aarjaneiro/planetmath huggingface dataset:
Peetre’s inequality

Peetre’s inequality

Theorem  [Peetre’s

inequality

]

[

1

,

2

]

If

t

is a

real number

and

x

,

y

are vectors in

ℝ

n

, then

(

1

+

|

x

|

2

1

+

|

y

|

2

)

t

≤

2

|

t

|

⁢

(

1

+

|

x

-

y

|

2

)

|

t

|

.

Proof.

(Following

[

1

]

.)
Suppose

b

and

c

are vectors in

ℝ

n

. Then, from

(

|

b

|

-

|

c

|

)

2

≥

0

, we obtain

2

⁢

|

b

|

⋅

|

c

|

≤

|

b

|

2

+

|

c

|

2

.

Using this inequality and the

Cauchy-Schwarz inequality

, we obtain

1

+

|

b

-

c

|

2

=

1

+

|

b

|

2

-

2

⁢

b

⋅

c

+

|

c

|

2

≤

1

+

|

b

|

2

+

2

⁢

|

b

|

⁢

|

c

|

+

|

c

|

2

≤

1

+

2

⁢

|

b

|

2

+

2

⁢

|

c

|

2

≤

2

⁢

(

1

+

|

b

|

2

+

|

c

|

2

+

|

b

|

2

⁢

|

c

|

2

)

=

2

⁢

(

1

+

|

b

|

2

)

⁢

(

1

+

|

c

|

2

)

Let us define

a

=

b

-

c

. Then for any vectors

a

and

b

, we have

1

+

|

a

|

2

1

+

|

b

|

2

≤

2

⁢

(

1

+

|

a

-

b

|

2

)

.

(1)

Let us now return to the given inequality.
If

t

=

0

, the claim is trivially true for all

x

,

y

in

ℝ

n

.
If

t

>

0

, then raising both sides in inequality

1

to
the power of

t

, using

t

=

|

t

|

, and setting

a

=

x

,

b

=

y

yields the result.
On the other hand, if

t

<

0

, then raising both sides in inequality

1

to the power to

-

t

, using

-

t

=

|

t

|

, and setting

a

=

y

,

b

=

x

yields the result.

□

References

1

J. Barros-Neta,

An introduction to the theory of distributions

,
Marcel Dekker, Inc., 1973.

2

F. Treves,

Introduction To Pseudodifferential and Fourier Integral Operators

,
Vol. I, Plenum Press, 1980.

Title

Peetre’s inequality

Canonical name

PeetresInequality

Date of creation

2013-03-22 13:55:26

Last modified on

2013-03-22 13:55:26

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

10

Author

Koro (127)

Entry type

Theorem

Classification

msc 15-00

Classification

msc 15A39

Generated on Fri Feb 9 18:15:41 2018 by

LaTeXML
pentadiagonal matrix

pentadiagonal matrix

An

n

×

n

pentadiagonal

matrix (with

n

≥

3

) is
a matrix of the form

(

c

1

d

1

e

1

0

⋯

⋯

0

b

1

c

2

d

2

e

2

⋱

⋮

a

1

b

2

⋱

⋱

⋱

⋱

⋮

0

a

2

⋱

⋱

⋱

e

n

-

3

0

⋮

⋱

⋱

⋱

⋱

d

n

-

2

e

n

-

2

⋮

⋱

a

n

-

3

b

n

-

2

c

n

-

1

d

n

-

1

0

⋯

⋯

0

a

n

-

2

b

n

-

1

c

n

)

.

It follows that a

pentadiagonal matrix

is determined by five vectors:
one

n

-vector

c

=

(

c

1

,

…

,

c

n

)

,
two

(

n

-

1

)

-vectors

b

=

(

b

1

,

…

,

b

n

-

1

)

and

d

=

(

d

1

,

…

,

d

n

-

1

)

,
and two

(

n

-

2

)

-vectors

a

=

(

a

1

,

…

,

a

n

-

2

)

and

e

=

(

e

1

,

…

,

e

n

-

2

)

.
It follows that a pentadiagonal matrix is completely determined
by

n

+

2

⁢

(

n

-

1

)

+

2

⁢

(

n

-

2

)

=

5

⁢

n

-

6

scalars.

