Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 28

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

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 28.4 . 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 28.3 .

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 Theorem 26.11 and Lemma 25.11 ). This is, because of Lemma 27.11 and Lemma 24.14 , 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 27.9 , 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 Lemma 26.8 . 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).

Multiplicities

For a more detailed investigation of eigenspaces, the following concepts are necessary. Let

$\varphi \colon V\longrightarrow V$

denote a linear mapping on a finite-dimensional vector space ${}V$, and ${}\lambda \in K$. Then the exponent of the linear polynomial ${}X-\lambda$ inside the characteristic polynomial ${}\chi _{\varphi }$ is called the algebraic multiplicity of ${}\lambda$, symbolized as ${}\mu _{\lambda }:=\mu _{\lambda }(\varphi )$. The dimension of the corresponding eigenspace, that is

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

is called the geometric multiplicity of ${}\lambda$. Because of Theorem 28.2 , the algebraic multiplicity is positive if and only if the geometric multiplicity is positive. In general, these multiplicities might be different, we have however always one estimate.

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

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

between the geometric and the

algebraic multiplicity.

Proof

Let ${}m=\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$ 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 Theorem 23.23 . 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 26.9 ) ${}(X-\lambda )^{m}\cdot \chi _{C}$, 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 ${}1$.

Diagonalizable mappings

The restriction of a linear mapping to an eigenspace is the homothety with the corresponding eigenvalue, so this is a quite simple linear mapping. If there are many eigenvalues with high-dimensional eigenspaces, then usually the linear mapping is simple in some sense. An extreme case are the so-called diagonalizable mappings.

For a diagonal matrix

${\begin{pmatrix}d_{1}&0&\cdots &\cdots &0\\0&d_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1}&0\\0&\cdots &\cdots &0&d_{n}\end{pmatrix}},$

the characteristic polynomial is just

$(X-d_{1})(X-d_{2})\cdots (X-d_{n}).$

If the number ${}d$ occurs ${}k$-times as a diagonal entry, then also the linear factor ${}X-d$ occurs with exponent ${}k$ inside the factorization of the characteristic polynomial. This is also true when we just have an upper triangular matrix. But in the case of a diagonal matrix, we can also read of immediately the eigenspaces, see Example 27.7 . The eigenspace for ${}d$ consists of all linear combinations of the standard vectors ${}e_{i}$, for which ${}d_{i}$ equals ${}d$. In particular, the dimension of the eigenspace equals the number how often ${}d$ occurs as a diagonal element. Thus, for a diagonal matrix, the algebraic and the geometric multiplicities coincide.

Definition

Let ${}K$ denote a field, let ${}V$ denote a vector space, and let

$\varphi \colon V\longrightarrow V$

denote a linear mapping. Then ${}\varphi$ is called diagonalizable, if ${}V$ has a basis consisting of eigenvectors

for ${}\varphi$.

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 the following statements are equivalent.

1. ${}\varphi$ is diagonalizable.
2. There exists a basis ${}{\mathfrak {v}}$ of ${}V$ such that the describing matrix ${}M_{\mathfrak {v}}^{\mathfrak {v}}(\varphi )$ is a diagonal matrix.
3. For every describing matrix ${}M=M_{\mathfrak {w}}^{\mathfrak {w}}(\varphi )$ with respect to a basis ${}{\mathfrak {w}}$, there exists an invertible matrix ${}B$ such that

   $BMB^{-1}$

   is a diagonal matrix.

Proof

The equivalence between (1) and (2) follows from the definition, from Example 27.7 , and the correspondence between linear mappings and matrices. The equivalence between (2) and (3) follows from Corollary 25.9 .

$\Box$

Corollary

Let ${}K$ denote a field, and let ${}V$ denote a finite-dimensional vector space. Let

$\varphi \colon V\longrightarrow V$

denote a linear mapping. Suppose that there exists ${}n$ different eigenvalues. Then ${}\varphi$ is

diagonalizable.

Proof

Because of Lemma 27.14 , there exist ${}n$ linearly independent eigenvectors. These form, due to Corollary 23.21 , a basis.

