Lecture 12
LINEAR TRANSFORMATIONS II
Lecture 12
LINEAR TRANSFORMATIONS II
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
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
Transformation Matrix Relative to Different Bases
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
.
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.
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
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
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:
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.
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
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.
IN TURN, WHEN A MATRIX IS NOT SIMILAR TO A
DIAGONAL MATRIX IT IS SIMILAR TO A NEARLY DIAGONAL
MATRIX, THE JORDAN BLOCK MATRIX
Jordan Blocks
.
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:
Any matrix may be transformed to Jordan normal form
using a similarity transformation.
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.
Example
The characteristic polynomial
The Jordan Form
4
0
0
0
1
4
0
0
0
0
2
0
0
0
0
1
J
The Jordan decomposition is, in general,
a computationally challenging task.
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
.
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