Title

pentadiagonal matrix

Canonical

name

PentadiagonalMatrix

Date of creation

2013-03-22 13:23:23

Last modified on

2013-03-22 13:23:23

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Definition

Classification

msc 15-00

Classification

msc 65-00

Synonym

penta-diagonal matrix

Related topic

TridiagonalMatrix

Generated on Fri Feb 9 18:15:43 2018 by

LaTeXML
point

point

In

The

, Euclid defines a point as that which has no part.

In a vector space, an

affine space

, or, more generally, an

incidence geometry

, a

point

is a zero (

http://planetmath.org/Zero

) dimensional (

http://planetmath.org/Dimension3

) .

In a

projective geometry

, a

point

is a one-dimensional

subspace

of the vector space underlying the projective geometry.

In a

topology

, a

point

is an element of a topological space.

In function theory, a

point

usually means a

complex number

as an element of the

complex plane

.

Note that there is also the possibility for a point-free approach to

geometry

in which points are not assumed as a

primitive

. Instead, points are defined by suitable abstraction processes. (See point-free geometry.)

Title

point

Canonical name

Point

Date of creation

2013-03-22 16:06:30

Last modified on

2013-03-22 16:06:30

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

16

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 15-00

Classification

msc 54-00

Classification

msc 51-00

Generated on Fri Feb 9 18:15:45 2018 by

LaTeXML
principal components analysis

principal components analysis

The

principal components analysis

is a mathematical way of determining that

linear transformation

of a sample of points in

N

-dimensional space which exhibits the properties of the sample most clearly along the

coordinate

axes. Along the new axes the

sample variances

are extremes (maxima and minima), and uncorrelated. The name comes from the principal axes of an ellipsoid (e.g. the ellipsoid of inertia), which are just the coordinate axes in question.

By their definition, the principal axes will include those along which the point sample has little or no spread (minima of

variance

). Hence, an analysis in terms of principal components can show (linear) interdependence in data. A point sample of

N

dimensions

for whose

N

coordinates

M

linear relations hold, will show only

(

N

-

M

)

axes along which the spread is non-zero. Using a cutoff on the spread along each axis, a sample may thus be reduced in its dimensionality (see [Bishop95]).

The principal axes of a point sample are found by choosing the origin at the “centre of gravity” and forming the dispersion matrix

t

i

⁢

j

=

(

1

/

N

)

⁢

∑

[

(

x

i

-

⟨

x

i

⟩

)

⁢

(

x

j

-

⟨

x

j

⟩

)

]

where the sum is over the

N

points of the sample and the

x

i

are the

i

th components of the point coordinates.

⟨

.

⟩

stands for the average of the parameter. The principal axes and the variance along each of them are then given by the

eigenvectors

and associated

eigenvalues

of the dispersion matrix.

Principal component analysis has in practice been used to reduce the dimensionality of problems, and to transform interdependent coordinates into significant and independent ones. An example used in several particle physics experiments is that of reducing redundant

observations

of a particle track in a detector to a low-dimensional

subspace

whose axes correspond to parameters describing the track. Another example is in image processing; where it can be used for color quantization. Principle components analysis is described in [O’Connel74].

References

•

Originally from The Data Analysis Briefbook
(

http://rkb.home.cern.ch/rkb/titleA.html

http://rkb.home.cern.ch/rkb/titleA.html)

Bishop95

C.M. Bishop, Neural Networks for Pattern Recognition, Oxford University Press, Oxford, 1995.

O’Connel74

M.J. O’Connel, Search Program for Significant Variables, Comp. Phys. Comm. 8 (1974) 49.