$\Box$

Example

We continue with Example 27.9 . There exists the two eigenvectors ${}{\begin{pmatrix}{\sqrt {5}}\\1\end{pmatrix}}$ and ${}{\begin{pmatrix}-{\sqrt {5}}\\1\end{pmatrix}}$ for the different eigenvalues ${}{\sqrt {5}}$ and ${}-{\sqrt {5}}$, so that the mapping is diagonalizable, due to Corollary 28.10 . With respect to the basis ${}{\mathfrak {u}}$, consisting of these eigenvectors, the linear mapping is described by the diagonal matrix

${\begin{pmatrix}{\sqrt {5}}&0\\0&-{\sqrt {5}}\end{pmatrix}}.$

The transformation matrix, from the basis ${}{\mathfrak {u}}$ to the standard basis ${}{\mathfrak {v}}$, consisting of ${}e_{1}$ and ${}e_{2}$, is simply

${}M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\,.$

The inverse matrix is

${}{\frac {1}{2{\sqrt {5}}}}{\begin{pmatrix}1&{\sqrt {5}}\\-1&{\sqrt {5}}\end{pmatrix}}={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}\,.$

Because of Corollary 25.9 , we have the relation

${}{\begin{aligned}{\begin{pmatrix}{\sqrt {5}}&0\\0&-{\sqrt {5}}\end{pmatrix}}&={\begin{pmatrix}{\frac {1}{2}}&{\frac {\sqrt {5}}{2}}\\{\frac {1}{2}}&{\frac {-{\sqrt {5}}}{2}}\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{2{\sqrt {5}}}}&{\frac {1}{2}}\\{\frac {-1}{2{\sqrt {5}}}}&{\frac {1}{2}}\end{pmatrix}}{\begin{pmatrix}0&5\\1&0\end{pmatrix}}{\begin{pmatrix}{\sqrt {5}}&-{\sqrt {5}}\\1&1\end{pmatrix}}.\end{aligned}}$

Multiplicities and diagonalizable matrices

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

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

holds.

Proof

This proof was not presented in the lecture.

$\Box$

The product of two diagonal matrices is again a diagonal matrix. The following example shows that the product of two diagonalizable matrices is in general not diagonalizable.

Example

Let ${}G_{1}$ and ${}G_{2}$ denote two lines in ${}\mathbb {R} ^{2}$ through the origin, and let ${}\varphi _{1}$ and ${}\varphi _{2}$ denote the reflections at these axes. A reflection at an axis is always diagonalizable, the axis and the line orthogonal to the axis are eigenlines (with eigenvalues ${}1$ and ${}-1$). The composition

${}\psi =\varphi _{2}\circ \varphi _{1}\,$

of the reflections is a plane rotation, the angle of rotation being twice the angle between the two lines. However, a rotation is only diagonalizable if the angle of rotation is ${}0$ or ${}180$ degree. If the angle between the axes is different from ${}0,90$ degree, then ${}\psi$ does not have any eigenvector.

Trigonalizable mappings

Definition

Let ${}K$ denote a field, and let ${}V$ denote a finite-dimensional vector space. A linear mapping ${}\varphi \colon V\rightarrow V$ is called trigonalizable, if there exists a basis such that the describing matrix of ${}\varphi$ with respect to this basis is an

upper triangular matrix.

Diagonalizable linear mappings are in particular trigonalizable. The reverse statement is not true, as Example 28.7 shows.

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 the following statements are equivalent.

1. ${}\varphi$ is trigonalizable.
2. The characteristic polynomial ${}\chi _{\varphi }$ has a factorization into linear factors.

If ${}\varphi$ is trigonalizable and is described by the matrix ${}M$ with respect to some basis, then there exists an invertible matrix

${}B\in \operatorname {Mat} _{n\times n}(K)$ such that ${}BMB^{-1}$ is an

upper triangular matrix.

Proof

This proof was not presented in the lecture.

$\Box$

Theorem

Let ${}M\in \operatorname {Mat} _{n\times n}(\mathbb {C} )$ denote a square matrix with complex entries. Then ${}M$ is

trigonalizable.

Proof

This follows from Theorem 28.15 and the Fundamental theorem of algebra.

$\Box$

Footnotes

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