aarjaneiro/planetmath huggingface dataset:
proof of the determinant condition for a sequence of vectors

proof of the determinant condition for a sequence of vectors

Theorem

.

Let

x

1

,

x

2

,

…

be a sequence of

d

dimensional vectors. Assume that there is

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

(1)

for every

n

∈

N

. Then

det

⁡

[

x

n

1

,

x

n

2

,

…

,

x

n

d

]

=

0

for all

(

n

1

,

…

,

n

d

)

∈

N

d

.

Proof.

Introduce a linear order over the set of ordered tuples:

(

n

1

,

n

2

,

…

,

n

d

)

≺

(

n

^

1

,

n

^

2

,

…

,

n

^

d

)

if

(

∑

i

=

1

d

n

i

,

n

^

d

,

n

^

d

-

1

,

…

,

n

^

1

)

precedes

(

∑

i

=

1

d

n

^

i

,

n

d

,

n

d

-

1

,

…

,

n

1

)

lexicographically. Let

(

n

1

,

n

2

,

…

,

n

d

)

be the minimal (according to the above order) ordered tuple for which

det

⁡

[

x

n

1

,

x

n

2

,

…

,

x

n

d

]

≠

0

.

(2)

Take another ordered tuple,

(

n

^

1

,

n

^

2

,

…

,

n

^

d

)

, such that

∑

i

=

1

d

n

i

=

∑

i

=

1

d

n

^

i

. By minimality, if

(

n

d

,

n

d

-

1

,

…

,

n

1

)

precedes

(

n

^

d

,

n

^

d

-

1

,

…

,

n

^

1

)

lexicographically then

det

⁡

[

x

n

^

1

,

x

n

^

2

,

…

,

x

n

^

d

]

=

0

. Otherwise, let

i

∈

{

0

,

1

,

…

,

d

-

1

}

be the first index such that

n

d

-

i

≠

n

^

d

-

i

(more specifically,

n

d

-

i

>

n

^

d

-

i

). Then,

n

^

d

-

j

=

n

d

-

j

for

j

=

0

,

…

,

i

-

1

and

n

^

d

-

j

<

n

d

-

i

for

j

=

i

,

…

,

d

-

1

. Therefore,

det

⁡

[

x

n

1

,

…

,

x

n

d

-

i

-

1

,

x

n

^

m

,

x

n

d

-

i

+

1

,

…

,

x

n

d

]

=

0

for all

m

=

1

,

2

,

…

,

d

(some because of repeated columns and the others
because

∑

j

=

1

d

n

j

-

n

d

-

i

+

n

^

m

<

∑

j

=

1

d

n

j

). Since the vectors

x

n

1

,

x

n

2

,

…

,

x

n

d

are

linearly independent

, we get
that

{

x

n

^

1

,

x

n

^

2

,

…

,

x

n

^

d

}

⊂

span

⁡

(

{

x

n

1

,

x

n

2

,

…

,

x

n

d

}

∖

{

x

n

d

-

i

}

)

.

In particular

det

⁡

[

x

n

^

1

,

x

n

^

2

,

…

,

x

n

^

d

]

=

0

. Therefore, (

1

) reduces to

det

⁡

[

x

n

1

,

x

n

2

,

…

,

x

n

d

]

=

0

which contradicts (

2

).

∎

Title

proof of the

determinant

condition for a sequence of vectors

Canonical name

ProofOfTheDeterminantConditionForASequenceOfVectors

Date of creation

2013-03-22 14:33:46

Last modified on

2013-03-22 14:33:46

Owner

GeraW (6138)

Last modified by

GeraW (6138)

Numerical id

5

Author

GeraW (6138)

Entry type

Proof

Classification

msc 15A15

Generated on Fri Feb 9 18:24:58 2018 by

LaTeXML
rule of Sarrus

rule of Sarrus

For calculating the value of a

determinant

D

=

|

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

|

with three rows, it is comfortable to use the

rule of Sarrus

(invented 1833 by the French mathematician P. F. Sarrus).

