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

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 ${}n\times n$-matrixMDLD/matrix ${}M$ with entries in a fieldMDLD/field ${}K$, the polynomialMDLD/polynomial (1)

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

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 functionsMDLD/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 23.3 . 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 23.24 .

The characteristic polynomial of the identity on an ${}n$-dimensional vector space is

${}\chi _{\operatorname {Id} }=\det {\left(XE_{n}-E_{n}\right)}=(X-1)^{n}=X^{n}-nX^{n-1}+{\binom {n}{2}}X^{n-2}-{\binom {n}{3}}X^{n-3}+\cdots \pm {\binom {n}{2}}X^{2}\mp nX\pm 1\,.$

TheoremTheorem 23.2 change

Let ${}K$ denote a field,MDLD/field and let ${}V$ denote an ${}n$-dimensionalMDLD/dimensional (fgvs) vector space.MDLD/vector space Let

$\varphi \colon V\longrightarrow V$

denote a linear mapping.MDLD/linear mapping Then ${}\lambda \in K$ is an eigenvalueMDLD/eigenvalue of ${}\varphi$ if and only if ${}\lambda$ is a zero of the characteristic polynomialMDLD/characteristic polynomial

${}\chi _{\varphi }$.

Proof

Let ${}M$ denote a describing matrixMDLD/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 bijectiveMDLD/bijective (and not injectiveMDLD/injective) (due to Theorem 16.11 and Lemma 12.5 ). This is, because of Lemma 22.1 and Lemma 11.4 , equivalent with

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

and this means that the eigenspaceMDLD/eigenspace for ${}\lambda$ is not the null space, thus ${}\lambda$ is an eigenvalue for ${}\varphi$.

$\Box$

Example Create referencenumber

We consider the real matrix ${}M={\begin{pmatrix}0&5\\1&0\end{pmatrix}}$. The characteristic polynomialMDLD/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 21.5 , without using the characteristic polynomial).

Example Create referencenumber

For the matrix

${}M={\begin{pmatrix}2&5\\-3&4\end{pmatrix}}\,,$

the characteristic polynomialMDLD/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 eigenvaluesMDLD/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 eigenspaceMDLD/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 Create referencenumber

For an upper triangular matrixMDLD/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 polynomialMDLD/characteristic polynomial is

${}\chi _{M}=(X-d_{1})(X-d_{2})\cdots (X-d_{n})\,,$

due to Lemma 16.4 . 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 eigenvaluesMDLD/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.MDLD/linear mapping A linear subspaceMDLD/linear subspace ${}U\subseteq V$ is called ${}\varphi$-invariant, if

${}\varphi (U)\subseteq U\,$

holds.

The zero-space and the total space are ${}\varphi$-invariant. Moreover, the eigenspaces for ${}\varphi$ are invariant.

LemmaLemma 26.10 change

Let ${}V$ be a finite-dimensionalMDLD/finite-dimensional ${}K$-vector space,MDLD/vector space and

$\varphi \colon V\longrightarrow V$

be a linear mapping.MDLD/linear mapping Let

${}V=U\oplus W\,$

denote a direct sum decompositionMDLD/direct sum decomposition in ${}\varphi$-invariantMDLD/invariant (linear mapping) 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 basisMDLD/basis (vs) 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 matrixMDLD/block matrix (2) ${}M={\begin{pmatrix}A&0\\0&B\end{pmatrix}}$, where ${}A$ describes the restriction ${}\varphi {|}_{U}$ and ${}B$ describes the restriction ${}\varphi {|}_{W}$. Then, using Exercise 16.23 , 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 mappingMDLD/linear mapping on a finite-dimensionalMDLD/finite-dimensional ${}K$-vector spaceMDLD/vector space ${}V$, and ${}\lambda \in K$. The exponent of the linear polynomial ${}X-\lambda$ in the characteristic polynomialMDLD/characteristic polynomial ${}\chi _{\varphi }$ is called the algebraic multiplicity of ${}\lambda$. It is denoted by

${}\mu _{\lambda }=\mu _{\lambda }(\varphi )\,.$

Recall that

$\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$

is called the geometric multiplicity of ${}\lambda$. We know, due to Theorem 23.2 , 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.

