Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 26/refcontrol

For further investigations of linear mappings, and, in particular, of trigonalizable mappings, we have first to discuss an important law in the polynomial ring over a field, the Lemma of Bezout.

The Lemma of Bezout

Recall that a polynomial ${}T\in K[X]$ divides a polynomial ${}P\in K[X]$, if there exists an polynomial ${}Q\in K[X]$ such that

${}P=TQ\,.$

This corresponds to the divisibility concept of the integers ${}\mathbb {Z}$. From there, also the concept of a common divisor should be known.

Definition

Let ${}P_{1},\ldots ,P_{n}\in K[X]$ denote polynomialsMDLD/polynomials (1) over a fieldMDLD/field ${}K$. We say that a polynomial ${}T\in K[X]$ is a common divisor of the given polynomials if ${}T$ dividesMDLD/divides (polynomial ring 1)

every ${}P_{i}$.

Definition

Let ${}P_{1},\ldots ,P_{n}\in K[X]$ denote polynomialsMDLD/polynomials (1) over a fieldMDLD/field ${}K$. We say that a polynomial ${}G\in K[X]$ is a greatest common divisor of the given polynomials, if ${}G$ is a common divisorMDLD/common divisor (polynomial ring 1) of the ${}P_{i}$, and if ${}G$ has among all common divisors of the ${}P_{i}$ maximal

degree.MDLD/degree (polynomial)

A greatest common divisor is not uniquely determined, since with ${}G$ also ${}cG$ for some constant ${}c\neq 0$ is also a greatest common divisor. However, if we restrict to normed polynomials, then the greatest common divisor is unique.

Definition

PolynomialsMDLD/Polynomials (1) ${}P_{1},\ldots ,P_{n}\in K[X]$ over a fieldMDLD/field ${}K$ are called coprime, if they have, with the exception of constants ${}c\neq 0$, no

common divisor.MDLD/common divisor (polynomial ring 1)

TheoremCreate referencenumber

Let ${}K$ be a fieldMDLD/field and let ${}P_{1},\ldots ,P_{n}$ denote polynomialsMDLD/polynomials (1) over ${}K$. Let ${}G$ be a greatest common divisorMDLD/greatest common divisor (polynomial ring) of the ${}P_{i}$. Then there exists a representation

${}G=Q_{1}P_{1}+\cdots +Q_{n}P_{n}\,$

with

${}Q_{1},\ldots ,Q_{n}\in K[X]$.

Proof

We consider the set of all linear combinations

${}I={\left\{Q_{1}P_{1}+\cdots +Q_{n}P_{n}\mid Q_{i}\in K[X]\right\}}\,.$

This is an idealMDLD/ideal of ${}K[X]$, as can be checked directly. Due to fact, this ideal is a principal ideal,MDLD/principal ideal hence,

${}I=(E)\,$

with a certain polynomial ${}E$. This ${}E$ is a common divisor of the ${}P_{i}$. Because of ${}P_{i}\in I=(E)$, we have

${}P_{i}=H_{i}E\,,$

that is ${}E$ is a factor of every ${}P_{i}$. A similar reasoning shows

${}P_{i}\in (G)\,$

for all ${}i$, and, therefore, also

${}(E)=I\subseteq (G)\,.$

Hence,

${}E=GF\,.$

By the condition, ${}G$ has the maximal degree among all common divisors. Therefore, ${}F\neq 0$ is a constant. Thus, we have

${}(G)=(E)=I\,,$

and, in particular, ${}G\in I$. Therefore, ${}G$ is a linear combination of the ${}P_{i}$.

$\Box$

CorollaryCreate referencenumber

Let ${}K$ be a fieldMDLD/field and let ${}P_{1},\ldots ,P_{n}$ denote coprime polynomialsMDLD/coprime polynomials over ${}K$. Then there exists a representation

${}Q_{1}P_{1}+\cdots +Q_{n}P_{n}=1\,$

with

${}Q_{1},\ldots ,Q_{n}\in K[X]$.

Proof

This follows directly from fact.

