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

The characteristic polynomial

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 28.4 . 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 28.3 .

## 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 Theorem 26.11 and Lemma 25.11 ). This is, because of Lemma 27.11 and Lemma 24.14 , 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 27.9 , 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 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 ${\displaystyle {}M}$, namely just the diagonal elements ${\displaystyle {}d_{1},d_{2},\ldots ,d_{n}}$ (which might not be all different).

Multiplicities

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

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

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

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

is called the geometric multiplicity of ${\displaystyle {}\lambda }$. 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 ${\displaystyle {}K}$ denote a field, and let ${\displaystyle {}V}$ denote a finite-dimensional vector space. Let

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

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

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

between the geometric and the algebraic multiplicity.

### Proof

Let ${\displaystyle {}m=\dim _{}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}}$ and let ${\displaystyle {}v_{1},\ldots ,v_{m}}$ be a basis of this eigenspace. We complement this basis with ${\displaystyle {}w_{1},\ldots ,w_{n-m}}$ to get a basis of ${\displaystyle {}V}$, using Theorem 23.23 . With respect to this basis, the describing matrix has the form

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

The characteristic polynomial equals therefore (using Exercise 26.9 ) ${\displaystyle {}(X-\lambda )^{m}\cdot \chi _{C}}$, so that the algebraic multiplicity is at least ${\displaystyle {}m}$.

${\displaystyle \Box }$

## Example

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

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

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

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

so that ${\displaystyle {}1}$ is the only eigenvalue of ${\displaystyle {}M}$. The corresponding eigenspace is

${\displaystyle {}\operatorname {Eig} _{1}{\left(M\right)}=\operatorname {ker} {\left({\begin{pmatrix}0&-a\\0&0\end{pmatrix}}\right)}\,.}$

From

${\displaystyle {}{\begin{pmatrix}0&-a\\0&0\end{pmatrix}}{\begin{pmatrix}r\\s\end{pmatrix}}={\begin{pmatrix}-as\\0\end{pmatrix}}\,,}$

we get that ${\displaystyle {}{\begin{pmatrix}1\\0\end{pmatrix}}}$ is an eigenvector, and in case ${\displaystyle {}a\neq 0}$, the eigenspace is one-dimensional (in case ${\displaystyle {}a=0}$, we have the identity and the eigenspace is two-dimensional). So in case ${\displaystyle {}a\neq 0}$, the algebraic multiplicity of the eigenvalue ${\displaystyle {}1}$ equals ${\displaystyle {}2}$, and the geometric multiplicity equals ${\displaystyle {}1}$.

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

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

the characteristic polynomial is just

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

If the number ${\displaystyle {}d}$ occurs ${\displaystyle {}k}$-times as a diagonal entry, then also the linear factor ${\displaystyle {}X-d}$ occurs with exponent ${\displaystyle {}k}$ 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 ${\displaystyle {}d}$ consists of all linear combinations of the standard vectors ${\displaystyle {}e_{i}}$, for which ${\displaystyle {}d_{i}}$ equals ${\displaystyle {}d}$. In particular, the dimension of the eigenspace equals the number how often ${\displaystyle {}d}$ occurs as a diagonal element. Thus, for a diagonal matrix, the algebraic and the geometric multiplicities coincide.

## Definition

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

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

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

for ${\displaystyle {}\varphi }$.

## Theorem

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

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

denote a

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

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 .

${\displaystyle \Box }$

## Corollary

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

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

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

### Proof

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

${\displaystyle \Box }$

## Example

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

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

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

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

The inverse matrix is

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

Because of Corollary 25.9 , we have the relation

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

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

denote a linear mapping. Then ${\displaystyle {}\varphi }$ is diagonalizable if and only if the characteristic polynomial ${\displaystyle {}\chi _{\varphi }}$ is a product of linear factors and if for every zero ${\displaystyle {}\lambda }$ with algebraic multiplicity ${\displaystyle {}\mu _{\lambda }}$, the identity

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

holds.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

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 ${\displaystyle {}G_{1}}$ and ${\displaystyle {}G_{2}}$ denote two lines in ${\displaystyle {}\mathbb {R} ^{2}}$ through the origin, and let ${\displaystyle {}\varphi _{1}}$ and ${\displaystyle {}\varphi _{2}}$ 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 ${\displaystyle {}1}$ and ${\displaystyle {}-1}$). The composition

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

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 ${\displaystyle {}0}$ or ${\displaystyle {}180}$ degree. If the angle between the axes is different from ${\displaystyle {}0,90}$ degree, then ${\displaystyle {}\psi }$ does not have any eigenvector.

Trigonalizable mappings

## Definition

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

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

denote a

linear mapping. Then the following statements are equivalent.
1. ${\displaystyle {}\varphi }$ is trigonalizable.
2. The characteristic polynomial ${\displaystyle {}\chi _{\varphi }}$ has a factorization into linear factors.
If ${\displaystyle {}\varphi }$ is trigonalizable and is described by the matrix ${\displaystyle {}M}$ with respect to some basis, then there exists an invertible matrix

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

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

## Theorem

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

### Proof

${\displaystyle \Box }$

Footnotes
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.