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Diagonalization
6
...
11 of eigenvectors of A
by collecting the basis vectors from the eigenspaces of each of the k eigenvalues
...
Suppose that A is an n x n matrix for which there is a basis {V1, •
...
Let the corresponding eigenvalues be denoted A1,
...
If
J, then we get
we let P =[v1
v1
1
AP=A[v1
vn
]
Avn
=[Av1
= [,,t1vi
J
AnVn
=[vi
vn
]
]
A1
0
0
A2
0
=PD
0
0
0
Recall that a square matrix D such that d;; = 0 for i t=
,
An
j
is said to be diagonal and
can be denoted by diag(d11,
d1111)
...
1 1, we can write AP = PD as
...
•
Definition
If there exists an invertible matrix P and diagonal matrix D such that p-l AP= D, then
Diagonalizable
we say that A is diagonalizable (some people prefer "diagonable")and that the matrix
P diagonalizes A to its diagonal form D
...
However, this isnot true in general since matrix multiplication is not
commutative
...
Theorem 1
If A and B are n x n matrices such that p-1AP = B for some invertible matrix P,
then A and B have
(1) The same determinant
(2) The same eigenvalues
(3) The same rank
11
(4) The same trace, where the trace of a matrix A is defined by tr A=
...
2
...
This theorem motivates the following definition
...
Thus, from our work above, if there is a basis of IR
...
On the other
hand, if at least one of the eigenvalues of A is deficient, then A will not haven linearly
independent eigenvectors
...
In this case, we say that A is not diagonalizable
...
Theorem 2
[Diagonalization Theorem]
An n x n matrix A can be diagonalized if and only if there exists a basis for JR11 of
eigenvectors of A
...
,A11 ), where
Ai
[v1
v11
)
is an eigenvalue
of A corresponding to v; for 1 ::S i ::S n
...
Corollary 3
A matrix A is diagonalizable if and only if every eigenvalue of a matrix A has its
geometric multiplicity equal to its algebraic multiplicity
...
Remark
Observe that it is possible for a matrix A with real entries to have non-real eigenvalues,
which will lead to non-real eigenvectors
...
In
Chapter 9, we will examine the case where complex eigenvalues and eigenvectors are
allowed
...
t
p- AP =
D, where
2
Solution: We need to find a basis for JR of eigenvectors of A
...
The characteristic poly nomial of
A is
Hence, the eigenvalues of A are
For
A1 = 5,
At = 5 and Az = -1
...
-1, we get
is an eigenvector for
{v1, v2}
A2= -1
and
{\12} is a basis for its eigenspace
...
we get
EXAMPLE 1
(continued)
EXAMPLE2
Note that wecould have instead taken P= [i12 vi] = r- � n which would have
given
2
0
3
Determine whether A = [-2 35 -2-01 is diagonalizable
...
Thus,
multipAliicsitdiy a1gonal
...
1
...
For /l1 = 2, we get A-/l1/ = [=-2� 3� -2=�]- [�0 -�0 2 �]·Thus,
a
basi
s
for
0
3
the eigenspace is { [ fl [ �]}
...
see that Ais diagonalizable
...
For /l2 = 1, we get A-/l2/ =[=�-2 !3 =�]- [�0 0� =�]·0 Therefore, {[�]} is a
basis for the eigenspace
...
D
Solution:
+
(-2
4)
+
+
+
1)
+
1
1
J1
-1
1
2
1
EXERCISE 1
Diagonalize A=U -� =il
7
EXAMPLE3
Is the matrix A
=
11
[=! =�]
-4
8
diagonalizable?
-3
Solution: The characteristic polynomial is
C(tl)
det(A
=
-(,,l - 3)(tl2 - 4,-l + 3)
- -tl
7
-4
0
11 - tl
- 3 + tl
1
11 - tl
-6
8
-3-tl
=
=
1-1)
=
1
...
Thus, a basis for
�
-4
1{[ [2]}·
-
-(tl- 3)(,-l - 3)(,-l
3 is an eigenvalue with algebraic multiplicity 2, and tl2
with algebraic multiplicity
For tl1
5
2
=
=
=
-
(-3 + tl)(-1)(6,,t + 6- 0) + (3 - tl)(tl2 - U + tl - 11 +
=
Thus, tl1
-
tll)
7
-1-tl
-4
-4
8
-6
0
0
0
Hence, the geometric multiplicity of
...