LemmaCreate referencenumber

Let ${}K$ denote a field,MDLD/field and let ${}V$ denote a finite-dimensionalMDLD/finite-dimensional (fgvs) vector space.MDLD/vector space Let

$\varphi \colon V\longrightarrow V$

denote a linear mappingMDLD/linear mapping and ${}\lambda \in K$. Then we have the estimate

${}\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\leq \mu _{\lambda }(\varphi )\,$

between the geometricMDLD/geometric (multiplicity) and the

algebraic multiplicity.MDLD/algebraic multiplicity

Proof

Let ${}m=\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$ and let ${}v_{1},\ldots ,v_{m}$ be a basisMDLD/basis (vs) of this eigenspace.MDLD/eigenspace We complement this basis with ${}w_{1},\ldots ,w_{n-m}$ to get a basis of ${}V$, using Theorem 8.10 . With respect to this basis, the describing matrixMDLD/describing matrix has the form

${\begin{pmatrix}\lambda E_{m}&B\\0&C\end{pmatrix}}.$

Ttherefore, the characteristic polynomialMDLD/characteristic polynomial equals (using Exercise 16.23 ) ${}(X-\lambda )^{m}\cdot \chi _{C}$, so that the algebraic multiplicityMDLD/algebraic multiplicity is at least ${}m$.

$\Box$

Example Create referencenumber

We consider the ${}2\times 2$-shearing matrix

${}M={\begin{pmatrix}1&a\\0&1\end{pmatrix}}\,,$

with ${}a\in K$. The characteristic polynomialMDLD/characteristic polynomial is

${}\chi _{M}=(X-1)(X-1)\,,$

so that ${}1$ is the only eigenvalueMDLD/eigenvalue of ${}M$. The corresponding eigenspaceMDLD/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,MDLD/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 multiplicityMDLD/algebraic multiplicity of the eigenvalue ${}1$ equals ${}2$, and the geometric multiplicityMDLD/geometric multiplicity equals ${}1$.

Multiplicities and diagonalizable mappings

TheoremCreate referencenumber

Let ${}K$ denote a field,MDLD/field and let ${}V$ denote a finite-dimensionalMDLD/finite-dimensional (fgvs) vector space.MDLD/vector space Let

$\varphi \colon V\longrightarrow V$

denote a linear mapping.MDLD/linear mapping Then ${}\varphi$ is diagonalizableMDLD/diagonalizable if and only if the characteristic polynomialMDLD/characteristic polynomial ${}\chi _{\varphi }$ is a product of linear factorsMDLD/linear factors (1K) and if for every zero ${}\lambda$ with algebraic multiplicityMDLD/algebraic multiplicity ${}\mu _{\lambda }$, the identity

${}\mu _{\lambda }=\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\,$

holds.

Proof

If ${}\varphi$ is diagonalizable,MDLD/diagonalizable (ev) then we can assume at once that ${}\varphi$ is described by a diagonal matrixMDLD/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 multiplicityMDLD/geometric multiplicity tells us. The characteristic polynomialMDLD/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

${}\mu _{i}:=\mu _{\lambda _{i}}(\varphi )=\dim _{K}{\left(\operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}\right)}\,$

denote the (geometric and algebraic) multiplicities. Due to the condition, the characteristic polynomial factors in linear factors. Therefore, the sum of these numbers equals ${}n=\dim _{K}{\left(V\right)}$. Because of Lemma 22.6 , 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 Lemma 22.11 , ${}\varphi$ is diagonalizable.

$\Box$

CorollaryCreate referencenumber

Let ${}K$ denote a field,MDLD/field and let ${}V$ denote a ${}K$-vector spaceMDLD/vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

be a linear mapping.MDLD/linear mapping Suppose that the characteristic polynomialMDLD/characteristic polynomial ${}\chi _{\varphi }$ factors into different linear factors.MDLD/linear factors (1K) Then ${}\varphi$ is

diagonalizable.MDLD/diagonalizable

Proof

See Exercise 23.28 .

$\Box$

This gives also a new proof for Corollary 22.10 .

Footnotes

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