Endomorphism/Characteristic polynomial/Introduction/Section

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

of ${}M$.

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).