The rule comprises that first one writes the two first columns of the determinant on the of the determinant (seeing thus a

3

×

5

matrix!):

|

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

|

⁢

a

11

a

12

a

21

a

22

a

31

a

32

Here one sums the products on all lines

parallel

to the main diagonal of

D

and subtracts the products on the lines parallel to the second

diagonal

of

D

. Accordingly, one obtains the

expression

a

11

⁢

a

22

⁢

a

33

+

a

12

⁢

a

23

⁢

a

31

+

a

13

⁢

a

21

⁢

a

32

-

a

13

⁢

a

22

⁢

a

31

-

a

11

⁢

a

23

⁢

a

32

-

a

12

⁢

a

21

⁢

a

33

,

which gives the value of the determinant

D

.

There is no corresponding rule for determinants with more or less rows.

Title

rule of Sarrus

Canonical

name

RuleOfSarrus

Date of creation

2013-03-22 17:32:49

Last modified on

2013-03-22 17:32:49

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Result

Classification

msc 15A15

Synonym

Sarrus rule

Synonym

Sarrus’ rule

Related topic

LaplaceExpansion

Generated on Fri Feb 9 18:25:00 2018 by

LaTeXML
Schur complement

Schur complement

Let A,B,C,D be matrices of sizes

p

×

p

,

p

×

q

,

q

×

p

and

q

×

q

respectively and suppose that

D

is

invertible

. Let

M

=

(

A

B

C

D

)

so that

M

is a

(

p

+

q

)

×

(

p

+

q

)

matrix.

Then the

Schur complement

of the block

D

of the matrix

M

is the

p

×

p

matrix,

A

-

B

⁢

D

-

1

⁢

C

. Analogously if

A

is invertible then the

Schur complement

of the block

A

of the matrix

M

is the

q

×

q

matrix,

D

-

C

⁢

A

-

1

⁢

B

.
In the first case, when

D

is invertible, the Schur complement arises as the result of performing a partial

Gaussian elimination

by multiplying the matrix

M

from the right with the

lower triangular

block matrix

,

T

=

(

I

O

-

D

-

1

⁢

C

D

-

1

)

where

I

is the

p

×

p

identity matrix

and

O

is the

p

×

q

zero matrix

. Analogously, in the second case, we take the Schur complement by multiplying the matrix

M

from the left with the lower triangular block matrix

T

=

(

A

-

1

O

-

C

⁢

A

-

1

I

)

see also:

•

Wikipedia,

http://en.wikipedia.org/wiki/Schur_complement

Schur complement

Title

Schur complement

Canonical name

SchurComplement

Date of creation

2013-03-22 15:27:11

Last modified on

2013-03-22 15:27:11

Owner

georgiosl (7242)

Last modified by

georgiosl (7242)

Numerical id

8

Author

georgiosl (7242)

Entry type

Definition

Classification

msc 15A15

Related topic

BlockDeterminants

Related topic

MatrixInversionLemma

Generated on Fri Feb 9 18:25:02 2018 by

LaTeXML
trace

trace

The

trace

Tr

⁡

(

A

)

of a

square matrix

A

is defined to be the sum of the

diagonal

entries of

A

. It satisfies the following formulas:

•

Tr

⁡

(

A

+

B

)

=

Tr

⁡

(

A

)

+

Tr

⁡

(

B

)

•

Tr

⁡

(

A

⁢

B

)

=

Tr

⁡

(

B

⁢

A

)

()

where

A

and

B

are square matrices of the same size.

The

trace

Tr

⁡

(

T

)

of a

linear transformation

T

:

V

⟶

V

from any finite dimensional

vector space

V

to itself is defined to be the trace of any

matrix representation

of

T

with respect to a basis of

V

. This scalar is

independent

of the choice of basis of

V

, and in fact is equal to the sum of the

eigenvalues

of

T

(over a

splitting field

of the

characteristic polynomial

), including

multiplicities

.

The following link presents some examples for calculating the trace of a matrix.