Title

principal components analysis

Canonical name

PrincipalComponentsAnalysis

Date of creation

2013-03-22 12:11:49

Last modified on

2013-03-22 12:11:49

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

7

Author

akrowne (2)

Entry type

Definition

Classification

msc 15-00

Classification

msc 65F30

Classification

msc 65-00

Synonym

Karhunen-Loeve transform

Synonym

Hotelling transform

Synonym

PCA

Generated on Fri Feb 9 18:15:46 2018 by

LaTeXML
proof of determinant lower bound of a strict diagonally dominant matrix

proof of determinant lower bound of a strict diagonally dominant matrix

Let’s define, for any

i

=

1

,

2

,

…

,

n

h

i

=

|

a

i

⁢

i

|

-

∑

j

=

1

,

j

≠

i

|

a

i

⁢

j

|

Then, by strict diagonally dominance, one has

h

i

>

0

∀

i

.
Let

D

=

d

⁢

i

⁢

a

⁢

g

⁢

{

(

h

1

)

-

1

,

(

h

2

)

-

1

,

…

,

(

h

n

)

-

1

}

and

B

=

D

⁢

A

, so that the i-th row of

B

matrix is equal to the corresponding row of

A

matrix multiplied by

(

h

i

)

-

1

. In this way , one has

d

i

=

|

b

i

⁢

i

|

-

∑

j

=

1

,

j

≠

i

|

b

i

⁢

j

|

=

|

a

i

⁢

i

|

h

i

-

∑

j

=

1

,

j

≠

i

|

a

i

⁢

j

|

h

i

=

1

Now, let

λ

be an

eigenvalue

of

B

, and

v

=

[

v

1

,

v

2

,

…

,

v

n

]

the corresponding

eigenvector

; let moreover

p

be the index of the maximal

component

of

v

, i.e.

|

v

p

|

≥

|

v

i

|

⁢

⁢

∀

i

Of course, by

definition

of eigenvector,

|

v

p

|

>

0

.
Writing the p-th

characteristic equation

, we have:

λ

⁢

v

p

=

∑

j

=

1

n

b

p

⁢

j

⁢

v

j

=

b

p

⁢

p

⁢

v

p

+

∑

j

=

1

,

j

≠

p

n

b

p

⁢

j

⁢

v

j

so that, being

|

v

j

v

p

|

≤

1

,

λ

=

b

p

⁢

p

+

∑

j

=

1

,

j

≠

p

n

b

p

⁢

j

⁢

v

j

v

p

|

λ

|

=

|

b

p

⁢

p

+

∑

j

=

1

,

j

≠

p

n

b

p

⁢

j

⁢

v

j

v

p

|

≥

|

|

b

p

⁢

p

|

-

|

∑

j

=

1

,

j

≠

p

n

b

p

⁢

j

⁢

v

j

v

p

|

|

≥

|

|

b

p

⁢

p

|

-

∑

j

=

1

,

j

≠

p

n

|

b

p

⁢

j

|

⁢

|

v

j

v

p

|

|

⁢

(*)

≥

||

b

p

⁢

p

⁢

|

-

∑

j

=

1

,

j

≠

p

n

|

⁢

b

p

⁢

j

||

⁢

(**)

=

|

b

p

⁢

p

|

-

∑

j

=

1

,

j

≠

p

n

|

b

p

⁢

j

|

=

d

p

=

1

In this way, we found that each eigenvalue of

B

is greater than one in

absolute value

; for this reason,

|

det

⁡

(

B

)

|

=

|

∏

i

=

1

n

λ

i

|

≥

1

Finally,

det

⁡

(

D

)

=

∏

i

=

1

n

(

h

i

)

-

1

=

(

∏

i

=

1

n

h

i

)

-

1

so that

1

≤

|

det

⁡

(

B

)

|

=

|

det

⁡

(

D

)

|

⁢

|

det

⁡

(

A

)

|

=

(

∏

i

=

1

n

h

i

)

-

1

⁢

|

det

⁡

(

A

)

|

whence the thesis.

Remark:

Perhaps it could be not immediately evident where the

hypothesis

of strict diagonally dominance is employed in this proof; in
fact,

inequality

(*) and (**) would be, in a general case,

not valid

; they can be stated only because we can assure, by virtue of strict
diagonally dominance, that the final

argument

of the absolute value (

|

b

p

⁢

p

|

-

∑

j

=

1

,

j

≠

p

n

|

b

p

⁢

j

|

) does remain

positive

.

