Example 10-29. The following is an example illustrating the Hamilton-Cayley
theorem. Let
A=
[
2 1
3 2
]
,
then the characteristic equation associated with the matrix Ais
C(λ) = det(A−λI ) =
∣∣
∣∣^2 −λ^1
3 2 −λ
∣∣
∣∣=λ^2 − 4 λ+ 1 = 0.
Replacing the scalar λby the matrix Aone obtains C(A) = A^2 − 4 A+I, where Iis the
2 × 2 identity matrix. The given matrix Awhen squared gives
A^2 =
[
2 1
3 2
][
2 1
3 2
]
=
[
7 4
12 7
]
.
Substituting I,Aand A^2 into C(A)gives
C(A) =
[
7 4
12 7
]
− 4
[
2 1
3 2
]
+
[
1 0
0 1
]
=
[
0 0
0 0
]
= [0]
and, hence, Asatisfies its own characteristic equation.
In order to prove the Hamilton-Cayley theorem, assume the n×nconstant square
matrix Ais given and it has associated with it the characteristic polynomial of the
form
C(λ) = |A−λI |=λn+α 1 λn−^1 +···+αn− 2 λ^2 +αn− 1 λ+αn
where α 1 , α 2 ,... , α n are appropriate scalar constants. Replace the scalar λ by the
matrix Aand replace the constant term αn by αnI, to obtain the Hamilton-Cayley
matrix equation
C(A) = An+α 1 An−^1 +··· +αn− 2 A^2 +αn− 1 A+αnI
To prove the Hamilton-Cayley theorem it must be demonstrated that C(A) = [0].
Toward this purpose replace the matrix Ain equation (10.23) by the matrix A−λI
to obtain
(A−λI )Adj(A−λI ) = |A−λI |I,
where the various elements of the matrix Adj(A−λI )are formed from A−λI by
deleting a certain row and column and then taking the determinant of the (n−1) ×
(n−1) system that remains. This implies λn−^1 is the highest power of λthat can be
