Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 25/refcontrol

Trigonalizable mappings

Definition

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

upper triangular matrix.MDLD/upper triangular matrix

Diagonalizable linear mappings are in particular trigonalizable. The inverse statement is not true, as Example 22.12 shows. We will see in Theorem 25.10 that a linear mapping is trigonalizable if and only if its characteristic polynomial factors into linear factors. A square matrix ${}M$ is called trigonalizable if the corresponding linear mapping ${}K^{n}\rightarrow K^{n}$ is trigonalizable. This means that there exists a basis such that the mapping is described by an upper triangular matrix with respect to this basis, or, that there exists an invertible matrix ${}B$ (the base change matrix) such that

$BMB^{-1}$

is an upper triangular matrix. Therefore, a matrix is trigonalizable if and only if it is similarMDLD/similar (matrix) to an upper triangular matrix. The process of finding such a basis and to perform this base change is called trigonalization.

Example Create referencenumber

We claim that the matrix

${}M={\begin{pmatrix}3&1\\-1&1\end{pmatrix}}\,$

is trigonalizable.MDLD/trigonalizable The matrix

${}B={\begin{pmatrix}3&2\\1&1\end{pmatrix}}\,$

is invertibleMDLD/invertible (matrix) with the inverse matrixMDLD/inverse matrix

${}B^{-1}={\begin{pmatrix}1&-2\\-1&3\end{pmatrix}}\,.$

A direct computation shows

${}BMB^{-1}={\begin{pmatrix}3&2\\1&1\end{pmatrix}}{\begin{pmatrix}3&1\\-1&1\end{pmatrix}}{\begin{pmatrix}1&-2\\-1&3\end{pmatrix}}={\begin{pmatrix}3&2\\1&1\end{pmatrix}}{\begin{pmatrix}2&-3\\-2&5\end{pmatrix}}={\begin{pmatrix}2&1\\0&2\end{pmatrix}}\,.$

In this verification of trigonalizability, the transformation matrix ${}B$ arises just like that. We get a more reasonable proof with the help of the characteristic polynomialMDLD/characteristic polynomial and Theorem 25.10 . The characteristic polynomial is

${}\chi _{M}=\det {\begin{pmatrix}X-3&-1\\1&X-1\end{pmatrix}}=(X-3)(X-1)+1=X^{2}-4X+4=(X-2)^{2}\,,$

and this factors into linear factors.

LemmaCreate referencenumber

Let ${}V_{1},\ldots ,V_{n}$ denote finite-dimensionalMDLD/finite-dimensional vector spacesMDLD/vector spaces over the fieldMDLD/field ${}K$, let

$\varphi _{i}\colon V_{i}\longrightarrow V_{i}$

denote linear mappings,MDLD/linear mappings and let

$\varphi =\varphi _{1}\times \cdots \times \varphi _{n}\colon V_{1}\times \cdots \times V_{n}\longrightarrow V_{1}\times \cdots \times V_{n}$

denote the product mapping.MDLD/product mapping Then ${}\varphi$ is trigonalizableMDLD/trigonalizable

if and only if this holds for all ${}\varphi _{i}$.

Proof

See Exercise 25.5 .

$\Box$

In particular, the preceding statement holds when

${}V=\bigoplus _{i\in I}V_{i}\,$

is a direct sumMDLD/direct sum (vs) of ${}\varphi$-invariant linear subspaces.MDLD/invariant linear subspaces

Invariant linear subspaces

A trigonalizable endomorphismMDLD/trigonalizable endomorphism is described, with respect to a suitable basis, by a matrix of the form

${}M={\begin{pmatrix}_{1}&\ast &\cdots &\cdots &\ast \\0&_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&_{n-1}&\ast \\0&\cdots &\cdots &0&_{n}\end{pmatrix}}\,.$