Title

proof of

determinant

lower bound

of a strict

diagonally dominant matrix

Canonical

name

ProofOfDeterminantLowerBoundOfAStrictDiagonallyDominantMatrix

Date of creation

2013-03-22 17:01:11

Last modified on

2013-03-22 17:01:11

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

13

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 15-00

Generated on Fri Feb 9 18:15:50 2018 by

LaTeXML
proof that d⁢e⁢te^A=e^{tr⁡A}

proof that

d

⁢

e

⁢

t

e

A

=

e

tr

⁡

A

According to

Schur decomposition

the matrix

A

can be written after a suitable

change of basis

as

A

=

D

+

N

where

D

is a

diagonal matrix

and

N

is a

strictly upper triangular matrix

.

The formula we aim to prove

det

⁡

e

A

=

e

tr

⁡

A

is

invariant

under a change of basis and thus we can carry out the computation of the

exponential

in any basis we choose.

By definition

e

A

=

∑

n

=

0

∞

A

n

n

!

(1)

By the properties of

diagonal

and strictly upper triangular matrices we know that both

D

⁢

N

and

N

⁢

D

will also be strictly upper triangular matrices and so will their sum.

Thus the powers of

A

are of the form:

A

=

(

D

+

N

)

=

D

+

N

1

(2)

A

2

=

(

D

+

N

)

⁢

(

D

+

N

)

=

D

2

+

N

2

(3)

A

3

=

(

D

+

N

)

⁢

(

D

2

+

N

2

)

=

D

3

+

N

3

(4)

⋮

(5)

A

k

=

D

k

+

N

k

(6)

⋮

(7)

where all the

N

i

matrices are strictly

upper triangular

.
Explicitly,

N

2

=

D

⁢

N

1

+

N

1

⁢

D

+

N

1

2

and by recursion

N

n

+

1

=

D

⁢

N

n

+

N

n

⁢

D

+

N

1

⁢

N

n

.

Using equation

1

we can write

e

A

=

e

D

+

N

~

(8)

where

N

~

=

∑

n

=

1

∞

N

n

n

!

is strictly upper triangular and

e

D

=

diag

⁡

(

e

λ

1

,

⋯

,

e

λ

n

)

, where

D

=

diag

⁡

(

λ

1

,

⋯

,

λ

n

)

.

e

A

will thus be an

upper triangular matrix

.
Since the

determinant

of an upper triangular matrix is just the product of the elements in its diagonal, we can write:

det

⁡

e

A

=

∏

i

=

1

n

e

λ

i

=

e

∑

i

=

1

n

λ

i

=

e

tr

⁡

A

(9)

Title

proof that

d

⁢

e

⁢

t

e

A

=

e

tr

⁡

A

Canonical name

ProofThatdetEAEoperatornametrA

Date of creation

2013-03-22 15:51:56

Last modified on

2013-03-22 15:51:56

Owner

cvalente (11260)

Last modified by

cvalente (11260)

Numerical id

7

Author

cvalente (11260)

Entry type

Proof

Classification

msc 15-00

Classification

msc 15A15

Related topic

SchurDecomposition

Generated on Fri Feb 9 18:15:53 2018 by

LaTeXML
properties of diagonally dominant matrix

properties of diagonally dominant matrix

1)(

Levy-Desplanques theorem

) A

strictly diagonally dominant matrix

is non-singular.

Proof.

Let

A

be a strictly diagonally dominant matrix and let’s assume

A

is

singular

, that is,

λ

=

0

∈

σ

⁢

(

A

)

. Then, by Gershgorin’s circle theorem, an index

i

exists such that:

∑

j

≠

i

|

a

i

⁢

j

|

≥

|

λ

-

a

i

⁢

i

|

=

|

a

i

⁢

i

|

,

which is in contrast with

strictly

diagonally dominance

definition

.
∎

2)()

|

det

⁡

(

A

)

|

≥

∏

i

=

1

n

(

|

a

i

⁢

i

|

-

∑

j

=

1

,

j

≠

i

|

a

i

⁢

j

|

)

(See here (

http://planetmath.org/ProofOfDeterminantLowerBoundOfAStrictDiagonallyDominantMatrix

) for a proof.)

