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\,.

This means that the matrix annihilates the characteristic polynomial.

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