Endomorphism/Multiplicities/Diagonalizability/Section

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 ${}\varphi$ is diagonalizable if and only if the characteristic polynomial ${}\chi _{\varphi }$ is a product of linear factors and if for every zero ${}\lambda$ with algebraic multiplicity ${}\mu _{\lambda }$, the identity

${}\mu _{\lambda }=\dim _{K}{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\,$

holds.

Proof

If ${}\varphi$ is diagonalizable, then we can assume at once that ${}\varphi$ is described by a diagonal matrix with respect to a basis of eigenvectors. The diagonal entries of this matrix are the eigenvalues, and these occur as often as their geometric multiplicity tells us. The characteristic polynomial can be read off directly from the diagonal matrix, every diagonal entry ${}\lambda$ constitutes a linear factor ${}X-\lambda$.

For the other direction, let ${}\lambda _{1},\ldots ,\lambda _{k}$ denote the different eigenvalues, and let

${}\mu _{i}:=\mu _{\lambda _{i}}(\varphi )=\dim _{K}{\left(\operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}\right)}\,$

denote the (geometric and algebraic) multiplicities. Due to the condition, the characteristic polynomial factors in linear factors. Therefore, the sum of these numbers equals ${}n=\dim _{K}{\left(V\right)}$. Because of fact, the sum of the eigenspaces

${}\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{k}}{\left(\varphi \right)}\subseteq V\,$

is direct. By the condition, the dimension on the left is also ${}n$, so that we have equality. Due to fact, ${}\varphi$ is diagonalizable.

$\Box$

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. Suppose that the characteristic polynomial ${}\chi _{\varphi }$ factors into different linear factors. Then ${}\varphi$ is

diagonalizable.

Proof

See exercise.

$\Box$

This gives also a new proof for fact.