A property which hold for such an upper triangular matrix and which can be described as a property of the linear mapping (being independent of a chosen basis), must hold for any trigonalizable mapping. We want to understand such properties. By an upper triangular matrix, the ${}j$-th standard vector ${}e_{j}$ is sent to

${}Me_{j}=a_{1j}e_{1}+\cdots +a_{jj}e_{j}\,.$

In particular, ${}e_{1}$ is a eigenvector with the eigenvalue ${}a_{11}$. It is typical for a trigonalizable mapping that the linear subspace

${}V_{j}=\langle e_{1},\ldots ,e_{j}\rangle \,$

is mapped by ${}M$ into itself. That is, the ${}V_{j}$ are ${}M$-invariantMDLD/invariant (endomorphism) linear subspaces, which are contained in each other and those dimensions are ${}j$. We will show, after some preparations, that these properties characterize trigonalizable mappings.

LemmaLemma 25.4 change

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

$\varphi \colon V\longrightarrow V$

be a linear mapping,MDLD/linear mapping and let ${}\lambda \in K$ be an eigenvalueMDLD/eigenvalue of ${}\varphi$. Then there exists a ${}\varphi$-invariantMDLD/invariant (linear subspace) linear subspaceMDLD/linear subspace ${}U\subset V$

of dimension ${}n-1$.

Proof

Because of the condition and Lemma 22.1 , the mapping ${}\varphi -\lambda \operatorname {Id} _{V}$ has a nontrival kernel.MDLD/kernel Hence, this mapping is not injectiveMDLD/injective and, due to Corollary 11.9 , also not surjective.MDLD/surjective Therefore,

${}B:=\operatorname {Im} {\left(\varphi -\lambda \operatorname {Id} _{V}\right)}\subset V\,$

is a strict linear subspace of ${}V$. It follows that there exists also a linear subspace ${}U\subset V$ of dimension ${}n-1$, which contains ${}B$. For ${}u\in U$, we have

${}\varphi (u)=\lambda u+(\varphi -\lambda \operatorname {Id} _{V})u\in U+B\subseteq U\,.$

Hence, the image of ${}U$ belongs to ${}U$, that is, ${}U$ is ${}\varphi$-invariant.

$\Box$

When ${}U\subseteq V$ is an ${}\varphi$-invariant linear subspace, and ${}P\in K[X]$ is a polynomial, then ${}U$ is also ${}P(\varphi )$-invariant, see Exercise 23.31 . In this situation, the following identity holds.

LemmaLemma 25.5 change

Let ${}K$ be a field,MDLD/field ${}V$ a ${}K$-vector spaceMDLD/vector space and

$\varphi \colon V\longrightarrow V$

a linear mapping.MDLD/linear mapping Let ${}U\subseteq V$ be an ${}\varphi$-invariant linear subspace.MDLD/invariant linear subspace Then, for every polynomial ${}P\in K[X]$, the relation

${}P{\left(\varphi {|}_{U}\right)}={\left(P(\varphi )\right)}{|}_{U}\,$

holds, where here ${}\varphi {|}_{U}$ denotes the restricted mapping

(with respect to range and target).

Proof

This can be checked directly for the powers ${}X^{n}$ and for linear combinations of powers.

$\Box$

CorollaryCorollary 25.6 change

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 Let ${}U\subseteq V$ be a ${}\varphi$-invariant linear subspaceMDLD/invariant linear subspace and

$\varphi {|}_{U}\colon U\longrightarrow U$

the restriction to ${}U$ (also in the target). Then the minimal polynomialMDLD/minimal polynomial (endomorphism)

of ${}\varphi$ is a multiple of the minimal polynomial of ${}\varphi {|}_{U}$.

Proof

Let ${}\mu$ be the minimal polynomial of ${}\varphi$. For ${}u\in U$, we have

${}\mu {\left(\varphi {|}_{U}\right)}(u)=\mu (\varphi )(u)=0\,,$

due to Lemma 25.5 . Therefore, ${}\mu$ annihilates the restricted endomorphism ${}\varphi {|}_{U}$, and so ${}\mu$ is a multiple of the minimal polynomial of ${}\varphi {|}_{U}$.