$\Box$

Remark

For given polynomials ${}P_{1},\ldots ,P_{n}\in K[X]$, we can determine explicitly their greatest common divisor ${}G$ and a representation

${}G=Q_{1}P_{1}+\cdots +Q_{n}P_{n}\,$

as stated in fact. For this, we restrict ourselves to the case ${}n=2$. Let the degree of ${}P_{1}$ be as large as the degree of ${}P_{2}$. The division with remainder yields

${}P_{1}=P_{2}H_{2}+R_{2}\,$

with a remaining polynomial, whose degree is smaller than the degree of ${}P_{2}$, or which is ${}0$. The main point is that the ideals

${}(P_{1},P_{2})=(P_{2},R_{2})\,$

are identical and, thus, the greatest common divisor of ${}P_{1}$ and ${}P_{2}$ and of ${}P_{2}$ and ${}R_{2}$ coincide. Now we perform again the division with remainder, dividing ${}P_{2}$ by ${}R_{2}$ with remainder ${}R_{3}$, and again the ideal ${}(R_{2},R_{3})$ coincides with the starting ideal. In this way, we obtain a sequence of remaining polynomials

$R_{0}=P_{1},R_{1}=P_{2},R_{2},\ldots ,R_{k}\neq 0,0,$

with the property that two adjacent polynomials generate the same ideal. Hence, ${}R_{k}$ (the last remaining polynomial different from ${}0$) is the greatest common divisor of ${}R_{0}$ and ${}R_{1}$. We can find a representation of ${}R_{k}$ as a linear combination of the ${}R_{0},R_{1}$, by working back this algorithm using the equations which describe the divisions with remainder.

The method described in the previous remark is called Euclidean algorithm. It holds accordingly also for the integers.

Example Create referencenumber

We want to determine the greatest common divisorMDLD/greatest common divisor (polynomial) for the polynomials ${}X^{2}+X+1$ and ${}X-1$ in ${}\mathbb {Q} [X]$. For this, we perform the division with remainder,MDLD/division with remainder (polynomial ring) and we obtain

${}X^{2}+X+1={\left(X+2\right)}{\left(X-1\right)}+3\,.$

Thus the two polynomials are coprime.MDLD/coprime (polynomial) A representation of the ${}1$ is

${}1={\frac {1}{3}}{\left(X^{2}+X+1\right)}-{\frac {1}{3}}{\left(X+2\right)}{\left(X-1\right)}\,.$

We mention the Lemma of Bezout for the integers.

TheoremCreate referencenumber

Every set of integers ${}a_{1},\ldots ,a_{n}$ has a greatest common divisor ${}d$, and this ${}d$ can be expressed as a linear combination of the ${}a_{1},\ldots ,a_{n}$, that is, there exist integers ${}r_{1},\ldots ,r_{n}$ such that

${}r_{1}a_{1}+r_{2}a_{2}+\cdots +r_{n}a_{n}=d\,.$

In particular, for coprime integers ${}a_{1},\ldots ,a_{n}$, there exists a representation of ${}1$.

Proof

See exercise.

$\Box$

Generalized eigenspaces

A trigonalizable mapping can be described by an upper triangular matrix. We want to understand whether there are even easier matrices to describe such a mapping. This will be done in two steps. In this lecture, we describe a trigonalizable mapping as a direct sum of mappings on so-called generalized eigenspaces. In the next two lectures, we will discuss endomorphisms on such generalized eigenspaces.

Definition

For a linear mappingMDLD/linear mapping ${}\varphi$ on a ${}K$-vector spaceMDLD/vector space ${}V$ and an eigenvalue ${}\lambda \in K$,

${}\operatorname {GeEig} _{\lambda }(\varphi )=\bigcup _{n\in \mathbb {N} }\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{n}\,$

is called generalized eigenspace

of ${}\varphi$ for this eigenvalue.

If ${}V$ is finite-dimensional, then the chain

${}\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}\subseteq \operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{2}\subseteq \operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{3}\subseteq ...\,$

becomes stationary, that is, there exists some ${}r\in \mathbb {N}$ such that

${}\operatorname {GeEig} _{\lambda }(\varphi )=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{r}\,.$

Generalized eigenspaces are, due to exercise, invariant under the linear mapping. By definition, we have

${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\subseteq \operatorname {GeEig} _{\lambda }(\varphi )\,,$

and for ${}\varphi$ diagonalizable we have equality, see exercise. We want to understand trigonalizable mappings via their generalized eigenspaces.

