Page 43 - Matrices theory and applications

2. Square Matrices
26
Theorem 2.5.1 Let M ∈ M n (K).Let
n
be its characteristic polynomial. Then the matrix
n−1
n
M + a 1 M
+ ··· + a n I n
equals 0 n. P M (X)= X + a 1 X n−1 + ··· + a n
One also writes P M (M) = 0. Though this formula looks trivial (obviously,
det(MI n − M) = 0), it is not. Actually, it must be understood in the
following way. Let us consider the expression XI n − M as a matrix with
entries in K[X]. When one substitutes a matrix N for the indeterminate
X in XI n − M, one obtains a matrix of M n (A), where A is the subring
of M n (K) spanned by I n and N (one denotes it by K(N)). The ring A is
commutative (but is not an integral domain in general), since it is the set
of the q(N)for q ∈ K[X]. Therefore,
 
N − m 11 I n
.
 . 
 . −m ij I n 
P M (N)=   .
.
 . 
 . 
N − m nn I n
The Cayley–Hamilton theorem expresses that the determinant (which is
an element of M n (K), rather than of K) of this matrix is zero.
Proof
Let R ∈ M n (K(X)) be the matrix XI n − M,and let S be the adjoint of
R.Each s ij is a polynomial of degree less than or equal to n − 1, because
the products arising in the calculation of the cofactors involve n − 1 linear
or constant terms. Thus we may write
S = S 0 X n−1 + ··· + S n−1 ,
where S j ∈ M n (K). Let us now write RS =(det R)I n = P M (X)I n :
n
(XI n − M)(S 0 X n−1 + ··· + S n−1 )= (X + a 1 X n−1 + ··· + a n )I n .
Identifying the powers of X,weobtain
S 0 = I n ,
S 1 − MS 0 = a 1 I n ,
.
. .

S j − MS j−1 = a j I n ,
.
.
.
= a n−1 I n ,
S n−1 − MS n−2
= a n I n .
−MS n−1