Diagonalizability and the sum of eigenspaces

Lemma. If $\lambda_1,\ldots,\lambda_r$ are distinct and $w_i\in E_{\lambda_i}$ satisfy $w_1+\cdots+w_r=0$, then all $w_i=0$.

*Proof, by induction on $r$.* The case $r=1$ is trivial. For $r\ge2$, apply $f-\lambda_r\,\mathrm{id}$: it maps $w_i$ to $(\lambda_i-\lambda_r)w_i\in E_{\lambda_i}$ and kills $w_r$. So $\sum_{i<r}(\lambda_i-\lambda_r)w_i=0$, and by induction $(\lambda_i-\lambda_r)w_i=0$, hence $w_i=0$ for $i<r$ (as $\lambda_i\neq\lambda_r$). Then $w_r=0$. $\square$

Consequently the sum of the eigenspaces is direct and $\sum_\lambda\dim E_\lambda=\dim\bigoplus_\lambda E_\lambda\le n$.

($\Rightarrow$) Let $(b_1,\ldots,b_n)$ be a basis of eigenvectors, with $f(b_j)=d_jb_j$. For each eigenvalue $\lambda$, let $n_\lambda$ be the number of $j$ with $d_j=\lambda$. These $b_j$ are independent vectors of $E_\lambda$, so $n_\lambda\le\dim E_\lambda$. Since $\sum_\lambda n_\lambda=n$, we get $n\le\sum_\lambda\dim E_\lambda\le n$, hence equality.

($\Leftarrow$) Suppose $\sum_\lambda\dim E_\lambda=n$. Choose a basis of each $E_\lambda$ and let $\mathcal{B}$ be the union of these bases: it has $n$ vectors. It is linearly independent: a vanishing combination groups into one vector $w_\lambda\in E_\lambda$ per eigenvalue with $\sum_\lambda w_\lambda=0$; by the Lemma every $w_\lambda=0$, and then all coefficients vanish because each family is a basis of $E_\lambda$. So $\mathcal{B}$ is a basis of $E$ made of eigenvectors, and $f$ is diagonalizable.

Conclusion. $f$ is diagonalizable iff $\sum_\lambda\dim E_\lambda=\dim E$.