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# 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 $M(A)=0$ ($\widehat{M}(A)=0$).

## Mathematics Subject Classification

15-01*no label found*

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