

n

⁢

times

→

K

The determinant of a matrix

𝐌

is then defined to be

det

⁡

(

𝐌

1

,

…

,

𝐌

n

)

,

where

𝐌

j

∈

K

n

denotes the

j

th

column of

𝐌

.

Starting with the definition

det

⁡

(

𝐌

1

,

…

,

𝐌

n

)

=

∑

π

∈

S

n

sgn

⁢

(

π

)

⁢

M

1

⁢

π

1

⁢

M

2

⁢

π

2

⁢

⋯

⁢

M

n

⁢

π

n

(1)

the following properties are easily established:

1.

the determinant is

multilinear

;

2.

the determinant is

anti-symmetric

;

3.

the determinant of the

identity matrix

is

1

.

These three properties uniquely characterize the determinant, and
indeed can — some would say should — be used as the definition of
the determinant operation.

Let us prove this. We proceed by representing elements of

K

n

as

linear combinations

of

𝐞

1

=

(

1

0

0

⋮

0

)

,

𝐞

2

=

(

0

1

0

⋮

0

)

,

…

𝐞

n

=

(

0

0

0

⋮

1

)

,

the

standard basis

of

K

n

. Let

𝐌

be an

n

×

n

matrix.
The

j

th

column is represented as

∑

i

M

i

⁢

j

⁢

𝐞

i

; whence
using multilinearity

det

⁡

(

𝐌

)

=

det

⁡

(

∑

i

M

i

⁢

1

⁢

𝐞

i

,

∑

i

M

i

⁢

2

⁢

𝐞

i

,

…

,

∑

i

M

i

⁢

n

⁢

𝐞

i

)

=

∑

i

1

,

…

,

i

n

=

1

n

M

i

1

⁢

1

⁢

M

i

2

⁢

2

⁢

⋯

⁢

M

i

n

⁢

n

⁢

det

⁡

(

𝐞

i

1

,

𝐞

i

2

,

…

,

𝐞

i

n

)

The anti-symmetry

assumption

implies that the expressions

det

⁡

(

𝐞

i

1

,

𝐞

i

2

,

…

,

𝐞

i

n

)

vanish if any two of the
indices

i

1

,

…

,

i

n

coincide. If all

n

indices are distinct,

det

⁡

(

𝐞

i

1

,

𝐞

i

2

,

…

,

𝐞

i

n

)

=

±

det

⁡

(

𝐞

1

,

…

,

𝐞

n

)

,

the sign in the above expression being
determined by the number of

transpositions

required to rearrange
the list

(

i

1

,

…

,

i

n

)

into the list

(

1

,

…

,

n

)

. The sign is
therefore the parity of the

permutation

(

i

1

,

…

,

i

n

)

. Since we
also assume that

det

⁡

(

𝐞

1

,

…

,

𝐞

n

)

=

1

,

we now recover the original definition
(

1

).

Title

determinant as a multilinear mapping

Canonical name

DeterminantAsAMultilinearMapping

Date of creation

2013-03-22 13:09:17

Last modified on

2013-03-22 13:09:17

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

5

Author

rmilson (146)

Entry type

Theorem

Classification

msc 15A15

Related topic

ExteriorAlgebra

Generated on Fri Feb 9 18:23:53 2018 by

LaTeXML
determinant condition for a sequence of vectors

determinant condition for a sequence of vectors

Let

x

1

,

x

2

,

…

be a sequence of

d

dimensional vectors. Assume that there is a

function

C

:

ℕ

d

→

ℝ

∖

{

0

}

such that

∑

n

1

+

⋯

+

n

d

=

n

0

<

n

1

<

⋯

<

n

d

C

⁢

(

n

1

,

…

,

n

d

)

⁢

det

⁡

[

x

n

1

,

x

n

2

,

…

,

x

n

d

]

=

0

for every

n

∈

ℕ

. Then the sequence is contained within a proper

linear subspace

.

Title

determinant

condition for a sequence of vectors

Canonical name

DeterminantConditionForASequenceOfVectors

Date of creation

2013-03-22 14:33:44

Last modified on

2013-03-22 14:33:44

Owner

GeraW (6138)

Last modified by

GeraW (6138)

Numerical id

10

Author

GeraW (6138)

Entry type

Result

Classification

msc 15A15

Generated on Fri Feb 9 18:23:55 2018 by

LaTeXML
determinant in terms of traces of powers

determinant in terms of traces of powers

It is possible to express the

determinant

of a matrix in of traces of
powers of a matrix.

The easiest way to derive these

expressions

is to specialize to the case of

diagonal matrices

. For instance, suppose we have a

2

×

2

matrix

M

=

diag

⁡

(

u

,

v

)

. Then

det

⁡

M

=

u

⁢

v

tr

⁡

M

=

u

+

v

tr

⁡

M

2

=

u

2

+

v

2

From the algebraic identity

(

u

+

v

)

2

=

u

2

+

v

2

+

2

⁢

u

⁢

v

, it can be concluded that

det

⁡

M

=

1

2

⁢

(

tr

⁡

M

)

2

-

1

2

⁢

tr

⁡

(

M

2

)

.

Likewise, one can derive expressions for the determinants of larger matrices from the identities for

elementary symmetric polynomials

in of power sums. For instance, from the identity

x

⁢

y

⁢

z

=

1

6

⁢

(

x

+

y

+

z

)

3

-

1

2

⁢

(

x

2

+

y

2

+

z

2

)

⁢

(

x

+

y

+

z

)

+

1

3

⁢

(

x

3

+

y

3

+

z

3

)

,

it can be concluded that

det

⁡

M

=

1

6

⁢

(

tr

⁡

M

)

3

-

1

2

⁢

(

tr

⁡

M

2

)

⁢

(

tr

⁡

M

)

+

1

3

⁢

tr

⁡

M

3

for a

3

×

3

matrix

M

.

Title

determinant in terms of traces of powers

Canonical

name

DeterminantInTermsOfTracesOfPowers

Date of creation

2013-03-22 15:57:08

Last modified on

2013-03-22 15:57:08

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

11

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 15A15

Generated on Fri Feb 9 18:23:57 2018 by

LaTeXML