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

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\zwischenueberschrift{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.
\inputdefinition { } {
For an $n \times n$-matrix $M$ with entries in a field $K$, the polynomial \mavergleichskettedisp {\vergleichskette { \chi_{ M } } { \defeq} {\det { \left( X \cdot E_{ n } - M \right) } } { } { } { } { } { } { } } is called the \definitionswort {characteristic polynomial}
For \mavergleichskette {\vergleichskette {M } { = }{ { \left( a_{ij} \right) }_{ij} } { }{ } { }{ } { }{ } }{,} this means \mavergleichskettedisp {\vergleichskette { \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 \mathl{K[X]}{.} But, since we can consider these elements also inside the field of rational functions \mathl{K(X)}{\zusatzfussnote {\mathlk{K(X)} is called the field of rational polynomials; it consists of all fractions \mathl{P/Q} for polynomials \mavergleichskette {\vergleichskette { P,Q } { \in }{ K [X] } { }{ } { }{ } { }{ } } with \mavergleichskette {\vergleichskette { Q } { \neq }{ 0 } { }{ } { }{ } { }{ } }{.} For \mavergleichskette {\vergleichskette { K } { = }{ \R } { }{ } { }{ } { }{ } } or $\C$, this field can be identified with the field of rational functions} {.} {,}} this is a useful definition. By definition, the determinant is an element in \mathl{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 \mavergleichskettedisp {\vergleichskette { \chi_{ M } } { =} { X^n + c_{n-1}X^{n-1} + \cdots + c_1 X+c_0 } { } { } { } { } { } { } }{.}
We have the important relation \mavergleichskettedisp {\vergleichskette { \chi_{ M } (\lambda) } { =} { \det { \left( \lambda E_{ n } - M \right) } } { } { } { } { } { } { } } for every \mavergleichskette {\vergleichskette { \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 \mathdisp {\varphi \colon V \longrightarrow V} { } on a finite-dimensional vector space, the \stichwort {characteristic polynomial} is defined by \mavergleichskettedisp {\vergleichskette { \chi_{ \varphi } } { \defeq} { \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 \mavergleichskettedisp {\vergleichskette { \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 n X \pm 1 } { } { } }{.}
\faktsituation {Let $K$ denote a field, and let $V$ denote an $n$-dimensional vector space. Let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping.} \faktfolgerung {Then \mavergleichskette {\vergleichskette { \lambda } { \in }{ K } { }{ } { }{ } { }{ } } is an eigenvalue of $\varphi$ if and only if $\lambda$ is a zero of the characteristic polynomial $\chi_{ \varphi }$.} \faktzusatz } {
Let $M$ denote a describing matrix for $\varphi$, and let \mavergleichskette {\vergleichskette { \lambda } { \in }{K } { }{ } { }{ } { }{ } } be given. We have \mavergleichskettedisp {\vergleichskette { \chi_{ M }\, (\lambda) } { =} { \det { \left( \lambda E_{ n } - M \right) } } { =} { 0 } { } { } { } { } }{,} if and only if the linear mapping \mathdisp {\lambda \operatorname{Id}_{ V } - \varphi} { } is not bijective \zusatzklammer {and not injective} \zusatzklammer {due to Theorem 26.11
and Lemma 25.11 } {.} This is, because of Lemma 27.11
and Lemma 24.14 , equivalent with \mavergleichskettedisp {\vergleichskette { \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 $\varphi$.
