5Eigenvalues and eigenvectors

IA Vectors and Matrices 5.7 Cayley-Hamilton Theorem
Theorem
(Cayley-Hamilton theorem)
.
Every
n × n
complex matrix satisfies
its own characteristic equation.
Proof.
We will only prove for diagonalizable matrices here. So suppose for our
matrix
A
, there is some
P
such that
D
=
diag
(
λ
1
, λ
2
, ··· , λ
n
) =
P
1
AP
. Note
that
D
i
= (P
1
AP )(P
1
AP ) ···(P
1
AP ) = P
1
A
i
P.
Hence
p
D
(D) = p
D
(P
1
AP ) = P
1
[p
D
(A)]P.
Since similar matrices have the same characteristic polynomial. So
p
A
(D) = P
1
[p
A
(A)]P.
However, we also know that D
i
= diag(λ
i
1
, λ
i
2
, ···λ
i
n
). So
p
A
(D) = diag(p
A
(λ
1
), p
A
(λ
2
), ··· , p
A
(λ
n
)) = diag(0, 0, ··· , 0)
since the eigenvalues are roots of
p
A
(
λ
) = 0. So 0 =
p
A
(
D
) =
P
1
p
A
(
A
)
P
and
thus p
A
(A) = 0.
There are a few things to note.
(i)
If
A
1
exists, then
A
1
p
A
(
A
) =
A
1
(
c
0
+
c
1
A
+
c
2
A
2
+
···
+
c
n
A
n
) = 0.
So
c
0
A
1
+
c
1
+
c
2
A
+
···
+
c
n
A
n1
. Since
A
1
exists,
c
0
=
±det A 6
= 0.
So
A
1
=
1
c
0
(c
1
+ c
2
A + ··· + c
n
A
n1
).
So we can calculate A
1
from positive powers of A.
(ii) We can define matrix exponentiation by
e
A
= I + A +
1
2!
A
2
+ ··· +
1
n!
A
n
+ ··· .
It is a fact that this always converges.
If
A
is diagonalizable with
P
with
D
=
P
1
AP
=
diag
(
λ
1
, λ
2
, ··· , λ
n
),
then
P
1
e
A
P = P
1
IP + P
1
AP +
1
2!
P
1
A
2
P + ···
= I + D +
1
2!
D
2
+ ···
= diag(e
λ
1
, e
λ
2
, ···e
λ
n
)
So
e
A
= P [diag(e
λ
1
, e
λ
2
, ··· , e
λ
n
)]P
1
.
(iii)
For 2
×
2 matrices which are similar to
B
=
λ 1
0 λ
We see that the
characteristic polynomial
p
B
(
z
) =
det
(
B zI
) = (
λ z
)
2
. Then
p
B
(
B
) =
(λI B)
2
=
0 1
0 0
2
=
0 0
0 0
.
Since we have proved for the diagonalizable matrices above, we now know
that any 2 × 2 matrix satisfies Cayley-Hamilton theorem.
In IB Linear Algebra, we will prove the Cayley Hamilton theorem properly for
all matrices without assuming diagonalizability.