generalized Bézout theorem on matrices

Generalized Bézout theorem 1.

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


Consider M[x] given by

M[x]=M0xm+M1xm-1++Mm,(M00). (1)

The polynomialMathworldPlanetmathPlanetmathPlanetmath can also be written as

M[x]=xmM0+xm-1M1++Mm. (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,




calling M(A) (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 matrixMathworldPlanetmath), we shall prove that the right (left) remainder R (R^) does not depend on x. In fact,

M[x]= M0xm+M1xm-1++Mm
= M0xm-1(xI-A)+(M0A+M1)xm-1+M2xm-2++Mm
= [M0xm-1+(M0A+M1)xm-2](xI-A)+(M0A2+M1A+M2)xm-2+M3xm-3++Mm
= [M0xm-1+(M0A+M1)xm-2+(M0A2+M1A+M2)xm-3](xI-A)
= [M0xm-1+(M0A+M1)xm-2+(M0A2+M1A+M2)xm-3+

whence we have found that


and analogously that


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 (M^(A)=0).

Title generalized Bézout theorem on matrices
Canonical name GeneralizedBezoutTheoremOnMatrices
Date of creation 2013-03-22 17:43:35
Last modified on 2013-03-22 17:43:35
Owner perucho (2192)
Last modified by perucho (2192)
Numerical id 8
Author perucho (2192)
Entry type Theorem
Classification msc 15-01