# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 23

    The characteristic polynomial
We want to determine, for a given endomorphism $\varphi \colon V\rightarrow V$, the eigenvalues and the eigenspaces. For this, the characteristic polynomial is decisive.

## Definition

For an $n\times n$-matrix $M$ with entries in a field $K$, the polynomial
    $\chi _{M}:=\det {\left(X\cdot E_{n}-M\right)}\,$
is called the characteristic polynomial
$M$
For $M={\left(a_{ij}\right)}_{ij}$, this means
    $\chi _{M}=\det {\begin{pmatrix}X-a_{11}&-a_{12}&\ldots &-a_{1n}\\-a_{21}&X-a_{22}&\ldots &-a_{2n}\\\vdots &\vdots &\ddots &\vdots \\-a_{n1}&-a_{n2}&\ldots &X-a_{nn}\end{pmatrix}}\,.$
In this definition, we use the determinant of a matrix, which we have only defined for matrices with entries in a field. The entries are now elements of the polynomial ring $K[X]$. But, since we can consider these elements also inside the field of rational functions $K(X)$, this is a useful definition. By definition, the determinant is an element in $K(X)$, but, because all entries of the matrix are polynomials, and because in the recursive definition of the determinant, only addition and multiplication is used, the characteristic polynomial is indeed a polynomial. The degree of the characteristic polynomial is $n$, and its leading coefficient is $1$, so it has the form
    $\chi _{M}=X^{n}+c_{n-1}X^{n-1}+\cdots +c_{1}X+c_{0}\,.$
We have the important relation
    $\chi _{M}(\lambda )=\det {\left(\lambda E_{n}-M\right)}\,$
for every $\lambda \in K$, see exercise. Here, on the left-hand side, the number $\lambda$ is inserted into the polynomial, and on the right-hand side, we have the determinant of a matrix which depends on $\lambda$.
For a linear mapping
    $\varphi \colon V\longrightarrow V$
on a finite-dimensional vector space, the characteristic polynomial is defined by
    $\chi _{\varphi }:=\chi _{M}\,,$
where $M$ is a describing matrix with respect to some basis. The multiplication theorem for the determinant shows that this definition is independent of the choice of the basis, see exercise.
The characteristic polynomial of the identity on an $n$-dimensional vector space is
    $\chi _{\operatorname {Id} }=\det {\left(XE_{n}-E_{n}\right)}=(X-1)^{n}=X^{n}-nX^{n-1}+{\binom {n}{2}}X^{n-2}-{\binom {n}{3}}X^{n-3}+\cdots \pm {\binom {n}{2}}X^{2}\mp nX\pm 1\,.$

## Theorem

Let $K$ denote a field, and let $V$ denote an $n$-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
denote a linear mapping. Then $\lambda \in K$ is an eigenvalue of $\varphi$ if and only if $\lambda$ is a zero of the characteristic polynomial
$\chi _{\varphi }$

### Proof

Let $M$ denote a describing matrix for $\varphi$, and let $\lambda \in K$ be given. We have
    $\chi _{M}\,(\lambda )=\det {\left(\lambda E_{n}-M\right)}=0\,,$
if and only if the linear mapping
    $\lambda \operatorname {Id} _{V}-\varphi$
is not bijective (and not injective) (due to fact and fact). This is, because of fact and fact, equivalent with
    $\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\operatorname {ker} {\left((\lambda \operatorname {Id} _{V}-\varphi )\right)}\neq 0\,,$
and this means that the eigenspace for $\lambda$ is not the null space, thus $\lambda$ is an eigenvalue for $\varphi$.
$\Box$

## Example

We consider the real matrix $M={\begin{pmatrix}0&5\\1&0\end{pmatrix}}$. The characteristic polynomial is
    ${\begin{aligned}\chi _{M}&=\det {\left(xE_{2}-M\right)}\\&=\det {\left(x{\begin{pmatrix}1&0\\0&1\end{pmatrix}}-{\begin{pmatrix}0&5\\1&0\end{pmatrix}}\right)}\\&=\det {\begin{pmatrix}x&-5\\-1&x\end{pmatrix}}\\&=x^{2}-5.\end{aligned}}$
The eigenvalues are therefore $x=\pm {\sqrt {5}}$ (we have found these eigenvalues already in example, without using the characteristic polynomial).

