L 12 Linear transformations II

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Lecture 12

LINEAR TRANSFORMATIONS II

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TRANSITION MARTIX

Let us consider a linear space V of order n and two different bases of this space:
B

1

= {v

1

,...,v

n

}, B

2

= {u

1

,...,u

n

}.

We write the basis vectors from B

2

as a linear combination of basis vectors B

1

:

u

1

= p

11

v

1

+ p

21

v

2

+...+ p

n1

v

n

;

.....
.....
.....

u

n

= p

1n

v

1

+ p

2n

v

2

+...+ p

nn

v

n

.

REVISION

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Definition
The transition matrix from basis B

1

to basis B

2

is matrix P = [ p

ij

]

in which the elements of the columns are the coordinates of vectors
taken from the basis B

2

relative to the basis B

1

nn

n

n

n

n

p

p

p

p

p

p

p

p

p

P

2

1

2

22

21

1

12

11

REVISION

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Transformation Matrix Relative to Different Bases

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DEPENDENCE OF TRANSITION MATRIX ON BASIS VECTORS

Let L: V V be a linear transformation of V of order n into oneself.

Recall that the matrix A of transformation L relative to basis B

1

=

{v

1

,...,v

n

}

is the matrix composed of the coordinates of vectors L(v

n

),..., L(v

n

)

relative to this basis set up in columns.

The essential question

is how the matrix of a

linear

transformation L

changes if instead of

considering the basis B

1

we consider another basis B

2

.

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Let A’ denote the matrix of transformation L relative to B

2

.

Theorem
The matrix of the linear transformation L relative to B

2

is


A’= P

-1

A P.

The Proof is a consequence of the fact that:
if P is a transition matrix from basis B

1

to basis B

2

, then P is invertible

and the inverse matrix P

-1

is the transition matrix from B

2

to B

1

,

and the fact that the transformation is linear.

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1

0

0

1

1

0

0

1

1

P

1

e

1

= 1u

1

;

e

2

= -1u

1

+ 1u

2

;

e

3

= -1u

2

+ 1u

3

.

It is easily checked that

P

1

= P

-1

.

Example

u

1

= 1 e

1

u

2

= 1 e

1

+ 1 e

2

u

3

= 1 e

1

+ 1 e

2

+ 1 e

3

We consider the following two bases:
B

1

= { e

1

, e

2

, e

3

} unit basis; old

B

2

= { u

1

, u

2

, u

3

} u

1

=(1,0,0), u

2

=(1,1,0), u

3

=(1,1,1)

new

.

1

0

0

1

1

0

1

1

1

P

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Let us consider the transformation f: R

3

R

3

, f(x)= A

f

x

with matrix A

f

relative to the unit basis of R

3

:

.

1

0

1

0

1

0

1

0

2

A

f

The transformation matrix relative to

B

2

is:

2

1

1

3

2

1

4

3

2

1

0

0

1

1

0

1

1

1

1

0

1

1

1

1

1

1

2

1

0

0

1

1

0

1

1

1

1

0

1

0

1

0

1

0

2

1

0

0

1

1

0

0

1

1

AP

P

1

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In the last lecture we found that if a linear transformation L has
n distinct eigenvalues then the eigenvectors constitute a basis
of this space. .

Let A be the matrix of the transformation L relative to an arbitrary basis
B

1

of V and

A’ be the matrix of this transformation relative the basis B

2

composed of

eigenvectors {v

1

,...,v

n

}.

,

'

A

n

0

0

2

1

where 

i

is an eigenvalue corressponding to eigenvector v

i

.

Then the matrix of L relative to the eigevector basis is:

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Definition
A square matrix with real elements can be

diagonalized (or it is similar to a diagonal matrix),

if there exists an invertible matrix P such that the
matrix
P

-1

AP is diagonal.

Theorem
The following conditions are equivalent:

matrix A is similar to a diagonal matrix;

the eigenvectors of A constitute a basis of
R

n

;

A = P D P

-1

, where D is a diagonal matrix

which has eigenvalues on its diagonal, while
the corresponding eigenvectors are the
columns of P.

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Example

Let us consider

:

.

C

A

1

0

1

1

1

4

1

1

The characteristic equation: det (A-I) = 0 is
(1-)

2

– 4 = 0 and (1- )

2

- 4 =( +1)( -3),

thus the eigenvalues 

1

= 3 and 

2

= -1 are distinct.

The corresponding eigenvectors are x

1

= (1,2) and x

2

= (1,-2)

Thus the matrix of the transformation f(x) = Ax relative to the
eigenvector basis has the following form:

.

/

/

/

/

2

2

1

1

1

4

1

1

4

1

2

1

4

1

2

1

1

0

0

3

1.

1

0

0

3

A

It is the diagonal matrix P

-1

A P

P

P

-1

A

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For matrix C the characteristic equation is: (1 - )

2

= 0, thus the

eigenvalue of C ( with multiplicity 2) is  =1.

From the equation Cx = x we conclude that x can be written
as (a, 0) , thus C has only one independent eigenvector, and
from (i) we know that C is not similar to a diagonal matrix.

2.

From the example it can be concluded that a matrix

is not similar to a diagonal matrix when an r–
multiple eigenvalue does not have r corresponding
independent eigenvectors.

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IN TURN, WHEN A MATRIX IS NOT SIMILAR TO A
DIAGONAL MATRIX IT IS SIMILAR TO A NEARLY DIAGONAL
MATRIX, THE JORDAN BLOCK MATRIX

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Jordan Blocks

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.

a

...

...

a

a

A

j

j

j

j

0

1

0

1

k

j

A

A

A

A

A

0

0

0

0

0

0

2

1

JORDAN BLOCK MATRIX

Definition
An upper triangular matrix A is a Jordan block matrix if:

,

The elements of the matrix A are matrices

The values a

1

, a

2

, ... ,a

k

in the Jordan block matrix are its

eigenvalues

.

The smallest possible Jordan submatrix is of order 1x1 consisting of one
eigenvalue.

3

0

0

0

1

0

0

1

1

B

.

Example:

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Any matrix may be transformed to Jordan normal form
using a similarity transformation.

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Example

1. The double eigenvalue

λ = 8

has only a single eigenvector, (1,0,0,0); as a result

λ = 8

appears only in a single block

J

1

.

2. The triple eigenvalue

λ = 0

has two eigenvectors v

1

, v

2

which correspond to

two Jordan blocks

J

2

, J

3

.

If A had 5 eigenvectors, all blocks would be 1 by 1 and

J

would be diagonal.

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Example

The characteristic polynomial

The Jordan Form

4

0

0

0

1

4

0

0

0

0

2

0

0

0

0

1

J

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The Jordan decomposition is, in general,

a computationally challenging task.

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Example
Find the Jordan block form for:

2

0

5

2

A

.

The eigenvalue :

1

=

2

= 2 is of multiplicity 2.

0

0

5

0

I

A

a

1

= 1 is the rank of

a

0

- a

1

= 1.

2

0

1

2

B

.

The Jordan block matrix is

Thus we have to calculate

a

0

= 2 - the rank of A,

There is one block

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.

A

7

7

6

8

7

4

4

3

1

Find the Jordan block matrix for

The eigenvalues:

1

=

2

= -1,

3

= 3.

3

0

0

0

1

0

0

1

1

B

a

0

= 3 - the rank of A .

.

I

A

8

7

6

8

6

4

4

3

2

a

1

= 2 - the rank of

=

The Jordan block matrix

Example


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