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 $\phi :V\to V$ is called trigonalizable, if there exists a basisMDLD/basis (vs) such that the describing matrix of $\phi$ 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 $K^n\to K^n$ 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

$BM{B}^{-1}$

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

$M=\left(\begin{array}{cc}3& 1\\ -1& 1\end{array}\right)$

is trigonalizable.MDLD/trigonalizable The matrix

$B=\left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)$

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

$B^{-1}=\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right).$

A direct computation shows

$BM{B}^{-1}=\left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}3& 1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& -2\\ -1& 3\end{array}\right)=\left(\begin{array}{cc}3& 2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}2& -3\\ -2& 5\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right).$

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

${\chi }_M=det\left(\begin{array}{cc}X-3& -1\\ 1& X-1\end{array}\right)=(X-3)(X-1)+1=X^2-4X+4=(X-2{)}^2,$

and this factors into linear factors.

LemmaCreate referencenumber

Let $V_1,\dots ,V_n$ denote finite-dimensionalMDLD/finite-dimensional vector spacesMDLD/vector spaces over the fieldMDLD/field $K$, let

${\phi }_i:V_i⟶V_i$

denote linear mappings,MDLD/linear mappings and let

$\phi ={\phi }_1\times \cdots \times {\phi }_n:V_1\times \cdots \times V_n⟶V_1\times \cdots \times V_n$

denote the product mapping.MDLD/product mapping Then $\phi$ is trigonalizableMDLD/trigonalizable

if and only if this holds for all ${\phi }_i$.

Proof

See Exercise 25.5 .

$◻$

In particular, the preceding statement holds when

$V=\underset{i\in I}{⨁}{V}_i$

is a direct sumMDLD/direct sum (vs) of $\phi$-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

$M=\left(\begin{array}{ccccc}{}_1& \ast & \cdots & \cdots & \ast \\ 0& {}_2& \ast & \cdots & \ast \\ ⋮& \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& {}_{n-1}& \ast \\ 0& \cdots & \cdots & 0& {}_n\end{array}\right).$

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

$M{e}_j=a_{1j}{e}_1+\cdots +a_{jj}{e}_j.$

In particular, $e_1$ is a eigenvector with the eigenvalue $a_{11}$. It is typical for a trigonalizable mapping that the linear subspace

$V_j=⟨e_1,\dots ,e_j⟩$

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

$\phi :V⟶V$

be a linear mapping,MDLD/linear mapping and let $\lambda \in K$ be an eigenvalueMDLD/eigenvalue of $\phi$. Then there exists a $\phi$-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 $\phi -\lambda {\mathrm{Id}}_V$ 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,

$B:=\mathrm{Im}(\phi -\lambda {\mathrm{Id}}_V)\subset V$

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

$\phi (u)=\lambda u+(\phi -\lambda {\mathrm{Id}}_V)u\in U+B\subseteq U.$

Hence, the image of $U$ belongs to $U$, that is, $U$ is $\phi$-invariant.

$◻$

When $U\subseteq V$ is an $\phi$-invariant linear subspace, and $P\in K[X]$ is a polynomial, then $U$ is also $P(\phi )$-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

$\phi :V⟶V$

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

$P\left(\phi {|}_U\right)=(P(\phi )){|}_U$

holds, where here $\phi {|}_U$ 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.

$◻$

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

$\phi :V⟶V$

be a linear mapping.MDLD/linear mapping Let $U\subseteq V$ be a $\phi$-invariant linear subspaceMDLD/invariant linear subspace and

$\phi {|}_U:U⟶U$

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

of $\phi$ is a multiple of the minimal polynomial of $\phi {|}_U$.

Proof

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

$\mu \left(\phi {|}_U\right)(u)=\mu (\phi )(u)=0,$

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

$◻$

Example Create referencenumber

We consider the permutation matrixMDLD/permutation matrix

$\left(\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right).$

The line $K\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$ is the eigenspaceMDLD/eigenspace for the eigenvalue $1$. Moreover,

$U=⟨\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)⟩$

is an invariant linear subspaceMDLD/invariant linear subspace (which, over $ℂ$, 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

$\left(\begin{array}{cc}0& -1\\ 1& -1\end{array}\right).$

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

$0=V_0\subset V_1\subset \dots \subset V_{n-1}\subset V_n=V$

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

$f:V⟶V$

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

$0=V_0\subset V_1\subset \dots \subset V_{n-1}\subset V_n$

is called $f$-invariant, if $f(V_i)\subseteq V_i$ holds for all

$i=0,1,\dots ,n-1,n$.

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

$\phi :V⟶V$

denote a

linear mapping.MDLD/linear mapping Then the following statements are equivalent.

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

Proof

Form (1) to (2). Let $v_1,\dots ,v_n$ be a basis with the property that the describing matrix of $\phi$ with respect to this matrix is an upper triangular form. The action of this matrix shows directly that the linear subspaces

$V_i:=⟨v_1,\dots ,v_i⟩$

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

From (2) to (1). Let

$0=V_0\subset V_1\subset \dots \subset V_{n-1}\subset V_n=V$

be a $\phi$-invariant flag.MDLD/invariant flag Due to Theorem 8.10 , there exists a basis $v_1,\dots ,v_n$ of $V$ such that

$V_i=⟨v_1,\dots ,v_i⟩.$

Since this flag is invariant, we have

$\phi (v_i)=b_{1i}{v}_1+b_{2i}{v}_2+\cdots +b_{ii}{v}_i.$

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

From (1) to (3). The characteristic polynomial of $\phi$ 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 $\phi$ 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 $\phi$-invariant.MDLD/invariant (linear) Due to Corollary 25.6 , the minimal polynomial of the restriction $\phi {|}_{V_{n-1}}$ divides the minimal polynomial of $\phi$, therefore, it also splits into linear factors. By the induction hypothesis, there exists an $\phi {|}_{V_{n-1}}$-invariant flag

$V_1\subset V_2\subset \dots \subset V_{n-2}\subset V_{n-1},$

and so this is also a $\phi$-invariant flag.

$◻$

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 $\phi {|}_{U_i}$ 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 $M\in {\mathrm{Mat}}_{n\times n}(ℂ)$ 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.

$◻$

Example Create referencenumber

We consider a real $2\times 2$-matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. The characteristic polynomialMDLD/characteristic polynomial is

$\begin{array}{cc}{\chi }_M& =det(x{E}_2-M)\\ & =det\left(\begin{array}{cc}x-a& -b\\ -c& x-d\end{array}\right)\\ & =(x-a)(x-d)-bc\\ & =x^2-(a+d)x+ad-bc\\ & ={(x-\frac{a+d}{2})}^2-{\left(\frac{a+d}{2}\right)}^2+ad-bc\\ & ={(x-\frac{a+d}{2})}^2-{\left(\frac{a-d}{2}\right)}^2-bc.\end{array}$

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

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