# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 24

    The Theorem of Cayley-Hamilton
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, see the 20th lecture. 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.
$\Box$
    Minimal polynomial and characteristic polynomial

## Corollary

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
be a linear mapping. Then the characteristic polynomial $\chi _{f}$ is a multiple of the minimal polynomial
$\mu _{f}$
$f$

### Proof

This follows directly from fact and fact.
$\Box$
In particular, the degree of the minimal polynomial of $\varphi \colon V\rightarrow V$ is bounded by the dimension of the vector space $V$. The minimal polynomial and the characteristic polynomial are related in several respects, for example, they have the same zeroes.

## Lemma

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
be a linear mapping. Let $v\in V$ be an eigenvector of $f$ with eigenvalue $\lambda$, and let $P\in K[X]$ denote a polynomial. Then
    $(P(f))(v)=P(\lambda )v\,.$
$v$
$P(f)$
$P(\lambda )$
$v\neq 0$
$P(f)$
$\lambda$
$P$

### Proof

We have
    $(f^{k})(v)=\lambda ^{k}v\,.$
This implies the statement, since the assignment $P\mapsto P(f)$ is compatible with addition and scalar multiplication.
$\Box$

## Lemma

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
be a linear mapping. Then the characteristic polynomial $\chi _{f}$ and the minimal polynomial
$\mu _{f}$

### Proof

It follows directly from Cayley-Hamilton that the zeroes of the minimal polynomial are also zeroes of the characteristic polynomial.
To prove the other implication, let $\lambda \in K$ be a zero of the characteristic polynomial, and let $v\in V$ denote an eigenvector of $f$ with eigenvalue $\lambda$, its existence is guaranteed by fact. We write the minimal polynomial as
    $\mu _{f}=(X-\lambda _{1})^{m_{1}}\cdots (X-\lambda _{k})^{m_{k}}Q\,,$
where $Q$ has no zero. Then
    ${\begin{aligned}0&=\mu _{f}(f)\\&={\left((X-\lambda _{1})^{m_{1}}\cdots (X-\lambda _{k})^{m_{k}}Q\right)}(f)\\&=(f-\lambda _{1}\operatorname {Id} _{V})^{m_{1}}\cdots (f-\lambda _{k}\operatorname {Id} _{V})^{m_{k}}Q(f).\end{aligned}}$
We apply this mapping to $v$. Because of fact, the factors send the vector $v$ to $(\lambda -\lambda _{i})^{m_{i}}v$ or to $Q(\lambda )v$, respectively. Altogether, $v$ is sent to
    $(\lambda -\lambda _{1})^{m_{1}}\cdots (\lambda -\lambda _{k})^{m_{k}}Q(\lambda )v.$
As the composed mapping is the zero mapping and $Q(\lambda )\neq 0$, we must have $\lambda _{i}=\lambda$ for some $i$.
$\Box$
    Further examples
For the moment, we will apply the following concept only for invertible matrices.

## Definition

Let $G$ be a group and $g\in G$ an element. Then we call the smallest positive number $n$ with $g^{n}=e_{G}$ the order of $g$. For this, we write $\operatorname {ord} \,(g)$. If all positive powers of $g$ are different from the neutral element, then we set
$\operatorname {ord} \,(g)=\infty$
We consider linear mappings
    $\varphi \colon V\longrightarrow V$
with the property that some power of it is the identity, say
    $\varphi ^{k}=\operatorname {Id} _{V}\,,$
that is, $\varphi$ has finite order. Typical examples are rotations around an angle of the form ${\frac {360}{k}}$ degree. The polynomial $X^{k}-1$ annihilates this endomorphism, and is, therefore, a multiple of the minimal polynomial.

## Definition

Let $K$ be a field, and $n\in \mathbb {N}$ A zero of the polynomial
    $X^{n}-1$
in $K$ is called an $n$-th root of unity
$K$

## Lemma

Let $n\in \mathbb {N} _{+}$. The zeroes of the polynomials $X^{n}-1$ over $\mathbb {C}$ are
    $e^{2\pi {\mathrm {i} }k/n}=\cos {\frac {2\pi k}{n}}+{\mathrm {i} }\sin {\frac {2\pi k}{n}},k=0,1,\ldots ,n-1.$
$\mathbb {C} [X]$
    $X^{n}-1=(X-1){\left(X-e^{2\pi {\mathrm {i} }/n}\right)}\cdots {\left(X-e^{2\pi {\mathrm {i} }(n-1)/n}\right)}\,.$

### Proof

The proof uses some basic facts about the complex exponential function. We have
    ${\left(e^{2\pi {\mathrm {i} }k/n}\right)}^{n}=e^{2\pi {\mathrm {i} }k}={\left(e^{2\pi {\mathrm {i} }}\right)}^{k}=1^{k}=1\,.$
Hence, the given complex numbers are indeed zeroes of the polynomial $X^{n}-1$. These zeroes are all different, because
    $e^{2\pi {\mathrm {i} }k/n}=e^{2\pi {\mathrm {i} }\ell /n}\,$
with $0\leq k\leq \ell \leq n-1$ implies, by considering the fraction, $e^{2\pi {\mathrm {i} }(\ell -k)/n}=1$ that
    $\ell -k=0\,$
holds. Therefore, there exist $n$ explicit zeroes and these are all the zeroes of the polynomial. The explicit description in coordinates follows from the Euler's formula.
$\Box$