\inputbeispiel {
We consider the real matrix \mavergleichskette {\vergleichskette {M } { = }{ \begin{pmatrix} 0 & 5 \\ 1 & 0 \end{pmatrix} } { }{ } { }{ } { }{ } }{.} The characteristic polynomial is \mavergleichskettealign {\vergleichskettealign { \chi_{ M } } { =} { \det { \left( x E_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 } }{.} The eigenvalues are therefore \mavergleichskette {\vergleichskette {x } { = }{ \pm \sqrt{5} } { }{ } { }{ } { }{ } } \zusatzklammer {we have found these eigenvalues already in Example 27.9 , without using the characteristic polynomial} {.}
\inputbeispiel {
For the matrix \mavergleichskettedisp {\vergleichskette {M } { =} { \begin{pmatrix} 2 & 5 \\ -3 & 4 \end{pmatrix} } { } { } { } { } { } { } }{,} the characteristic polynomial is \mavergleichskettedisp {\vergleichskette { \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 \mavergleichskettedisp {\vergleichskette { (X-3)^2 } { =} { -23 +9 } { =} { -14 } { } { } { } { } }{,} which has no solution over $\R$, so that the matrix has no eigenvalues over $\R$. However, considered over the complex numbers $\Complex$, we have the two eigenvalues \mathkor {3+\sqrt{14} { \mathrm i}} {and} {3 - \sqrt{14} { \mathrm i}} {.} For the eigenspace for \mathl{3+\sqrt{14} { \mathrm i}}{,} we have to determine \mavergleichskettealign {\vergleichskettealign { \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) }
} { } { } { } { } }{,} a basis vector \zusatzklammer {hence an eigenvector} of this is \mathl{\begin{pmatrix} 5 \\1+ \sqrt{14} { \mathrm i} \end{pmatrix}}{.} Analogously, we get \mavergleichskettedisp {\vergleichskette { \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 } { } { } { } { } }{.}
\inputbeispiel {
For an upper triangular matrix \mavergleichskettedisp {\vergleichskette {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 \mavergleichskettedisp {\vergleichskette { \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 \mathl{d_1,d_2 , \ldots , d_n} \zusatzklammer {which might not be all different} {.}
\zwischenueberschrift{Multiplicities}
For a more detailed investigation of eigenspaces, the following concepts are necessary. Let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping on a finite-dimensional vector space $V$, and \mavergleichskette {\vergleichskette { \lambda } { \in }{ K } { }{ } { }{ } { }{ } }{.} Then the exponent of the linear polynomial \mathl{X - \lambda} inside the characteristic polynomial $\chi_{ \varphi }$ is called the \stichwort {algebraic multiplicity} of $\lambda$, symbolized as \mavergleichskette {\vergleichskette { \mu_\lambda } { \defeq }{ \mu_\lambda(\varphi) } { }{ } { }{ } { }{ } }{.} The dimension of the corresponding eigenspace, that is \mathdisp {\dim_{ } { \left( \operatorname{Eig}_{ \lambda } { \left( \varphi \right) } \right) }} { , } is called the \stichwort {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.
\faktsituation {Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping and \mavergleichskette {\vergleichskette { \lambda } { \in }{ K } { }{ } { }{ } { }{ } }{.}} \faktfolgerung {Then we have the estimate \mavergleichskettedisp {\vergleichskette { \dim_{ } { \left( \operatorname{Eig}_{ \lambda } { \left( \varphi \right) } \right) } } { \leq} { \mu_\lambda(\varphi) } { } { } { } { } { } { } } between the geometric and the algebraic multiplicity.} \faktzusatz } {
Let \mavergleichskette {\vergleichskette {m } { = }{ \dim_{ } { \left( \operatorname{Eig}_{ \lambda } { \left( \varphi \right) } \right) } } { }{ } { }{ } { }{ } } and let \mathl{v_1 , \ldots , v_m} be a basis of this eigenspace. We complement this basis with \mathl{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 \mathdisp {\begin{pmatrix} \lambda E_m & B \\ 0 & C \end{pmatrix}} { . } Ttherefore, the characteristic polynomial equals \zusatzklammer {using Exercise 26.9 } \mathl{(X- \lambda)^m \cdot \chi_{ C }}{,} so that the algebraic multiplicity is at least $m$.
