Endomorphism/Diagonalizable/Algebraic and geometric multiplicity/Fact/Proof

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.

To fact