## Example

For the matrix
    $M={\begin{pmatrix}2&5\\-3&4\end{pmatrix}}\,,$
the characteristic polynomial is
    $\chi _{M}=\det {\begin{pmatrix}X-2&-5\\3&X-4\end{pmatrix}}=(X-2)(X-4)+15=X^{2}-6X+23\,.$
Finding the zeroes of this polynomial leads to the condition
    $(X-3)^{2}=-23+9=-14\,,$
which has no solution over $\mathbb {R}$, so that the matrix has no eigenvalues over $\mathbb {R}$. However, considered over the complex numbers $\mathbb {C}$, we have the two eigenvalues $3+{\sqrt {14}}{\mathrm {i} }$ and $3-{\sqrt {14}}{\mathrm {i} }$. For the eigenspace for $3+{\sqrt {14}}{\mathrm {i} }$, we have to determine
    ${\begin{aligned}\operatorname {Eig} _{3+{\sqrt {14}}{\mathrm {i} }}{\left(M\right)}&=\operatorname {ker} {\left({\left({\left(3+{\sqrt {14}}{\mathrm {i} }\right)}E_{2}-M\right)}\right)}\\&=\operatorname {ker} {\left({\begin{pmatrix}1+{\sqrt {14}}{\mathrm {i} }&-5\\3&-1+{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\right)},\end{aligned}}$
a basis vector (hence an eigenvector) of this is ${\begin{pmatrix}5\\1+{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}$. Analogously, we get
    $\operatorname {Eig} _{3-{\sqrt {14}}{\mathrm {i} }}{\left(M\right)}=\operatorname {ker} {\left({\begin{pmatrix}1-{\sqrt {14}}{\mathrm {i} }&-5\\3&-1-{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\right)}=\langle {\begin{pmatrix}5\\1-{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\rangle \,.$

## Example

For an upper triangular matrix
    $M={\begin{pmatrix}d_{1}&\ast &\cdots &\cdots &\ast \\0&d_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1}&\ast \\0&\cdots &\cdots &0&d_{n}\end{pmatrix}}\,,$
the characteristic polynomial is
    $\chi _{M}=(X-d_{1})(X-d_{2})\cdots (X-d_{n})\,,$
due to fact. In this case, we have directly a factorization of the characteristic polynomial into linear factors, so that we can see immediately the zeroes and the eigenvalues of $M$, namely just the diagonal elements $d_{1},d_{2},\ldots ,d_{n}$ (which might not be all different).
    Invariant linear subspaces

## Definition

Let $K$ be a field, $V$ a vector space over $K$ and
    $\varphi \colon V\longrightarrow V$
a linear mapping. A linear subspace $U\subseteq V$ is called $\varphi$-invariant, if
    $\varphi (U)\subseteq U\,$
The zero-space and the total space are $\varphi$-invariant. Moreover, the eigenspaces for $\varphi$ are invariant.

## Lemma

Let $V$ be a finite-dimensional $K$-vector space, and
    $\varphi \colon V\longrightarrow V$
be a linear mapping. Let
    $V=U\oplus W\,$
denote a direct sum decomposition in $\varphi$-invariant linear subspaces. Then the characteristic polynomial fulfills the relation
    $\chi _{\varphi }=\chi _{\varphi {|}_{U}}\cdot \chi _{\varphi {|}_{W}}\,.$

### Proof

Let $u_{1},\ldots ,u_{k}$ be a basis of $U$ and $w_{1},\ldots ,w_{m}$ be a basis of $W$; together they form a basis of $V$. With respect to this basis, $\varphi$ is described by the block matrix $M={\begin{pmatrix}A&0\\0&B\end{pmatrix}}$, where $A$ describes the restriction $\varphi {|}_{U}$ and $B$ describes the restriction $\varphi {|}_{W}$. Then, using exercise, we get
    $\chi _{M}=\det {\left(t\operatorname {Id} -M\right)}=\det {\left(t\operatorname {Id} -A\right)}\det {\left(t\operatorname {Id} -B\right)}=\chi _{\varphi {|}_{U}}\cdot \chi _{\varphi {|}_{W}}\,.$
$\Box$
    Algebraic multiplicity
For a more detailed investigation of the eigenspaces, the following concept is useful.

## Definition

Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping on a finite-dimensional $K$-vector space $V$, and $\lambda \in K$. The exponent of the linear polynomial $X-\lambda$ in the characteristic polynomial $\chi _{\varphi }$ is called the algebraic multiplicity of $\lambda$. It is denoted by
    $\mu _{\lambda }=\mu _{\lambda }(\varphi )\,.$
Recall that
    $\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$
is called the geometric multiplicity of $\lambda$. We know, due to fact, that one of these multiplicities is positive if and only if this is true for the other multiplicity, and this is the case if and only if $\lambda$ is an eigenvalue.
In general, the multiplicities can be different, we have, however, an estimate between them.