## Definition

For a permutation $\pi$ on $\{1,\ldots ,n\}$, the $n\times n$-matrix
    $M_{\pi }={\left(a_{ij}\right)}\,,$
where
    $a_{\pi (j),j}=1\,$
and all other entries are $0$, is called a
We want to determine the characteristic polynomial of a permutation matrix. Here, we use that a permutation is a product of cycles. For a cycle of the form $1\mapsto 2\mapsto 3\mapsto \ldots \mapsto k\mapsto 1$, the corresponding permutation matrix is
    ${\begin{pmatrix}0&0&\ldots &0&1\\1&0&0&\ldots &0\\0&1&0&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&1&0\end{pmatrix}}.$
Every cycle can be brought (by renumbering) into this form.

## Lemma

The characteristic polynomial of a permutation matrix $M_{\rho }$ for a cycle $\rho \in S_{n}$ of
order $k$ is
    $\chi _{M}=(X-1)^{n-k}(X^{k}-1)\,.$

### Proof

We may assume that the cycle has the form $1\mapsto 2\mapsto 3\mapsto \ldots \mapsto k\mapsto 1$. The corresponding permutation matrix $M_{\rho }$ looks with respect to $e_{k+1},\ldots ,e_{n}$ like the identity matrix and has, with respect to the first $k$ standard vectors, the form
    ${\begin{pmatrix}0&0&\ldots &0&1\\1&0&0&\ldots &0\\0&1&0&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&1&0\end{pmatrix}}.$
The determinant of $XE_{n}-M_{\rho }$ is $(X-1)^{n-k}$ multiplied with the determinant of
    ${\begin{pmatrix}X&0&\ldots &0&-1\\-1&X&0&\ldots &0\\0&-1&X&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&-1&X\end{pmatrix}}.$
The expansion with respect to the first row yields
    $X^{k}+(-1)^{k+1}(-1)(-1)^{k-1}=X^{k}-1\,.$
$\Box$

## Lemma

For a permutation matrix $M_{\rho }$ over $\mathbb {C}$ for a cycle $\rho \in S_{n}$ with $\rho :i_{1}\mapsto i_{2}\mapsto \ldots \mapsto i_{k}\mapsto i_{1}$ and a $k$-th root of unity $\zeta$, the vectors
    $v_{\zeta }:=\zeta ^{k-1}e_{i_{1}}+\zeta ^{k-2}e_{i_{2}}+\cdots +\zeta e_{i_{k-1}}+e_{i_{k}}\,$
are eigenvectors of $M_{\rho }$ for the eigenvalue $\zeta$. In particular, a permutation matrix of a cycle over $\mathbb {C}$ is

### Proof

We have
    ${\begin{aligned}M_{\rho }(v_{\zeta })&=M_{\rho }(\zeta ^{k-1}e_{i_{1}}+\zeta ^{k-2}e_{i_{2}}+\cdots +\zeta e_{i_{k-1}}+e_{i_{k}})\\&=\zeta ^{k-1}M_{\rho }(e_{i_{1}})+\zeta ^{k-2}M_{\rho }(e_{i_{2}})+\cdots +\zeta M_{\rho }(e_{i_{k-1}})+M_{\rho }(e_{i_{k}})\\&=\zeta ^{k-1}e_{i_{2}}+\zeta ^{k-2}e_{i_{3}}+\cdots +\zeta e_{i_{k}}+e_{i_{1}}\\&=e_{i_{1}}+\zeta ^{k-1}e_{i_{2}}+\zeta ^{k-2}e_{i_{3}}+\cdots +\zeta e_{i_{k}}\\&=\zeta (\zeta ^{k-1}e_{i_{1}}+\zeta ^{k-2}e_{i_{2}}+\cdots +\zeta e_{i_{k-1}}+e_{i_{k}})\\&=\zeta v_{\zeta }.\end{aligned}}$
Since there are $k$ different $k$-th roots of unity in $\mathbb {C}$, these vectors are linearly independent due to fact, and they generate a $k$-dimensional linear subspace $U$ of $K^{n}$. In fact, we have
    $U=\langle e_{i_{j}},\,j=1,\ldots ,k\rangle \,.$
Since the vectors $e_{i}$, $i\neq {i_{j}}$, are fixed vectors, the $v_{\zeta }$ together with the $e_{i}$, $i\neq {i_{j}}$, form a basis consisting of eigenvectors of $M_{\rho }$. Hence, $M_{\rho }$ is diagonalizable.
$\Box$

## Theorem

A permutation matrix over $\mathbb {C}$ is

### Proof

$\Box$