Let $\gamma:V\rightarrow V$ linear. with $\lambda_1,\ldots,\lambda_k$ his eigenvalues. and $C_i=\ker(\lambda_i I_d - \gamma)$ for all $1\leq i\leq k$. Prove if $C=C_1+\cdots+C_k$ then $\dim(C)=\dim(C_1)+\cdots+\dim(C_k)$
Definition: $\gamma:V\rightarrow V$ is diagonalizable if the matrix associated to $\gamma$ is diagonal.
I have a idea of how prove this. Basically i need show
$\;\sum_{i=1}^k C_i=\bigoplus_{i=1}^k C_i$
I'm trying to generalize the idea with this:
Suppose that for $\;1\le i\neq j\le n\;$ , we have
$$x\in\ker (\lambda_iI-\gamma)\cap\ker(\lambda_jI-\gamma)\implies \gamma x=\lambda_ix=\lambda_jx$$
and since $\;\lambda_i\neq\lambda_j\;$ , we get that $\;x=0\;$ .
Moreover, as $C=C_1+C_2$ then $C=C_1\oplus C_2$
but i have serious problem trying to generalize that. can someone help me?