generalized Bézout theorem on matrices

Generalized Bézout theorem 1.

Let $M[x]$ be an arbitrary matrix polynomial of order $n$ and $A$ a square matrix of the same order. Then, when the matrix polynomial is divided on the right (left) by the characteristic polynomial $xI-A$ , the remainder is $M(A)$ ( $\\widehat{M}(A)$ ).

Proof.

Consider $M[x]$ given by

M[x]=M_{0}x^{m}+M_{1}x^{m-1}+\\cdots+M_{m},\\qquad(M_{0}\eq 0).

(1)

The polynomial can also be written as

M[x]=x^{m}M_{0}+x^{m-1}M_{1}+\\cdots+M_{m}.

(2)

We are now substituting the scalar argument (real or complex) $x$ by the matrix $A$ and therefore (1) and (2) will, in general, be distinct, as the powers of $A$ need not be permutable with the polynomial matrix coefficients. So that,

M(A)=M_{0}A^{m}+M_{1}A^{m-1}+\\cdots+M_{m}

and

\\widehat{M}(A)=A^{m}M_{0}+A^{m-1}M_{1}+\\cdots+M_{m},

calling $M(A)$ ( $\\widehat{M}(A)$ ) the right (left) value of $M[x]$ on substitution of $A$ for $x$ .
If we divide $M[x]$ by the binomial $xI-A$ ( $I$ is the correspondent identity matrix), we shall prove that the right (left) remainder $R$ ( $\\widehat{R}$ ) does not depend on $x$ . In fact,

	$\\displaystyle M[x]=$	$\\displaystyle M_{0}x^{m}+M_{1}x^{m-1}+\\cdots+M_{m}$
	$\\displaystyle=$	$\\displaystyle M_{0}x^{m-1}(xI-A)+(M_{0}A+M_{1})x^{m-1}+M_{2}x^{m-2}+\\cdots+M_{m}$
	$\\displaystyle=$	$\\displaystyle[M_{0}x^{m-1}+(M_{0}A+M_{1})x^{m-2}](xI-A)+(M_{0}A^{2}+M_{1}A+M_{%2})x^{m-2}+M_{3}x^{m-3}+\\cdots+M_{m}$
	$\\displaystyle=$	$\\displaystyle[M_{0}x^{m-1}+(M_{0}A+M_{1})x^{m-2}+(M_{0}A^{2}+M_{1}A+M_{2})x^{m%-3}](xI-A)$
		$\\displaystyle+(M_{0}A^{3}+M_{1}A^{2}+M_{2}A+M_{3})x^{m-3}+M_{4}x^{m-4}+\\cdots+%M_{m}$
	$\\displaystyle=$	$\\displaystyle[M_{0}x^{m-1}+(M_{0}A+M_{1})x^{m-2}+(M_{0}A^{2}+M_{1}A+M_{2})x^{m%-3}+\\cdots$
		$\\displaystyle+(M_{0}A^{m-1}+M_{1}A^{m-2}+\\cdots+M_{m-1})](xI-A)+M_{0}A^{m}+M_{%1}A^{m-1}+\\cdots+M_{m},$

whence we have found that

R=M_{0}A^{m}+M_{1}A^{m-1}+\\cdots+M_{m}\\equiv M(A),

and analogously that

\\widehat{R}=A^{m}M_{0}+A^{m-1}M_{1}+\\cdots+M_{m}\\equiv\\widehat{M}(A),

which proves the theorem.∎

From this theorem we have the following

Corollary 1.

A polynomial $M[x]$ is divisible by the characteristic polynomial $xI-A$ on the right (left) without remainderiff $M(A)=0$ ( $\\widehat{M}(A)=0$ ).