3) A

Hermitian

diagonally dominant matrix

with real nonnegative

diagonal

entries is

positive semidefinite

.

Proof.

Let

A

be a Hermitian diagonally dominant matrix with real nonnegative diagonal entries; then its

eigenvalues

are real and, by Gershgorin’s circle theorem, for each eigenvalue an index

i

exists such that:

λ

∈

[

a

i

⁢

i

-

∑

j

≠

i

|

a

i

⁢

j

|

,

a

i

⁢

i

+

∑

i

≠

j

|

a

i

⁢

j

|

]

,

which implies, by definition of diagonally dominance,

λ

≥

0

.

∎

Title

properties of diagonally dominant matrix

Canonical

name

PropertiesOfDiagonallyDominantMatrix

Date of creation

2013-03-22 15:34:32

Last modified on

2013-03-22 15:34:32

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

15

Author

Andrea Ambrosio (7332)

Entry type

Result

Classification

msc 15-00

Generated on Fri Feb 9 18:15:55 2018 by

LaTeXML
pseudoinverse

pseudoinverse

The

inverse

A

-

1

of a matrix

A

exists only if

A

is square and has full rank. In this case,

A

⁢

x

=

b

has the solution

x

=

A

-

1

⁢

b

.

The

pseudoinverse

A

+

(beware, it is often denoted otherwise) is a generalization of the inverse, and exists for any

m

×

n

matrix. We assume

m

>

n

. If

A

has full rank (

n

) we define:

A

+

=

(

A

T

⁢

A

)

-

1

⁢

A

T

and the solution of

A

⁢

x

=

b

is

x

=

A

+

⁢

b

.

More accurately, the above is called the

Moore-Penrose pseudoinverse

.

1

Calculation

The best way to compute

A

+

is to use

singular value decomposition

. With

A

=

U

⁢

S

⁢

V

T

, where

U

and

V

(both

n

×

n

)

orthogonal

and

S

(

m

×

n

) is

diagonal

with real, non-negative

singular values

σ

i

,

i

=

1

,

…

,

n

. We find

A

+

=

V

⁢

(

S

T

⁢

S

)

-

1

⁢

S

T

⁢

U

T

If the rank

r

of

A

is smaller than

n

, the inverse of

S

T

⁢

S

does not exist, and one uses only the first

r

singular values;

S

then becomes an

r

×

r

matrix and

U

,

V

shrink accordingly. see also

Linear Equations

.

2

Generalization

The term “pseudoinverse” is actually used for any operator

pinv

satisfying

M

⁢

pinv

⁡

(

M

)

⁢

M

=

M

for a

m

×

n

matrix

M

. Beyond this, pseudoinverses can be
defined on any reasonable matrix identity.

References

•

Originally from The Data Analysis Briefbook
(

http://rkb.home.cern.ch/rkb/titleA.html

http://rkb.home.cern.ch/rkb/titleA.html)

Title

pseudoinverse

Canonical

name

Pseudoinverse

Date of creation

2013-03-22 12:07:21

Last modified on

2013-03-22 12:07:21

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

6

Author

akrowne (2)

Entry type

Definition

Classification

msc 15-00

Classification

msc 65-00

Synonym

pseudo-inverse

Synonym

Moore-Penrose pseudoinverse

Related topic

MoorePenroseGeneralizedInverse

Generated on Fri Feb 9 18:15:57 2018 by

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rotation matrix

rotation matrix

Definition 1

.

A

rotation matrix

is a
(real)

orthogonal matrix

whose

determinant

is

+

1

.
All

n

×

n

rotation matrices form a group called
the

special orthogonal group

and it is denoted by

SO

⁡

(

n

)

.

Examples

1.

The

identity matrix

in

ℝ

n

is a rotation matrix.

2.

