# Linear mapping/Characteristic polynomial/Introduction/Section

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

## Definition

For an ${\displaystyle {}n\times n}$-matrix ${\displaystyle {}M}$ with entries in a field ${\displaystyle {}K}$, the polynomial

${\displaystyle {}\chi _{M}:=\det {\left(X\cdot E_{n}-M\right)}\,}$

is called the characteristic polynomial[1]

of ${\displaystyle {}M}$.

For ${\displaystyle {}M={\left(a_{ij}\right)}_{ij}}$, this means

${\displaystyle {}\chi _{M}=\det {\begin{pmatrix}X-a_{11}&-a_{12}&\ldots &-a_{1n}\\-a_{21}&X-a_{22}&\ldots &-a_{2n}\\\vdots &\vdots &\ddots &\vdots \\-a_{n1}&-a_{n2}&\ldots &X-a_{nn}\end{pmatrix}}\,.}$

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 ${\displaystyle {}K[X]}$. But, since we can consider these elements also inside the field of rational functions ${\displaystyle {}K(X)}$,[2] this is a useful definition. By definition, the determinant is an element in ${\displaystyle {}K(X)}$, 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 ${\displaystyle {}n}$, and its leading coefficient is ${\displaystyle {}1}$, so it has the form

${\displaystyle {}\chi _{M}=X^{n}+c_{n-1}X^{n-1}+\cdots +c_{1}X+c_{0}\,.}$

We have the important relation

${\displaystyle {}\chi _{M}(\lambda )=\det {\left(\lambda E_{n}-M\right)}\,}$

for every ${\displaystyle {}\lambda \in K}$, see exercise. Here, on the left-hand side, the number ${\displaystyle {}\lambda }$ is inserted into the polynomial, and on the right-hand side, we have the determinant of a matrix which depends on ${\displaystyle {}\lambda }$.

For a linear mapping

${\displaystyle \varphi \colon V\longrightarrow V}$

on a finite-dimensional vector space, the characteristic polynomial is defined by

${\displaystyle {}\chi _{\varphi }:=\chi _{M}\,,}$

where ${\displaystyle {}M}$ 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.

## Theorem

Let ${\displaystyle {}K}$ denote a field, and let ${\displaystyle {}V}$ denote an ${\displaystyle {}n}$-dimensional vector space. Let

${\displaystyle \varphi \colon V\longrightarrow V}$

denote a linear mapping. Then ${\displaystyle {}\lambda \in K}$ is an eigenvalue of ${\displaystyle {}\varphi }$, if and only if ${\displaystyle {}\lambda }$ is a zero of the characteristic polynomial ${\displaystyle {}\chi _{\varphi }}$.

### Proof

Let ${\displaystyle {}M}$ denote a describing matrix for ${\displaystyle {}\varphi }$, and let ${\displaystyle {}\lambda \in K}$ be given. We have

${\displaystyle {}\chi _{M}\,(\lambda )=\det {\left(\lambda E_{n}-M\right)}=0\,,}$

if and only if the linear mapping

${\displaystyle \lambda \operatorname {Id} _{V}-\varphi }$

is not bijective (and not injective) (due to fact and fact). This is, because of fact and fact, equivalent with

${\displaystyle {}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\operatorname {ker} {\left((\lambda \operatorname {Id} _{V}-\varphi )\right)}\neq 0\,,}$

and this means that the eigenspace for ${\displaystyle {}\lambda }$ is not the nullspace, thus ${\displaystyle {}\lambda }$ is an eigenvalue for ${\displaystyle {}\varphi }$.

${\displaystyle \Box }$

## Example

We consider the real matrix ${\displaystyle {}M={\begin{pmatrix}0&5\\1&0\end{pmatrix}}}$. The characteristic polynomial is

{\displaystyle {}{\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 ${\displaystyle {}x=\pm {\sqrt {5}}}$ (we have found these eigenvalues already in example, without using the characteristic polynomial).

## Example

For the matrix

${\displaystyle {}M={\begin{pmatrix}2&5\\-3&4\end{pmatrix}}\,,}$
${\displaystyle {}\chi _{M}=\det {\begin{pmatrix}X-2&-5\\3&X-4\end{pmatrix}}=(X-2)(X-4)+15=X^{2}-6X+23\,.}$

Finding the zeroes of this polynomial leads to the condition

${\displaystyle {}(X-3)^{2}=-23+9=-14\,,}$

which has no solution over ${\displaystyle {}\mathbb {R} }$, so that the matrix has no eigenvalues over ${\displaystyle {}\mathbb {R} }$. However, considered over the complex numbers ${\displaystyle {}\mathbb {C} }$, we have the two eigenvalues ${\displaystyle {}3+{\sqrt {14}}{\mathrm {i} }}$ and ${\displaystyle {}3-{\sqrt {14}}{\mathrm {i} }}$. For the eigenspace for ${\displaystyle {}3+{\sqrt {14}}{\mathrm {i} }}$, we have to determine

{\displaystyle {}{\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 ${\displaystyle {}{\begin{pmatrix}5\\1+{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}}$. Analogously, we get

${\displaystyle {}\operatorname {Eig} _{3-{\sqrt {14}}{\mathrm {i} }}{\left(M\right)}=\operatorname {ker} {\left({\begin{pmatrix}1-{\sqrt {14}}{\mathrm {i} }&-5\\3&-1-{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\right)}=\langle {\begin{pmatrix}5\\1-{\sqrt {14}}{\mathrm {i} }\end{pmatrix}}\rangle \,.}$

## Example

For an upper triangular matrix

${\displaystyle {}M={\begin{pmatrix}d_{1}&\ast &\cdots &\cdots &\ast \\0&d_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1}&\ast \\0&\cdots &\cdots &0&d_{n}\end{pmatrix}}\,,}$
${\displaystyle {}\chi _{M}=(X-d_{1})(X-d_{2})\cdots (X-d_{n})\,,}$

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 ${\displaystyle {}M}$, namely just the diagonal elements ${\displaystyle {}d_{1},d_{2},\ldots ,d_{n}}$ (which might not be all different).

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