Cayley-Hamilton/Matrix version/Fact/Proof

Proof

We consider the matrix ${}XE_{n}-M$ as a matrix whose entries are in the field ${}K(X)$. The adjugate matrix

$(XE_{n}-M)^{\operatorname {adj} }$

belongs also to ${}\operatorname {Mat} _{n}(K(X))$. The entries of the adjugate matrix are by definition the determinants of ${}(n-1)\times (n-1)$-submatrices of ${}XE_{n}-M$. In the entries of this matrix, the variable ${}X$ occurs at most in its first power, so that, in the entries of the adjugate matrix, the variable occurs at most in its ${}(n-1)$-th power. We write

$(XE_{n}-M)^{\operatorname {adj} }=X^{n-1}A_{n-1}+X^{n-2}A_{n-2}+\cdots +XA_{1}+A_{0}\,$

with matrices

${}A_{i}\in \operatorname {Mat} _{n}(K)\,,$

that is, we write the entries as polynomials, and we collect all coefficients referring to ${}X^{i}$ into a matrix. Because of fact, we have

${}{\begin{aligned}\chi _{M}E_{n}&=(XE_{n}-M)\circ (XE_{n}-M)^{\operatorname {adj} }\\&=(XE_{n}-M)\circ (X^{n-1}A_{n-1}+X^{n-2}A_{n-2}+\cdots +XA_{1}+A_{0})\\&=X^{n}A_{n-1}+X^{n-1}(A_{n-2}-M\circ A_{n-1})+X^{n-2}(A_{n-3}-M\circ A_{n-2})+\cdots +X^{1}(A_{0}-M\circ A_{1})-M\circ A_{0}.\end{aligned}}$

We can write the matrix on the left according to the powers of ${}X$ and we get

$\chi _{M}E_{n}=X^{n}E_{n}+X^{n-1}c_{n-1}E_{n}+X^{n-2}c_{n-2}E_{n}+\cdots +X^{1}c_{1}E_{n}+c_{0}E_{n}\,.$

Since these polynomials coincide, their coefficients coincide. That is, we have a system of equations

${\begin{matrix}E_{n}&=&A_{n-1}\\c_{n-1}E_{n}&=&A_{n-2}-M\circ A_{n-1}\\c_{n-2}E_{n}&=&A_{n-3}-M\circ A_{n-2}\\\vdots &\vdots &\vdots \\c_{1}E_{n}&=&A_{0}-M\circ A_{1}\\c_{0}E_{n}&=&-M\circ A_{0}\,.\end{matrix}}$

We multiply these equations from the left from top down with ${}M^{n},M^{n-1},M^{n-2},\ldots ,M^{1},E_{n}$, yielding the system of equations

${\begin{matrix}M^{n}&=&M^{n}\circ A_{n-1}\\c_{n-1}M^{n-1}&=&M^{n-1}\circ A_{n-2}-M^{n}\circ A_{n-1}\\c_{n-2}M^{n-2}&=&M^{n-2}\circ A_{n-3}-M^{n-1}\circ A_{n-2}\\\vdots &\vdots &\vdots \\c_{1}M^{1}&=&MA_{0}-M^{2}\circ A_{1}\\c_{0}E_{n}&=&-M\circ A_{0}\,.\end{matrix}}$

If we add the left-hand side of this system, then we just get ${}\chi _{M}\,(M)$. If we add the right-hand side, then we get ${}0$, because every partial summand ${}M^{i+1}\circ A_{i}$ occurs once positively and once negatively. Hence, we have ${}\chi _{M}\,(M)=0$.

To fact