Trigonalizable mapping/Eigentheorie/No proof/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 reverse statement is not true, as example shows.

Theorem

Let ${}K$ denote a field, and let ${}V$ denote an 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. The characteristic polynomial ${}\chi _{\varphi }$ has a factorization into linear factors.

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

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

Proof

This proof was not presented in the lecture.

$\Box$

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.

$\Box$