Linear mapping/Diagonalizable/Direct sum of eigenspaces/Fact/Proof

Proof

If ${}\varphi$ is diagonalizable, then there exists a basis ${}v_{1},\ldots ,v_{n}$ of ${}V$ consisting of eigenvectors. Hence,

${}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\langle v_{i},\,{\text{the eigenvalue of }}v_{i}{\text{ is }}\lambda \rangle \,.$

Therefore,

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

That the sum is direct was noticed in fact. If, the other way round,

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

holds, then we can choose in every eigenspace a basis. These bases consist of eigenvectors and yield together a basis of ${}V$.

To fact