5Eigenvalues and eigenvectors

IA Vectors and Matrices 5.5 Diagonalizable matrices
Definition
(Diagonalizable matrices)
.
An
n × n
matrix
A
is diagonalizable if
it is similar to a diagonal matrix. We showed above that this is equivalent to
saying the eigenvectors form a basis of C
n
.
The requirement that matrix
A
has
n
distinct eigenvalues is a sufficient
condition for diagonalizability as shown above. However, it is not necessary.
Consider the second example in Section 5.2,
A =
2 2 3
2 1 6
1 2 0
We found three linear eigenvectors
˜
e
1
=
1
2
1
,
˜
e
2
=
2
1
0
,
˜
e
3
=
3
0
1
If we let
P =
1 2 3
2 1 0
1 0 1
, P
1
=
1
8
1 2 3
2 4 6
1 2 5
,
then
˜
A = P
1
AP =
5 0 0
0 3 0
0 0 3
,
so A is diagonalizable.
Theorem.
Let
λ
1
, λ
2
, ··· , λ
r
, with
r n
be the distinct eigenvalues of
A
. Let
B
1
, B
2
, ···B
r
be the bases of the eigenspaces
E
λ
1
, E
λ
2
, ··· , E
λ
r
correspondingly.
Then the set B =
r
[
i=1
B
i
is linearly independent.
This is similar to the proof we had for the case where the eigenvalues are
distinct. However, we are going to do it much concisely, and the actual meat of
the proof is actually just a single line.
Proof.
Write
B
1
=
{x
(1)
1
, x
(1)
2
, ···x
(1)
m(λ
1
)
}
. Then
m
(
λ
1
) =
dim
(
E
λ
1
), and simi-
larly for all B
i
.
Consider the following general linear combination of all elements in
B
. Con-
sider the equation
r
X
i=1
m(λ
i
)
X
j=1
α
ij
x
(i)
j
= 0.
The first sum is summing over all eigenspaces, and the second sum sums over
the basis vectors in B
i
. Now apply the matrix
Y
k=1,2,··· ,
¯
K,··· ,r
(A λ
k
I)
to the above sum, for some arbitrary K. We obtain
m(λ
K
)
X
j=1
α
Kj
Y
k=1,2,··· ,
¯
K,··· ,r
(λ
K
λ
k
)
x
(K)
j
= 0.
Since the
x
(K)
j
are linearly independent (
B
K
is a basis),
α
Kj
= 0 for all
j
. Since
K was arbitrary, all α
ij
must be zero. So B is linearly independent.
Proposition. A is diagonalizable iff all its eigenvalues have zero defect.