LemmaCreate referencenumber

Let ${}\varphi \colon V\rightarrow V$ be a linear mappingMDLD/linear mapping on a finite-dimensionalMDLD/finite-dimensional ${}K$-vector spaceMDLD/vector space ${}V$, and let

${}\chi _{\varphi }=P\cdot Q\,$

be a factorization of the characteristic polynomialMDLD/characteristic polynomial in coprimeMDLD/coprime (polynomial) polynomials ${}P,Q\in K[X]$. Then we have the direct sum decompositionMDLD/direct sum decomposition

${}V=\operatorname {kern} P(\varphi )\oplus \operatorname {kern} Q(\varphi )\,,$

where these linear subspaces are ${}\varphi$-invariant.MDLD/invariant (linear)

The restriction of ${}P(\varphi )$ onto ${}\operatorname {kern} Q(\varphi )$ is bijective.

Proof

Due to the Lemma of Bezout, there exist polynomials ${}S,T\in K[T]$ such that

${}SP+TQ=1\,.$

Set ${}U=\operatorname {kern} P(\varphi )$ and ${}W=\operatorname {kern} Q(\varphi )$. Let ${}v\in V$. Due to the Theorem of Cayley-Hamilton , we have

${}0=\chi _{\varphi }(\varphi )=(P(\varphi )\circ Q(\varphi ))(v)=P(\varphi )(Q(\varphi )(v))\,.$

Therefore, the image of ${}Q(\varphi )$ belongs to the kernel of ${}P(\varphi )$ and vice versa. From

${}{\begin{aligned}v&=\operatorname {Id} _{V}(v)\\&=(SP+TQ)(\varphi )(v)\\&=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))\\&=P(\varphi )(S(\varphi )(v))+Q(\varphi )(T(\varphi )(v))\end{aligned}}$

we can read off that the left-hand summand belongs to ${}\operatorname {Im} P(\varphi )\subseteq \operatorname {kern} Q(\varphi )$ and the right-hand summand belongs to ${}\operatorname {Im} Q(\varphi )\subseteq \operatorname {kern} P(\varphi )$. Therefore, we have a sum decomposition, which is direct, since ${}P(\varphi )(v)=Q(\varphi )(v)=0$ implies ${}v=0$. For the ${}\varphi$-invarianceMDLD/invariance (endomorphism) of these spaces, see exercise. For ${}v\in \operatorname {kern} Q(\varphi )$, we have

${}v=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))=S(\varphi )(P(\varphi )(v))=P(\varphi )(S(\varphi )(v))\,,$

that is, we have ${}\operatorname {Im} P(\varphi )=\operatorname {kern} Q(\varphi )$. Therefore, the restriction of ${}P(\varphi )$ to the kernel of ${}Q(\varphi )$ is surjective, thus bijective.

$\Box$

Example Create referencenumber

We consider the permutation matrixMDLD/permutation matrix

${\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}$

over ${}\mathbb {R}$, the characteristic polynomialMDLD/characteristic polynomial is

${}\chi _{M}=X^{3}-1=(X-1){\left(X^{2}+X+1\right)}=P\cdot Q\,,$

where the two factors are coprime.MDLD/coprime (polynomial) We want to check fact in this example. We have

${}P(M)=M-E_{3}={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}\,$

with

${}\operatorname {kern} P(M)=\operatorname {Eig} _{1}{\left(M\right)}=\mathbb {R} {\begin{pmatrix}1\\1\\1\end{pmatrix}}\,,$

and

${}Q(M)={\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}}+{\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}}+{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}={\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}\,$

with

${}\operatorname {kern} Q(M)=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,.$

We have

${}\mathbb {R} ^{3}=\mathbb {R} {\begin{pmatrix}1\\1\\1\end{pmatrix}}\oplus \langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,.$