A

trace

on a

C

*

-

algebra

A

is a

positive linear functional

ϕ

:

A

→

ℂ

that has the .

Title

trace

Canonical

name

Trace

Date of creation

2013-03-22 12:17:57

Last modified on

2013-03-22 12:17:57

Owner

mhale (572)

Last modified by

mhale (572)

Numerical id

10

Author

mhale (572)

Entry type

Definition

Classification

msc 15A15

Classification

msc 15A04

Related topic

FrobeniusMatrixNorm

Generated on Fri Feb 9 18:25:03 2018 by

LaTeXML
Van der Waerden’s permanent conjecture

Van der Waerden’s permanent conjecture

Let

A

be any

doubly stochastic

n

×

n

matrix (i.e. nonnegative real entries, each row sums to 1, each column too, hence square).

Let

A

∘

be the one where all entries are equal (i.e. they are

1

n

). Its

permanent

works out to

per

A

∘

=

n

!

⁢

(

1

n

)

n

and Van der Waerden conjectured in 1926 that this is the smallest value for the permanent of any doubly stochastic

A

, and is attained only for

A

=

A

∘

:

per

A

>

n

!

⁢

(

1

n

)

n

(for

A

≠

A

∘

).

It was finally proven independently by Egorychev and by Falikman, in 1979/80.

References

1

Hal86

Marshall J. Hall, Jr.

,

Combinatorial Theory

(2nd ed.),

Wiley 1986, repr. 1998,

ISBN  0 471 09138 3

and

0 471 31518 4

has a proof of the permanent conjecture.

Title

Van der Waerden’s permanent conjecture

Canonical name

VanDerWaerdensPermanentConjecture

Date of creation

2013-03-22 15:10:51

Last modified on

2013-03-22 15:10:51

Owner

marijke (8873)

Last modified by

marijke (8873)

Numerical id

5

Author

marijke (8873)

Entry type

Theorem

Classification

msc 15A15

Classification

msc 15A51

Synonym

permanent conjecture

Generated on Fri Feb 9 18:25:05 2018 by

LaTeXML
algebraic and geometric multiplicity do not coincide

algebraic and geometric multiplicity do not coincide

Zero is an

eigenvalue

of

A

=

(

0

1

0

0

)

with

algebraic multiplicity

2

and geometric multiplicity

1

.

Indeed, as

det

⁡

(

A

-

λ

⁢

I

)

=

λ

2

it follows that

0

is an eigenvalue of

A

with algebraic multiplicity

2

.
To find the geometric multiplicity of

A

we need to calculate

ker

⁡

A

. Thus, suppose

(

0

1

0

0

)

⁢

(

a

b

)

=

(

0

0

)

.

This implies

b

=

0

, so

ker

⁡

A

=

span

⁡

(

1

0

)

,

and the geometric multiplicity of

0

is

1

.

Title

algebraic and geometric multiplicity do not coincide

Canonical name

AlgebraicAndGeometricMultiplicityDoNotCoincide

Date of creation

2013-03-22 15:15:18

Last modified on

2013-03-22 15:15:18

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

5

Author

matte (1858)

Entry type

Example

Classification

msc 15A18

Generated on Fri Feb 9 18:25:07 2018 by

LaTeXML
characteristic polynomial

characteristic polynomial

Characteristic Polynomial of a Matrix

Let

A

be a

n

×

n

matrix over some field

k

. The

characteristic polynomial

p

A

⁢

(

x

)

of

A

in an indeterminate

x

is defined by the

determinant

:

p

A

⁢

(

x

)

:=

det

⁡

(

A

-

x

⁢

I

)

=

|

a

11

-

x

a

12

⋯

a

1

⁢

n

a

21

a

22

-

x

⋯

a

2

⁢

n

⋮

⋮

⋱

⋮

a

n

⁢

1

a

n

⁢

2

⋯

a

n

⁢

n

-

x

|

Remarks

•

The

polynomial

p

A

⁢

(

x

)

is an

n

th-degree polynomial over

k

.