The most general rotation matrix in

ℝ

2

can be written as

(

cos

⁡

θ

-

sin

⁡

θ

sin

⁡

θ

cos

⁡

θ

)

,

where

θ

∈

ℝ

.

Multiplication

(from the left) with this matrix
rotates a vector (in

ℝ

2

)

θ

radians in the anti-clockwise

direction

.

Properties

1.

Suppose

v

∈

ℝ

n

is a

unit vector

.
Then there exists a rotation matrix

R

such that

R

⋅

v

=

(

1

,

0

,

…

,

0

)

.

2.

In fact, for

v

∈

ℝ

n

,

n

≥

3

, there are many rotation matrices

𝐑

∈

SO

⁡

(

n

)

such that

R

⋅

v

=

(

1

,

0

,

…

,

0

)

T

.
To see this, let

f

be the mapping

f

:

SO

⁡

(

n

-

1

)

→

SO

⁡

(

n

)

,
defined as

f

⁢

(

Q

)

=

(

1

0

1

×

n

-

1

0

n

-

1

×

1

Q

n

-

1

×

n

-

1

)

.

Then for each

Q

∈

SO

⁡

(

n

-

1

)

,

f

⁢

(

Q

)

maps

(

1

,

0

,

…

,

0

)

T

onto itself. Thus, if

R

0

∈

SO

⁡

(

n

)

satisfies

R

⋅

v

=

(

1

,

0

,

…

,

0

)

T

,
then

f

⁢

(

Q

)

⋅

R

satisfies the same

property

for all

Q

∈

SO

⁡

(

n

-

1

)

.

Title

rotation matrix

Canonical

name

RotationMatrix

Date of creation

2013-03-22 15:03:57

Last modified on

2013-03-22 15:03:57

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

17

Author

matte (1858)

Entry type

Definition

Classification

msc 15-00

Synonym

rotational matrix

Related topic

OrthogonalMatrices

Related topic

ExampleOfRotationMatrix

Related topic

DecompositionOfOrthogonalOperatorsAsRotationsAndReflections

Related topic

DerivationOfRotationMatrixUsingPolarCoordinates

Related topic

DerivationOf2DReflectionMatrix

Related topic

TransitionToSkewAngledCoordinates

Generated on Fri Feb 9 18:15:59 2018 by

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Schur decomposition

Schur decomposition

If

A

is a complex

square matrix

of order n

(i.e.

A

∈

Mat

n

⁢

(

C

)

)

, then there exists a

unitary matrix

Q

∈

Mat

n

⁢

(

ℂ

)

such that

Q

H

⁢

A

⁢

Q

=

T

=

D

+

N

where

H

is the

conjugate transpose

,

D

=

diag

⁡

(

λ

1

,

…

,

λ

n

)

(the

λ

i

are

eigenvalues

of

A

)

, and

N

∈

Mat

n

⁢

(

ℂ

)

is

strictly upper triangular matrix

. Furthermore,

Q

can be chosen such that the eigenvalues

λ

i

appear in any order along the

diagonal

.

[

GVL

]

References

GVL

Golub, H. Gene, Van Loan F. Charles: Matrix Computations

(Third Edition)

. The Johns Hopkins University Press, London, 1996.

Title

Schur decomposition

Canonical name

SchurDecomposition

Date of creation

2013-03-22 13:42:12

Last modified on

2013-03-22 13:42:12

Owner

Daume (40)

Last modified by

Daume (40)

Numerical id

8

Author

Daume (40)

Entry type

Theorem

Classification

msc 15-00

Related topic

AnExampleForSchurDecomposition

Related topic

ProofThatDetEAEoperatornametrA

Generated on Fri Feb 9 18:16:00 2018 by

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Schur decomposition, proof of

Schur decomposition, proof of

The columns of the

unitary matrix

Q

in Schur’s

decomposition

theorem

form an

orthonormal basis

of

ℂ

n

. The matrix

A

takes the upper-triangular form

D

+

N

on this basis.

Conversely

, if

v

1

,

…

,

v

n

is an orthonormal basis for which

A

is of this form then the matrix

Q

with

v

i

as its

i

-th column

satisfies

the theorem.