Moreover, we have

${}{\begin{aligned}P(M){\begin{pmatrix}1\\-1\\0\end{pmatrix}}&={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}{\begin{pmatrix}1\\-1\\0\end{pmatrix}}\\&={\begin{pmatrix}-1\\2\\-1\end{pmatrix}}\\&=-{\begin{pmatrix}1\\-1\\0\end{pmatrix}}+{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\end{aligned}}$

and

${}{\begin{aligned}P(M){\begin{pmatrix}0\\1\\-1\end{pmatrix}}&={\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\\&={\begin{pmatrix}-1\\-1\\2\end{pmatrix}}\\&=-{\begin{pmatrix}1\\-1\\0\end{pmatrix}}-2{\begin{pmatrix}0\\1\\-1\end{pmatrix}}.\end{aligned}}$

From this, we can read off that the restriction of ${}P(M)$ to ${}\operatorname {kern} Q(M)$ is bijective. The representation of the ${}1$ from example yields the matrix equation

${}{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}={\frac {1}{3}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}}-{\frac {1}{3}}{\begin{pmatrix}2&0&1\\1&2&0\\0&1&2\end{pmatrix}}{\begin{pmatrix}-1&0&1\\1&-1&0\\0&1&-1\end{pmatrix}}\,.$

TheoremCreate referencenumber

Let

$\varphi \colon V\longrightarrow V$

be an endomorphismMDLD/endomorphism on the finite-dimensionalMDLD/finite-dimensional ${}K$-vector spaceMDLD/vector space ${}V$, and let ${}\lambda \in K$. Then the dimensionMDLD/dimension (vs) of the generalized eigenspaceMDLD/generalized eigenspace ${}\operatorname {GeEig} _{\lambda }(\varphi )$ equals the algebraic multiplicityMDLD/algebraic multiplicity

of ${}\lambda$.

Proof

We write the characteristic polynomialMDLD/characteristic polynomial of ${}\varphi$ as

${}\chi _{\varphi }=(X-\lambda )^{k}Q\,,$

where ${}(X-\lambda )$ does not occur in ${}Q$ as a linear factor, that is, ${}k$ is the algebraic multiplicity of ${}\lambda$. Then, ${}P=(X-\lambda )^{k}$ and ${}Q$ are coprime,MDLD/coprime (polynomial) and, due to fact, we have the decompositon

${}V=\operatorname {kern} P(\varphi )\oplus \operatorname {kern} Q(\varphi )\,,$

and

$P(\varphi )={\left(\varphi -\lambda \operatorname {Id} \right)}^{k}\colon \operatorname {kern} Q(\varphi )\longrightarrow \operatorname {kern} Q(\varphi )$

is a bijection. Moreover,

${}H:=\operatorname {GeEig} _{\lambda }(\varphi )=\operatorname {kern} P(\varphi )\,,$

where the inclusion ${}\supseteq$ is clear, and the other inclusion follows from the fact that higher powerse of ${}X-\lambda$ do not annihilate further elements, by the bijectivity on ${}\operatorname {kern} Q(\varphi )$ just mentioned. For the characteristic polynomial, we have, due to the direct sum decomposition according to fact, the relation

${}\chi _{\varphi }=\chi _{1}\cdot \chi _{2}\,,$

where ${}\chi _{1}$ is the characteristic polynomial of ${}\varphi {|}_{H}$ and ${}\chi _{2}$ is the characteristic polynomial of ${}\varphi {|}_{\operatorname {kern} Q(\varphi )}$. Since ${}(\varphi -\lambda )^{k}$ restricted to ${}H$ is the zero mapping, the minimal polynomial of ${}\varphi {|}_{H}$ and, hence, also the characteristic polynomial ${}\chi _{1}$ are some power of ${}(X-\lambda )$, say

${}\chi _{1}=(X-\lambda )^{d}\,,$

where

${}d=\dim _{K}{\left(H\right)}\,.$

In particular, ${}d\leq k$, as ${}\chi _{1}$ is a divisor of ${}\chi _{\varphi }$. Assume that ${}d<k$. Then ${}\lambda$ is a zero of ${}\chi _{2}$ and ${}\lambda$ is an eigenvalue of ${}\varphi {|}_{\operatorname {kern} Q(\varphi )}$. But this is a contradiction to the fact that ${}P(\varphi )$ is a bijection on this space.