•

If

A

and

B

are

similar matrices

, then

p

A

⁢

(

x

)

=

p

B

⁢

(

x

)

, because

p

A

⁢

(

x

)

=

det

⁡

(

A

-

x

⁢

I

)

=

det

⁡

(

P

-

1

⁢

B

⁢

P

-

x

⁢

I

)

=

det

⁡

(

P

-

1

⁢

B

⁢

P

-

P

-

1

⁢

x

⁢

I

⁢

P

)

=

det

⁡

(

P

-

1

)

⁢

det

⁡

(

B

-

x

⁢

I

)

⁢

det

⁡

(

P

)

=

det

⁡

(

P

)

-

1

⁢

det

⁡

(

B

-

x

⁢

I

)

⁢

det

⁡

(

P

)

=

det

⁡

(

B

-

x

⁢

I

)

=

p

B

⁢

(

x

)

for some

invertible matrix

P

.

•

The

characteristic equation

of

A

is the

equation

p

A

⁢

(

x

)

=

0

, and the solutions to which are the

eigenvalues

of

A

.

Characteristic Polynomial of a Linear Operator

Now, let

T

be a

linear operator

on a

vector space

V

of

dimension

n

<

∞

. Let

α

and

β

be any two

ordered bases

for

V

. Then we may form the matrices

[

T

]

α

and

[

T

]

β

. The two

matrix representations

of

T

are similar matrices, related by a

change of bases

matrix. Therefore, by the second remark above, we define the

characteristic polynomial

of

T

, denoted by

p

T

⁢

(

x

)

, in the indeterminate

x

, by

p

T

⁢

(

x

)

:=

p

[

T

]

α

⁢

(

x

)

.

The characteristic equation of

T

is defined accordingly.

Title

characteristic polynomial

Canonical name

CharacteristicPolynomial

Date of creation

2013-03-22 12:17:47

Last modified on

2013-03-22 12:17:47

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Definition

Classification

msc 15A18

Related topic

Equation

Defines

characteristic equation

Generated on Fri Feb 9 18:25:09 2018 by

LaTeXML
characteristic polynomial of algebraic number

characteristic polynomial of algebraic number

Let

ϑ

be an

algebraic number

of degree

n

,

f

⁢

(

x

)

its

minimal polynomial

and

ϑ

1

=

ϑ

,

ϑ

2

,

…

,

ϑ

n

its

algebraic conjugates

.

Let

α

be an element of the

number field

ℚ

⁢

(

ϑ

)

and

r

⁢

(

x

)

:=

c

0

+

c

1

⁢

x

+

…

+

c

n

-

1

⁢

x

n

-

1

the

canonical polynomial

of

α

with respect to

ϑ

.  We consider the numbers

r

⁢

(

ϑ

1

)

=

α

:=

α

(

1

)

,

r

⁢

(

ϑ

2

)

:=

α

(

2

)

,

…

,

r

⁢

(

ϑ

n

)

:=

α

(

n

)

(1)

and form the equation

g

⁢

(

x

)

:=

∏

i

=

1

n

[

x

-

r

⁢

(

ϑ

i

)

]

=

(

x

-

α

(

1

)

)

⁢

(

x

-

α

(

2

)

)

⁢

⋯

⁢

(

x

-

α

(

n

)

)

=

x

n

+

g

1

⁢

x

n

-

1

+

…

+

g

n

=

0

,

the roots of which are the numbers (1) and only these.  The

coefficients

g

i

of the

polynomial

g

⁢

(

x

)

are

symmetric polynomials

in the numbers

ϑ

1

,

ϑ

2

,

…

,

ϑ

n

and also symmetric polynomials in the numbers

α

(

i

)

.  The

fundamental theorem of symmetric polynomials

implies now that the symmetric polynomials

g

i

in the roots

ϑ

i

of the equation

f

⁢

(

x

)

=

0

belong to the ring determined by the coefficients of the equation and of the canonical polynomial

r

⁢

(

x

)

; thus the numbers

g

i

are

rational

(whence the degree of

α

is at most equal to

n

).

