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

Arthur Cayley (1821-1895)

William Hamilton (1805-1865)

${}P=3X^{2}-5X+2\,$

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 ist (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 what happens when we insert a matrix in its characteristic polynomial einsetzt.

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

Cayley-Hamilton/Matrix version/Fact/Proof

$\Box$

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.

$\Box$