Endomorphism/Trigonalizable/Characterizations/Section

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 fact, 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 fact, the characteristic polynomial is the product of the linear factors arising from the diagonal entries.

From (3) to (4). Due to fact, 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 fact and fact, the mapping ${}\varphi$ has an eigenvalue. Because of fact, there exists an ${}(n-1)$-dimensional linear subspace

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

which is ${}\varphi$-invariant. Due to fact, 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 fact 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 fact and which rests on fact, 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 fact 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 fact.