Linear mapping/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.

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 nullspace, 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).