\inputbeispiel {
We consider the \mathl{2\times 2}{-}\stichwort {shearing matrix}
\mavergleichskettedisp {\vergleichskette { M } { =} { \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} } { } { } { } { } { } { } }{,} with \mavergleichskette {\vergleichskette {a } { \in }{K } { }{ } { }{ } { }{ } }{.} The characteristic polynomial is \mavergleichskettedisp {\vergleichskette { \chi_{ M } } { =} {(X-1)(X-1) } { } { } { } { } { } { } }{,} so that $1$ is the only eigenvalue of $M$. The corresponding eigenspace is \mavergleichskettedisp {\vergleichskette { \operatorname{Eig}_{ 1 } { \left( M \right) } } { =} { \operatorname{ker} { \left( \begin{pmatrix} 0 & -a \\ 0 & 0 \end{pmatrix} \right) } } { } { } { } { } { } { } }{.} From \mavergleichskettedisp {\vergleichskette { \begin{pmatrix} 0 & -a \\ 0 & 0 \end{pmatrix} \begin{pmatrix} r \\s \end{pmatrix} } { =} { \begin{pmatrix} -as \\0 \end{pmatrix} } { } { } { } { } { } { } }{,} we get that \mathl{\begin{pmatrix} 1 \\0 \end{pmatrix}} is an eigenvector, and in case \mavergleichskette {\vergleichskette {a } { \neq }{0 } { }{ } { }{ } { }{ } }{,} the eigenspace is one-dimensional \zusatzklammer {in case \mavergleichskette {\vergleichskette { a } { = }{ 0 } { }{ } { }{ } { }{ } }{,} we have the identity and the eigenspace is two-dimensional} {.} So in case \mavergleichskette {\vergleichskette {a } { \neq }{0 } { }{ } { }{ } { }{ } }{,} the algebraic multiplicity of the eigenvalue $1$ equals $2$, and the geometric multiplicity equals $1$.
\zwischenueberschrift{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 \mathdisp {\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 \mathdisp {(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 \mathl{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.
\inputdefinition { } {
Let $K$ denote a field, let $V$ denote a vector space, and let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping. Then $\varphi$ is called \definitionswort {diagonalizable}{,} if $V$ has a basis consisting of eigenvectors
\faktsituation {Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping.} \faktuebergang {Then the following statements are equivalent.} \faktfolgerung {\aufzaehlungdrei {$\varphi$ is diagonalizable. } {There exists a basis $\mathfrak{ v }$ of $V$ such that the describing matrix \mathl{M_ \mathfrak{ v }^ \mathfrak{ v }(\varphi)} is a diagonal matrix. } {For every describing matrix \mavergleichskette {\vergleichskette {M } { = }{ M_ \mathfrak{ w }^ \mathfrak{ w }(\varphi) } { }{ } { }{ } { }{ } } with respect to a basis $\mathfrak{ w }$, there exists an invertible matrix $B$ such that \mathdisp {B M B^{-1}} { } is a diagonal matrix. }} \faktzusatz } {
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 .
\faktsituation {Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping.} \faktvoraussetzung {Suppose that there exists $n$ different eigenvalues.} \faktfolgerung {Then $\varphi$ is diagonalizable.} \faktzusatz } {
Because of Lemma 27.14 , there exist $n$ linearly independent eigenvectors. These form, due to Corollary 23.21 , a basis.
