# Trigonalizable mapping/Eigentheorie/No proof/Section

## Definition

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

upper triangular matrix.

Diagonalizable linear mappings are in particular trigonalizable. The reverse statement is not true, as example shows.

## Theorem

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

${\displaystyle \varphi \colon V\longrightarrow V}$

denote a

linear mapping. Then the following statements are equivalent.
1. ${\displaystyle {}\varphi }$ is trigonalizable.
2. The characteristic polynomial ${\displaystyle {}\chi _{\varphi }}$ has a factorization into linear factors.
If ${\displaystyle {}\varphi }$ is trigonalizable and is described by the matrix ${\displaystyle {}M}$ with respect to some basis, then there exists an invertible matrix

${\displaystyle {}B\in \operatorname {Mat} _{n\times n}(K)}$ such that ${\displaystyle {}BMB^{-1}}$ is an upper triangular matrix.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

## Theorem

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

### Proof

This follows from fact and the Fundamental theorem of algebra.

${\displaystyle \Box }$