$\Box$

Example Create referencenumber

We consider the permutation matrixMDLD/permutation matrix

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

The line ${}K{\begin{pmatrix}1\\1\\1\end{pmatrix}}$ is the eigenspaceMDLD/eigenspace for the eigenvalue ${}1$. Moreover,

${}U=\langle {\begin{pmatrix}1\\-1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\-1\end{pmatrix}}\rangle \,$

is an invariant linear subspaceMDLD/invariant linear subspace (which, over ${}\mathbb {C}$, according to Lemma 24.11 , can be decomposed further into smaller eigenspaces). With respect to the given basis, the restriction of the linear mapping to ${}U$ has the describing matrix

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

Therefore, the characteristic polynomialMDLD/characteristic polynomial of this matrix is

${}X(X+1)+1=X^{2}+X+1\,.$

This is also the minimal polynomialMDLD/minimal polynomial (matrix) of the restriction. The minimal polynomial of the permutation matrix is ${}X^{3}-1$, and indeed we have

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

in accordance with Corollary 25.6 .

Characterizations for trigonalizable mappings

A flag consists of the base point, the flagpole, the fabric and the space in which the fabric blows.

Definition

Let ${}K$ denote a field,MDLD/field and let ${}V$ denote a finite-dimensionalMDLD/finite-dimensional (fgvs) vector spaceMDLD/vector space of dimension ${}n=\dim _{K}{\left(V\right)}$. Then a chain of linear subspacesMDLD/linear subspaces

${}0=V_{0}\subset V_{1}\subset \ldots \subset V_{n-1}\subset V_{n}=V\,$

is called a flag in ${}V$.

Hence, a flag is a chain of linear subspaces, contained in each other, and where the dimension increase by ${}1$ in each step.

Definition

Let ${}V$ be a vector spaceMDLD/vector space of dimensionMDLD/dimension (vs) ${}n$, and let

$f\colon V\longrightarrow V$

be a linear mapping.MDLD/linear mapping A flagMDLD/flag

${}0=V_{0}\subset V_{1}\subset \ldots \subset V_{n-1}\subset V_{n}\,$

is called ${}f$-invariant, if ${}f(V_{i})\subseteq V_{i}$ holds for all

${}i=0,1,\ldots ,n-1,n$.

TheoremTheorem 25.10 change

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 the following statements are equivalent.

1. ${}\varphi$ is trigonalizable.MDLD/trigonalizable
2. There exists a ${}\varphi$-invariant flag.MDLD/invariant flag
3. The characteristic polynomialMDLD/characteristic polynomial ${}\chi _{\varphi }$ splits into linear factors.MDLD/linear factors (1K)
4. The minimal polynomialMDLD/minimal polynomial (linear) ${}\mu _{\varphi }$ splits into linear factors.MDLD/linear factors (1K)

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

invertible matrixMDLD/invertible matrix (set ${}n=\dim _{K}{\left(V\right)}$) ${}B\in \operatorname {Mat} _{n\times n}(K)$ such that ${}BMB^{-1}$ is an

upper triangular matrix.MDLD/upper triangular matrix

Proof

Form (1) to (2). Let ${}v_{1},\ldots ,v_{n}$ be a basis with the property that the describing matrix of ${}\varphi$ with respect to this matrix is an upper triangular form. The action of this matrix shows directly that the linear subspaces

${}V_{i}:=\langle v_{1},\ldots ,v_{i}\rangle \,$

are ${}\varphi$-invariantMDLD/invariant (linear) and that they form an invariant flag.

