Endomorphism/Trigonalizable/Introduction/Section

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 shows. We will see in fact 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 fact. 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.

$\Box$

In particular, the preceding statement holds when

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

is a direct sum of ${}\varphi$-invariant linear subspaces.