$\Box$

CorollaryCreate referencenumber

For a linear mappingMDLD/linear mapping ${}\varphi \colon V\rightarrow V$ on a finite-dimensionalMDLD/finite-dimensional ${}K$-vector spaceMDLD/vector space ${}V$ and two eigenvaluesMDLD/eigenvalues ${}\lambda \neq \delta$, the corresponding generalized eigenspaceMDLD/generalized eigenspace have trivial intersection, that is

${}\operatorname {GeEig} _{\lambda }(\varphi )\cap \operatorname {GeEig} _{\delta }(\varphi )=0\,.$

Proof

The characteristic polynomial of ${}\varphi$ has the form

${}\chi _{\varphi }=(X-\lambda )^{k}(X-\delta )^{\ell }F\,,$

where neither ${}\lambda$ nor ${}\delta$ is a zero of ${}F$. Because of fact, applied to ${}Q=(X-\delta )^{\ell }F$, we have

${}\operatorname {GeEig} _{\lambda }(\varphi )\cap \operatorname {kern} Q(\varphi )=0\,.$

Because of ${}\operatorname {GeEig} _{\delta }(\varphi )\subseteq \operatorname {kern} Q(\varphi )$, this implies immediately

${}\operatorname {GeEig} _{\lambda }(\varphi )\cap \operatorname {GeEig} _{\delta }(\varphi )=0\,.$

$\Box$

TheoremCreate referencenumber

Let

$\varphi \colon V\longrightarrow V$

be a trigonalizableMDLD/trigonalizable ${}K$-endomorphismMDLD/endomorphism on the finite-dimensionalMDLD/finite-dimensional ${}K$-vector spaceMDLD/vector space ${}V$. Then ${}V$ is the direct sumMDLD/direct sum (vs) of the generalized eigenspaces,MDLD/generalized eigenspaces that is,

${}V=\operatorname {GeEig} _{\lambda _{1}}(\varphi )\oplus \cdots \oplus \operatorname {GeEig} _{\lambda _{m}}(\varphi )\,,$

where ${}\lambda _{1},\ldots ,\lambda _{m}$ are the different eigenvaluesMDLD/eigenvalues of ${}\varphi$, and ${}\varphi$ is the direct sumMDLD/direct sum (mapping) of the restrictions

$\varphi _{i}=\varphi {|}_{H_{i}}\colon H_{i}\longrightarrow H_{i}$

on the generalized eigenspaces.

Proof

Let

${}\chi _{\varphi }=(X-\lambda _{1})^{k_{1}}\cdots (X-\lambda _{m})^{k_{m}}\,$

be the characteristic polynomial,MDLD/characteristic polynomial which splits into linear factors according to fact, where the ${}\lambda _{i}$ are different. We do induction over ${}m$. For ${}m=1$, there is only one eigenvalue ${}\lambda$ and only one generalized eigenspace. Due to fact, the minimal polynomial is of the form ${}(X-\lambda )^{s}$ and thus ${}V=\operatorname {GeEig} _{\lambda }(\varphi )$. Suppose that the statement is already proven for smaller ${}m$. We set ${}P=(X-\lambda _{1})^{k_{1}}$ and ${}Q=(X-\lambda _{2})^{k_{2}}\cdots (X-\lambda _{m})^{k_{m}}$. We are then in the situation of fact and fact. Therefore, we have a direct sum decomposition into ${}\varphi$-invariantMDLD/invariant (endomorphism) linear subspaces

${}V=\operatorname {GeEig} _{\lambda _{1}}(\varphi )\oplus \operatorname {kern} Q(\varphi )\,.$

The characteristic polynomial is, according to fact, the product of the characteristic polynomials of the restrictions to the spaces. Because of fact, the polynomial ${}(X-\lambda _{1})^{k_{1}}$ is the characteristic polynomial of the restriction to the first generalized eigenspace, hence, ${}Q$ is the characteristic polynomial of the restriction to ${}\operatorname {kern} Q(P)$. In particular, this restriction is also trigonalizable. By the induction hypothesis, ${}\operatorname {kern} Q(P)$ is the direct sum of the generalized eigenspaces for ${}\lambda _{2},\ldots ,\lambda _{m}$. Altogether, this implies the direct sum decomposition of ${}V$ and of ${}\varphi$.

$\Box$

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