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

Trigonalizable mappings

Definition

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

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 similar to an upper triangular matrix. The process of finding such a basis and to perform this base change is called trigonalization.

Example

We claim that the matrix

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

is trigonalizable. The matrix

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

is invertible with the 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 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.

Lemma

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

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

denote 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. Then ${}\varphi$ is 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 sum of ${}\varphi$-invariant linear subspaces.

Invariant linear subspaces

A 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 ${}M$-invariant 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.

Lemma

Let ${}K$ denote a field, and let ${}V$ be a ${}n$-dimensional vector space. Let

$\varphi \colon V\longrightarrow V$

be a linear mapping, and let ${}\lambda \in K$ be an eigenvalue of ${}\varphi$. Then there exists a ${}\varphi$-invariant 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. Hence, this mapping is not injective and, due to Corollary 11.9 , also not 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.

Lemma

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

$\varphi \colon V\longrightarrow V$

a linear mapping. Let ${}U\subseteq V$ be an ${}\varphi$-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.

$\Box$

Corollary

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

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

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

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

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

We consider the 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 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 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 polynomial of this matrix is

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

This is also the minimal polynomial 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, and let ${}V$ denote a finite-dimensional vector space of dimension ${}n=\dim _{K}{\left(V\right)}$. Then a chain of 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 space of dimension ${}n$, and let

$f\colon V\longrightarrow V$

be a linear mapping. A 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

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

Theorem

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

$\varphi \colon V\longrightarrow V$

denote a

linear mapping. Then the following statements are equivalent.

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

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

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.

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 ${}\varphi$-invariant 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. 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 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 subspace

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

which is ${}\varphi$-invariant. 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.

Theorem

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

trigonalizable.

Proof

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

$\Box$

Example

We consider a real ${}2\times 2$-matrix ${}M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}$. The 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 trigonalizable exactly in this case, due to Theorem 25.10 .

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