It is not hard to show (see the entry

degree of algebraic number

) of that the degree

k

of

α

divides

n

and that the numbers (1) consist of

α

and its algebraic conjugates

α

2

,

…

,

α

k

, each of which appears in (1) exactly

n

k

=

m

times.  In fact,

g

⁢

(

x

)

=

[

a

⁢

(

x

)

]

m

where

a

⁢

(

x

)

is the minimal polynomial of

α

(consequently, the coefficients

g

i

are

integers

if

α

is an

algebraic integer

).

The polynomial

g

⁢

(

x

)

is the

characteristic polynomial

(in German

Hauptpolynom

) of the element

α

of the

algebraic number field

ℚ

⁢

(

ϑ

)

and the equation

g

⁢

(

x

)

=

0

the

characteristic equation

(

Hauptgleichung

) of

α

.  See the independence of characteristic polynomial on

primitive element

.

So, the roots of the characteristic equation of

α

are

α

(

1

)

,

α

(

2

)

,

…

,

α

(

n

)

.  They are called the

ℚ

⁢

(

ϑ

)

-

conjugates

of

α

; they all are algebraic conjugates of

α

.

Title

characteristic polynomial of algebraic number

Canonical name

CharacteristicPolynomialOfAlgebraicNumber

Date of creation

2013-03-22 19:08:41

Last modified on

2013-03-22 19:08:41

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Definition

Classification

msc 15A18

Classification

msc 12F05

Classification

msc 11R04

Related topic

RationalIntegersInIdeals

Related topic

DegreeOfAlgebraicNumber

Defines

characteristic polynomial

Defines

characteristic equation

Defines

ℚ

⁢

(

ϑ

)

-conjugates

Defines

K

-conjugates

Generated on Fri Feb 9 18:25:11 2018 by

LaTeXML
eigenspace

eigenspace

Let

V

be a

vector space

over a field

k

. Fix a

linear transformation

T

on

V

. Suppose

λ

is an

eigenvalue

of

T

. The set

{

v

∈

V

∣

T

⁢

v

=

λ

⁢

v

}

is called the

eigenspace

(of

T

) corresponding to

λ

. Let us write this set

W

λ

.

Below are some basic

properties

of eigenspaces.

1.

W

λ

can be viewed as the kernel of the linear transformation

T

-

λ

⁢

I

. As a result,

W

λ

is a

subspace

of

V

.

2.

The

dimension

of

W

λ

is called the

geometric multiplicity

of

λ

. Let us denote this by

g

λ

. It is easy to see that

1

≤

g

λ

, since the existence of an eigenvalue means the existence of a non-zero

eigenvector

corresponding to the eigenvalue.

3.

W

λ

is an

invariant subspace

under

T

(

T

-

invariant

).

4.

W

λ

1

∩

W

λ

2

=

0

iff

λ

1

≠

λ

2

.

5.

In fact, if

W

λ

′

is the sum of eigenspaces corresponding to eigenvalues of

T

other than

λ

, then

W

λ

∩

W

λ

′

=

0

.

From now on, we assume

V

finite-dimensional.

Let

S

T

be the set of all eigenvalues of

T

and let

W

=

⊕

λ

∈

S

W

λ

. We have the following properties:

1.

If

m

λ

is the algebraic multiplicity of

λ

, then

g

λ

≤

m

λ

.

2.

Suppose the

characteristic polynomial

p

T

⁢

(

x

)

of

T

can be factored into linear terms, then

T

is

diagonalizable

iff

m

λ

=

g

λ

for every

λ

∈

S

T

.

3.

In other words, if

p

T

⁢

(

x

)

splits over

k

, then

T

is diagonalizable iff

V

=

W

.

For example, let

T

:

ℝ

2

→

ℝ

2

be given by

T

⁢

(

x

,

y

)

=

(

x

,

x

+

y

)

. Using the

standard basis

,

T

is represented by the matrix

M

T

=

(

1

1

0

1

)

.

