Cayley-Hamilton/Matrix version/Fact/Proof

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

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

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

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

with matrices

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

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

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

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

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

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

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

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

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

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