From (2) to (1). Let

${}0=V_{0}\subset V_{1}\subset \ldots \subset V_{n-1}\subset V_{n}=V\,$

be a ${}\varphi$-invariant flag.MDLD/invariant flag Due to Theorem 8.10 , there exists a basis ${}v_{1},\ldots ,v_{n}$ of ${}V$ such that

${}V_{i}=\langle v_{1},\ldots ,v_{i}\rangle \,.$

Since this flag is invariant, we have

${}\varphi (v_{i})=b_{1i}v_{1}+b_{2i}v_{2}+\cdots +b_{ii}v_{i}\,.$

Therefore, the describing matrixMDLD/describing matrix of ${}\varphi$ with respect to this basis is upper triangular.

From (1) to (3). The characteristic polynomial of ${}\varphi$ is equal to the characteristic polynomial ${}\chi _{M}$, where ${}M$ denotes a describing matrix with respect to an arbitrary basis. We may assume that ${}M$ is an upper triangular matrix. Then, according to Lemma 16.4 , the characteristic polynomial is the product of the linear factors arising from the diagonal entries.

From (3) to (4). Due to Corollary 24.3 , the minimal polynomial divides the characteristic polynomial.

From (4) to (2). We prove the statement by induction over ${}n$, the cases

${}n=0,1\,$

being clear. Due to the condition and Corollary 24.3 and Theorem 23.2 , the mapping ${}\varphi$ has an eigenvalue. Because of Lemma 25.4 , there exists an ${}(n-1)$-dimensional linear subspaceMDLD/linear subspace

${}V_{n-1}\subset V\,$

which is ${}\varphi$-invariant.MDLD/invariant (linear) Due to Corollary 25.6 , the minimal polynomial of the restriction ${}\varphi {|}_{V_{n-1}}$ divides the minimal polynomial of ${}\varphi$, therefore, it also splits into linear factors. By the induction hypothesis, there exists an ${}\varphi {|}_{V_{n-1}}$-invariant flag

${}V_{1}\subset V_{2}\subset \ldots \subset V_{n-2}\subset V_{n-1}\,,$

and so this is also a ${}\varphi$-invariant flag.

The supplement follows as follows. Suppose that the trigonalizable mapping ${}\varphi$ is described by the matrix ${}M$ with respect to the basis ${}{\mathfrak {u}}$, and by the upper triangular matrix ${}T$ with respect to the basis ${}{\mathfrak {v}}$ Then, because of Corollary 11.12 the relation ${}T=BMB^{-1}$ holds, where ${}B$ is the transformation matrix of the base change.

$\Box$

Remark

The method to find an invariant flag, which is described in the proof of the implication (4) ${}\Rightarrow$ (2) of Theorem 25.10 and which rests on Lemma 25.4 , is constructive. If the restriction ${}\varphi _{|}U_{i}$ to some invariant linear subspace ${}U_{i}$ already constructed does not have an eigenvalue, then we know that the linear mapping is not trigonalizable.

TheoremCreate referencenumber

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

trigonalizable.MDLD/trigonalizable

Proof

This follows from Theorem 25.10 and the Fundamental theorem of algebra.

$\Box$

Example Create referencenumber

We consider a real ${}2\times 2$-matrix ${}M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}$. The characteristic polynomialMDLD/characteristic polynomial is

${}{\begin{aligned}\chi _{M}&=\det {\left(xE_{2}-M\right)}\\&=\det {\begin{pmatrix}x-a&-b\\-c&x-d\end{pmatrix}}\\&=(x-a)(x-d)-bc\\&=x^{2}-(a+d)x+ad-bc\\&={\left(x-{\frac {a+d}{2}}\right)}^{2}-{\left({\frac {a+d}{2}}\right)}^{2}+ad-bc\\&={\left(x-{\frac {a+d}{2}}\right)}^{2}-{\left({\frac {a-d}{2}}\right)}^{2}-bc.\end{aligned}}$

This polynomial splits into (real) linear factors if and only if ${}{\left({\frac {a-d}{2}}\right)}^{2}+bc\geq 0$. The matrix is trigonalizableMDLD/trigonalizable exactly in this case, due to Theorem 25.10 .

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