From this matrix, it is easy to see that

p

T

⁢

(

x

)

=

(

x

-

1

)

2

is the characteristic polynomial of

T

and

1

is the only eigenvalue of

T

with

m

1

=

2

. Also, it is not hard to see that

T

⁢

(

x

,

y

)

=

(

x

,

y

)

only when

y

=

0

. So

W

1

is a one-dimensional subspace of

ℝ

2

generated by

(

1

,

0

)

. As a result,

T

is not diagonalizable.

Title

eigenspace

Canonical

name

Eigenspace

Date of creation

2013-03-22 17:23:07

Last modified on

2013-03-22 17:23:07

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Definition

Classification

msc 15A18

Generated on Fri Feb 9 18:25:29 2018 by

LaTeXML
eigenvalue

eigenvalue

Let

V

be a

vector space

over

k

and

T

a

linear operator

on

V

.
An

eigenvalue

for

T

is an scalar

λ

(that is, an element of

k

) such that

T

⁢

(

z

)

=

λ

⁢

z

for some nonzero vector

z

∈

V

.
Is that case, we also say that

z

is an

eigenvector

of

T

.

This can also be expressed as follows:

λ

is an eigenvalue for

T

if the kernel of

A

-

λ

⁢

I

is non trivial.

A linear operator can have several eigenvalues (but no more than the

dimension

of the space). Eigenvectors corresponding to different eigenvalues are

linearly independent

.

Title

eigenvalue

Canonical

name

Eigenvalue1

Date of creation

2013-03-22 14:01:53

Last modified on

2013-03-22 14:01:53

Owner

drini (3)

Last modified by

drini (3)

Numerical id

8

Author

drini (3)

Entry type

Definition

Classification

msc 15A18

Related topic

LinearTransformation

Related topic

Scalar

Related topic

Vector

Related topic

Kernel

Related topic

Dimension2

Defines

eigenvector

Generated on Fri Feb 9 18:25:33 2018 by

LaTeXML
example of bounded operator with no eigenvalues

example of bounded operator with no eigenvalues

In this entry we show that there are

operators

with no

eigenvalues

. Moreover, we exhibit an operator

T

in a

Hilbert space

which is

bounded

,

self-adjoint

, has a non-empty

spectrum

but no eigenvalues.

Consider the Hilbert space

L

2

⁢

(

[

0

,

1

]

)

(

http://planetmath.org/L2SpacesAreHilbertSpaces

) and let

f

:

[

0

,

1

]

⟶

ℂ

be the

function

f

⁢

(

t

)

=

t

.

Let

T

:

L

2

⁢

(

[

0

,

1

]

)

⟶

L

2

⁢

(

[

0

,

1

]

)

be the operator of

multiplication

(

http://planetmath.org/MultiplicationOperatorOnMathbbL22

) by

f

T

⁢

(

φ

)

=

f

⁢

φ

,

φ

∈

L

2

⁢

(

[

0

,

1

]

)

Thus,

T

is a

bounded operator

, since it is a

multiplication operator

(see this entry (

http://planetmath.org/OperatorNormOfMultiplicationOperatorOnL2

)). Also, it is easily seen that

T

is self-adjoint.

We now prove that

T

has no eigenvalues: suppose

λ

∈

ℂ

is an eigenvalue of

T

and

φ

is an

eigenvector

. Then,

T

⁢

φ

=

λ

⁢

φ

This means that

(

f

-

λ

)

⁢

φ

=

0

, but this is impossible for

φ

≠

0

since

f

-

λ

has at most one zero. Hence,

T

has no eigenvalues.

