# 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

 $M[x]=M_{0}x^{m}+M_{1}x^{m-1}+\cdots+M_{m},\qquad(M_{0}\neq 0).$ (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 remainder iff $M(A)=0$ ($\widehat{M}(A)=0$).

Title generalized Bézout theorem on matrices GeneralizedBezoutTheoremOnMatrices 2013-03-22 17:43:35 2013-03-22 17:43:35 perucho (2192) perucho (2192) 8 perucho (2192) Theorem msc 15-01