## Lemma

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
denote a linear mapping and $\lambda \in K$. Then we have the estimate
    $\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\leq \mu _{\lambda }(\varphi )\,$
between the geometric and the

### Proof

Let $m=\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$ and let $v_{1},\ldots ,v_{m}$ be a basis of this eigenspace. We complement this basis with $w_{1},\ldots ,w_{n-m}$ to get a basis of $V$, using fact. With respect to this basis, the describing matrix has the form
    ${\begin{pmatrix}\lambda E_{m}&B\\0&C\end{pmatrix}}.$
Ttherefore, the characteristic polynomial equals (using exercise) $(X-\lambda )^{m}\cdot \chi _{C}$, so that the algebraic multiplicity is at least $m$.
$\Box$

## Example

We consider the $2\times 2$-shearing matrix
    $M={\begin{pmatrix}1&a\\0&1\end{pmatrix}}\,,$
with $a\in K$. The characteristic polynomial is
    $\chi _{M}=(X-1)(X-1)\,,$
so that $1$ is the only eigenvalue of $M$. The corresponding eigenspace is
    $\operatorname {Eig} _{1}{\left(M\right)}=\operatorname {ker} {\left({\begin{pmatrix}0&-a\\0&0\end{pmatrix}}\right)}\,.$
From
    ${\begin{pmatrix}0&-a\\0&0\end{pmatrix}}{\begin{pmatrix}r\\s\end{pmatrix}}={\begin{pmatrix}-as\\0\end{pmatrix}}\,,$
we get that ${\begin{pmatrix}1\\0\end{pmatrix}}$ is an eigenvector, and in case $a\neq 0$, the eigenspace is one-dimensional (in case $a=0$, we have the identity and the eigenspace is two-dimensional). So in case $a\neq 0$, the algebraic multiplicity of the eigenvalue $1$ equals $2$, and the geometric multiplicity equals $1$.
    Multiplicities and diagonalizable mappings

## Theorem

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
denote a linear mapping. Then $\varphi$ is diagonalizable if and only if the characteristic polynomial $\chi _{\varphi }$ is a product of linear factors and if for every zero $\lambda$ with algebraic multiplicity $\mu _{\lambda }$, the identity
    $\mu _{\lambda }=\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\,$

### Proof

If $\varphi$ is diagonalizable, then we can assume at once that $\varphi$ is described by a diagonal matrix with respect to a basis of eigenvectors. The diagonal entries of this matrix are the eigenvalues, and these occur as often as their geometric multiplicity tells us. The
characteristic polynomial can be read off directly from the diagonal matrix, every diagonal entry $\lambda$ constitutes a linear factor $X-\lambda$.
For the other direction, let $\lambda _{1},\ldots ,\lambda _{k}$ denote the different eigenvalues, and let
    $\mu _{i}:=\mu _{\lambda _{i}}(\varphi )=\dim _{\left(\operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}\right)}\,$
denote the (geometric and algebraic) multiplicities. Due to the condition, the characteristic polynomial factors in linear factors. Therefore, the sum of these numbers equals $n=\dim _{\left(V\right)}$. Because of fact, the sum of the eigenspaces
    $\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{k}}{\left(\varphi \right)}\subseteq V\,$
is direct. By the condition, the dimension on the left is also $n$, so that we have equality. Due to fact, $\varphi$ is diagonalizable.
$\Box$

## Corollary

Let $K$ denote a field, and let $V$ denote a $K$-vector space of finite dimension. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping. Suppose that the characteristic polynomial $\chi _{\varphi }$ factors into different linear factors. Then $\varphi$ is

### Proof

$\Box$
This gives also a new proof for fact.
    Footnotes
* ↑ Some authors define the characteristic polynomial as the determinant of $M-X\cdot E_{n}$, instead of $X\cdot E_{n}-M$. This does only change the sign.
* ↑ $K(X)$ is called the field of rational polynomials; it consists of all fractions $P/Q$ for polynomials $P,Q\in K[X]$ with $Q\neq 0$. For $K=\mathbb {R}$ or $\mathbb {C}$, this field can be identified with the field of rational functions.