Of course, since the Hilbert space is complex, the spectrum of

T

is non-empty (see this entry (

http://planetmath.org/SpectrumIsANonEmptyCompactSet

)). Moreover, the spectrum of

T

can be easily computed and seen to be the whole

interval

[

0

,

1

]

, as we explain now:

It is known that an operator of multiplication by a

continuous function

g

is

invertible

if and only if

g

is invertible. Thus, for every

λ

∈

ℂ

,

T

-

λ

⁢

I

is easily seen to be the operator of multiplication by

(

f

-

λ

)

. Hence,

T

-

λ

⁢

I

is not invertible if and only if

λ

∈

[

0

,

1

]

, i.e.

σ

⁢

(

T

)

=

[

0

,

1

]

.

Title

example of bounded operator with no eigenvalues

Canonical

name

ExampleOfBoundedOperatorWithNoEigenvalues

Date of creation

2013-03-22 17:57:53

Last modified on

2013-03-22 17:57:53

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

7

Author

asteroid (17536)

Entry type

Example

Classification

msc 15A18

Classification

msc 47A10

Generated on Fri Feb 9 18:25:35 2018 by

LaTeXML
finding eigenvalues

finding eigenvalues

This example investigates

eigenvalues

and the

similarity transformation

used
to

diagonalize

matrices. We seek the eigenvalues of the matrix

A

below. Afterward,
we can transform this matrix into a

diagonal matrix

which has many useful applications.

A

=

(

2

1

1

2

)

Here, we need to solve the corresponding

matrix equation

;

(

2

1

1

2

)

⁢

(

x

1

x

2

)

=

λ

⁢

(

x

1

x

2

)

or

A

⁢

X

=

λ

⁢

X

rearranging gives

A

⁢

X

-

λ

⁢

X

=

0

or

(

A

-

λ

⁢

I

)

⁢

X

=

0

We seek the values for

λ

and

X

.
First, we need to solve the

characteristic equation

of

A

. We do this by finding

d

⁢

e

⁢

t

⁢

(

A

-

λ

⁢

I

)

. First, calculating

A

-

λ

⁢

I

gives;

A

-

λ

⁢

I

=

(

2

-

λ

1

1

2

-

λ

)

Next, calculating

d

⁢

e

⁢

t

⁢

(

A

-

λ

⁢

I

)

yields

d

⁢

e

⁢

t

⁢

(

A

-

λ

⁢

I

)

=

(

2

-

λ

)

2

-

1

=

λ

2

-

4

⁢

λ

+

3

=

(

λ

-

1

)

⁢

(

λ

-

3

)

=

0

Substituting

λ

=

1

into

(

A

-

λ

⁢

I

)

⁢

X

gives…

{

x

1

+

x

2

=

0

x

1

+

x

2

=

0

so that

x

2

=

-

x

1

and the corresponding

eigenvector

is

(

t

-

t

)

=

t

⁢

(

1

-

1

)

where

t

≠

0

.

Substituting

λ

=

3

gives…

{

-

x

1

+

x

2

=

0

x

1

-

x

2

=

0

so that

x

2

=

x

1

and the corresponding eigenvector is

(

t

t

)

=

t

⁢

(

1

1

)

where

t

≠

0

.

Finally, to diagonalize

A

we let the eigenvectors be the columns of a new matrix

P

=

(

1

-

1

1

1

)

and then since our eigenvectors are

linearly independent

we can also find;

P

-

1

=

1

2

⁢

(

1

1

-

1

1

)

then we create a diagonal matrix as follows…

D

=

P

-

1

⁢

A

⁢

P

=

(

1

0

0

3

)

Computing powers of

A

is a very useful application of

D

. Solving for

A

lets us compute powers of

A

A

=

P

⁢

D

⁢

P

-

1

so that

A

n

=

P

⁢

D

n

⁢

P

-

1

or

A

n

=

P

⁢

(

1

n

0

0

3

n

)

⁢

P

-

1

Title

finding eigenvalues

Canonical name

FindingEigenvalues

Date of creation

2013-03-22 15:52:35

Last modified on

2013-03-22 15:52:35

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

6

Author

PrimeFan (13766)

Entry type

Example

Classification

msc 15A18

Generated on Fri Feb 9 18:25:37 2018 by

LaTeXML