\inputbeispiel {
We continue with Example 27.9 . There exists the two eigenvectors \mathkor {\begin{pmatrix} \sqrt{5} \\1 \end{pmatrix}} {and} {\begin{pmatrix} -\sqrt{5} \\1 \end{pmatrix}} for the different eigenvalues \mathkor {\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 \mathdisp {\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 \mathkor {e_1} {and} {e_2} {,} is simply \mavergleichskettedisp {\vergleichskette { M^{ \mathfrak{ u } }_{ \mathfrak{ v } } } { =} { \begin{pmatrix} \sqrt{5} & - \sqrt{5} \\ 1 & 1 \end{pmatrix} } { } { } { } { } { } { } }{.} The inverse matrix is \mavergleichskettedisp {\vergleichskette { \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 \mavergleichskettealign {\vergleichskettealign { \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} } { } { } { } }{.}
\zwischenueberschrift{Multiplicities and diagonalizable matrices}
\faktsituation {Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping.} \faktfolgerung {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 \mavergleichskettedisp {\vergleichskette { \mu_\lambda } { =} { \dim_{ } { \left( \operatorname{Eig}_{ \lambda } { \left( \varphi \right) } \right) } } { } { } { } { } { } { } } holds.} \faktzusatz } {
If $\varphi$ is diagonalizable, then we can assume at once that $\varphi$ is described by a 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 multiplicity tells us. The
characteristic polynomial can be read off directly from the diagonal matrix, every diagonal entry $\lambda$ constitutes a linear factor \mathl{X- \lambda}{.}
For the other direction, let \mathl{\lambda_1 , \ldots , \lambda_k} denote the different eigenvalues, and let \mavergleichskettedisp {\vergleichskette { \mu_i } { \defeq} { \mu_{\lambda_i}(\varphi) } { =} { \dim_{ } { \left( \operatorname{Eig}_{ \lambda_i } { \left( \varphi \right) } \right) } } { } { } { } { } } denote the \zusatzklammer {geometric and algebraic} multiplicities. Due to the condition, the characteristic polynomial factors in linear factors. Therefore, the sum of these numbers equals \mavergleichskette {\vergleichskette { n } { = }{ \dim_{ } { \left( V \right) } } { }{ } { }{ } { }{ } }{.} Because of Fact *****, the sum of the eigenspaces \mavergleichskettedisp {\vergleichskette { \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 Fact *****, $\varphi$ is diagonalizable.
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. \inputbeispiel {
Let \mathkor {G_1} {and} {G_2} denote two lines in $\R^2$ through the origin, and let \mathkor {\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 \zusatzklammer {with eigenvalues $1$ and $-1$} {.} The composition \mavergleichskettedisp {\vergleichskette { \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 \mathkor {0} {or} {180} degree. If the angle between the axes is different from \mathl{0,90} degree, then $\psi$ does not have any eigenvector.
\zwischenueberschrift{Trigonalizable mappings}
\inputdefinition { } {
Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. A linear mapping $\varphi \colon V \rightarrow V$
is called \definitionswort {trigonalizable}{,} if there exists a basis such that the describing matrix of $\varphi$ with respect to this basis is an
Diagonalizable linear mappings are in particular trigonalizable. The reverse statement is not true, as Example 28.7
shows.
\faktsituation {Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let \mathdisp {\varphi \colon V \longrightarrow V} { } denote a linear mapping.} \faktuebergang {Then the following statements are equivalent.} \faktfolgerung {\aufzaehlungzwei {$\varphi$ is trigonalizable. } {The characteristic polynomial $\chi_{ \varphi }$ has a factorization into linear factors. }} \faktzusatz {If $\varphi$ is trigonalizable and is described by the matrix $M$ with respect to some basis, then there exists an invertible matrix \mavergleichskette {\vergleichskette {B } { \in }{ \operatorname{Mat}_{ n \times n } (K) } { }{ } { }{ } { }{ } } such that \mathl{BMB^{-1}} is an upper triangular matrix.} } {Linear mapping/Trigonalizable/Characterization with characteristic polynomial/Fact/Proof
\faktsituation {Let \mavergleichskette {\vergleichskette {M } { \in }{\operatorname{Mat}_{ n \times n } (\Complex) } { }{ } { }{ } { }{ } } denote a square matrix with complex entries.} \faktfolgerung {Then $M$ is trigonalizable.} \faktzusatz } {
This follows from Theorem 28.15
and the Fundamental theorem of algebra.