EXERCISE 2
EXAMPLE4
Show that A
=
[� �]
Show that matrix A
=
is not diagonalizable
...
2bxy
+
cy2
=
d
...
This problem will be discussed in Section 8
...
physical application related to these geometrical applications is the analysis of
the deformation of a solid
...
The change of shape in the block
can be described in terms of a 3 x 3 strain matrix
...
This application is discussed in
Section 8
...
Diagonalization is also an important tool for studying systems of linear difference
equations, which arise in many settings
...
For some situations, the change from month to month can be described
by saying that the vector jJ changes according to the rule
jJ(n + 1) =AjJ(n)
whereA is some known 2 x 2 matrix
...
We are often inter
ested in understanding what happens to the population "in the long run
...
This problem is easy to deal with if we can diagonal
izeA
...
3
...
This application is
discussed in Section 6
...
In Section 4
...
11 � JR
...
Examples 5 and 7 in
Section 4
...
Hence, our
diagonalization process is a method for finding such a geometrically natural basis
...
11 of eigen
vectors forms the geometrically natural basis
...
2
Practice Problems
Al By checking whether columns of P are eigenvec
tors ofA, determine whether P diagonalizesA
...
[ ]
[� -�l
11
(a)A= 9
(b)A=
6
4
'
[� � ]
[� n
P=
p=
-
(c) A=
(d) A=
'-8
[� �]
[� -7]
]
2
n
-� :J
[: 2
4 4
4
p=
4,
(f) A=
P=
A2 For the following matrices, determine the eigenval
ues and corresponding eigenvectors and determine
whether each matrix is diagonalizable over R If it
is diagonalizable, give a matrix P and diagonal ma
trix D such that p-1 AP = D
...
[-2-� 2: -=!2]
[2� �2 �1
-[-�1 -�2 �i
[�� �� -�:1
0
3
Homework Problems
Bl By checking whether columns of P are eigenvec-
tors of A, determine whether P diagonalizes A
...
(a) A=
(b) A=
(c) A=
(d) A
=
p=
1
-[ 8 -1 1 [-! -12 :
18 11 � -2l
-6
P=
is diagonalizable, give a matrix P and diagonal ma(a) A=
p=
4,
whether each matrix is diagonalizable over R If it
ues and corresponding eigenvectors and determine
-[ 2 ;]
trix D such that p-1 AP = D
...
(a) A=
(b) A=
(c) A=
[=� � � ]
[� � =�1
�[
; �1
[�� -� -��]
2
[ � �1
[ � -�1
[ ! -�1
A=
(f)
-
�)
A=
5
1
1
0
-1
0
0
-2
_
Conceptual Problems
Dl Prove that if A and B are similar, then A and B have
the same eigenvalues
...
D3 (a) Let A and B be
n x n matrices
...
(b) Use the result of part (a) to prove that if A and
B are similar, then tr A = tr B
...
eigenvalues 2 and 3 with corresponding eigenand
[H
(c) Determine a matrix that has eigenvalues 2, -2,
[
i]
...
12
(b) Use the result of part (a) to determine, by in
spection, the algebraic and geometric multi
[:
]
plicities of all of the eigenvalues of
a b
a
:
a b
a
:
a+b
·
D7 (a) Suppose that A is diagonalizable
...
tic polynomial is det A
...
)?)
(c) Without assuming that A is diagonalizable,
show that det A is equal to the product of the
roots of the characteristic equation of A (in
cluding any repeated roots and complex roots)
...
Show that Ak = PDkp-1,
A=
Theorem 1
...
and 3 , whh correspond;ng dgenvectors
A (including repeated eigenvalues) by using
A=
D4 (a) Suppose that P diagonalizes A and that the di
1
agonal form is D
...
Prove that
8
is the matrix from Problem
that equation
...
Prove that A is invertible
if and only if A does not have 0 as an eigenvalue
...
)
D9 Suppose that A is diagonalized by the matrix P and
that the eigenvalues of A are A
...
, A
...
• ,An -A1• (Hint: A-Ail is diagonalized by P
...