To find such a basis we proceed by

induction

on

n

. For

n

=

1

we can simply take

Q

=

1

. If

n

>

1

then let

v

∈

ℂ

n

be an

eigenvector

of

A

of unit length and let

V

=

v

⟂

be its

orthogonal complement

. If

π

denotes the

orthogonal projection

onto the line spanned by

v

then

(

1

-

π

)

⁢

A

maps

V

into

V

.

By induction there is an orthonormal basis

v

2

,

…

,

v

n

of

V

for which

(

1

-

π

)

⁢

A

takes the desired form on

V

. Now

A

=

π

⁢

A

+

(

1

-

π

)

⁢

A

so

A

⁢

v

i

≡

(

1

-

π

)

⁢

A

⁢

v

i

(

mod

v

)

for

i

∈

{

2

,

…

,

n

}

. Then

v

,

v

2

,

…

,

v

n

can be used as a basis for the

Schur decomposition

on

ℂ

n

.

Title

Schur decomposition, proof of

Canonical name

SchurDecompositionProofOf

Date of creation

2013-03-22 14:04:01

Last modified on

2013-03-22 14:04:01

Owner

mps (409)

Last modified by

mps (409)

Numerical id

6

Author

mps (409)

Entry type

Proof

Classification

msc 15-00

Generated on Fri Feb 9 18:16:02 2018 by

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singular value decomposition

singular value decomposition

Any real

m

×

n

matrix

A

can be decomposed into

A

=

U

⁢

S

⁢

V

T

where

U

is an

m

×

m

orthogonal matrix

,

V

is an

n

×

n

orthogonal matrix, and

S

is a unique

m

×

n

diagonal matrix

with real, non-negative elements

σ

i

,

i

=

1

,

…

,

min

⁡

(

m

,

n

)

, in

descending order

:

σ

1

≥

σ

2

≥

…

≥

σ

min

⁡

(

m

,

n

)

≥

0

The

σ

i

are the

singular values

of

A

and the first

min

⁡

(

m

,

n

)

columns of

U

and

V

are the left and right (respectively)

singular vectors

of

A

.

S

has the form:

[

Σ

0

]

⁢

if

⁡

m

≥

n

⁢

and

⁡

[

Σ

0

]

⁢

if

⁡

m

<

n

,

where

Σ

is a diagonal matrix with the

diagonal

elements

σ

1

,

σ

2

,

…

,

σ

min

⁡

(

m

,

n

)

. We assume now

m

≥

n

. If

r

=

rank

⁡

(

A

)

<

n

, then

σ

1

≥

σ

2

≥

⋯

≥

σ

r

>

σ

r

+

1

=

⋯

=

σ

n

=

0

.

If

σ

r

≠

0

and

σ

r

+

1

=

⋯

=

σ

n

=

0

, then

r

is the rank of

A

. In this case,

S

becomes an

r

×

r

matrix, and

U

and

V

shrink accordingly. SVD can thus be used for rank determination.

The SVD provides a numerically robust solution to the

least-squares problem

. The matrix-algebraic phrasing of the least-squares solution

x

is

x

=

(

A

T

⁢

A

)

-

1

⁢

A

T

⁢

b

Then utilizing the SVD by making the replacement

A

=

U

⁢

S

⁢

V

T

we have

x

=

V

⁢

[

Σ

-

1

0

]

⁢

U

T

⁢

b

.

References

•

Originally from The Data Analysis Briefbook (

http://rkb.home.cern.ch/rkb/titleA.html

http://rkb.home.cern.ch/rkb/titleA.html)

Title

singular value decomposition

Canonical name

SingularValueDecomposition

Date of creation

2013-03-22 12:05:17

Last modified on

2013-03-22 12:05:17

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

9

Author

akrowne (2)

Entry type

Definition

Classification

msc 15-00

Classification

msc 65-00

Synonym

SVD

Synonym

singular value

Synonym

singular vector

Related topic

Eigenvector

Related topic

Eigenvalue

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