# Endomorphism/Cayley-Hamilton/Section

One highlight of the linear algebra is the Theorem of Cayley-Hamilton. In order to formulate this theorem, recall that we can plug in a square matrix into a polynomial. Here, the variable $X$ is everywhere replaced by the matrix $M$, the powers $M^{i}$ are the $i$-th matrix product of $M$ with itself, and the addition is the (componentwise) addition of matrices. A scalar $a$ has to be interpreted as the $a$-fold of the identity matrix. For the polynomial
    $P=3X^{2}-5X+2\,$
and the matrix
    $M={\begin{pmatrix}2&4\\3&1\end{pmatrix}}\,,$
we get
    ${\begin{aligned}P(M)&=3{\begin{pmatrix}2&4\\3&1\end{pmatrix}}^{2}-5{\begin{pmatrix}2&4\\3&1\end{pmatrix}}+2\\&={\begin{pmatrix}3&0\\0&3\end{pmatrix}}{\begin{pmatrix}16&12\\9&13\end{pmatrix}}+{\begin{pmatrix}-5&0\\0&-5\end{pmatrix}}{\begin{pmatrix}2&4\\3&1\end{pmatrix}}+{\begin{pmatrix}2&0\\0&2\end{pmatrix}}\\&={\begin{pmatrix}40&16\\12&36\end{pmatrix}}.\end{aligned}}$
For a fixed matrix $M\in \operatorname {Mat} _{n}(K)$, we have the substitution mapping
    $K[X]\longrightarrow \operatorname {Mat} _{n}(K),P\longmapsto P(M).$
This is (like the substitution mapping for an element $a\in K$), a ring homomorphism, that is, the relations (see also fact)
    $(P+Q)(M)=P(M)+Q(M),\,(P\cdot Q)(M)=P(M)\circ Q(M){\text{ and }}1(M)=E_{n}$
hold. The Theorem of Cayley-Hamilton answers the question of what happens when we insert a matrix in its characteristic polynomial.

## Theorem

Let $K$ be a field, and let $M$ be an $n\times n$-matrix. Let
    $\chi _{M}=X^{n}+c_{n-1}X^{n-1}+\cdots +c_{1}X+c_{0}\,$
denote the characteristic polynomial of $M$. Then
    $\chi _{M}\,(M)=M^{n}+c_{n-1}M^{n-1}+\cdots +c_{1}M+c_{0}=0\,.$

### 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 $\chi _{M}\,(M)=0$.
$\Box$

## Theorem

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
denote a linear mapping. Then the characteristic polynomial of $f$ fulfills the relation
    $\chi _{f}(f)=0\,.$

### Proof

This follows immediately from fact.
$\Box$