$\newcommand{\End}{\operatorname{End}}\newcommand{\id}{\operatorname{id}}\newcommand{\im}{\operatorname{im}}\newcommand{\family}[1]{\left(#1\right)}\newcommand{\mat}[1]{\boldsymbol{#1}}$Let $V$ be a finite-dimensional vector space over some algebraically closed field $\mathbb{F}$ and $\Phi \in \End(V)$.
$\Phi$ is diagonalisable iff $V = \ker(\Phi – \lambda\id) \oplus \im(\Phi – \lambda\id)$ for all $\lambda \in \mathbb{F}$.
Proof: If $\Phi$ has a diagonal matrix w.r.t. $\family{e_k}_{k=0}^{n-1}$, then $\Phi – \lambda\id$ is also diagonal w.r.t. $\family{e_k}_{k=0}^{n-1}$. Hence (invoking a known proposition) $V = \ker(\Phi – \lambda\id) \oplus \im(\Phi – \lambda\id)$.
Conversely, since $\mathbb F$ is algebraically closed, there exists a basis $\family{e_k}_{k=0}^{n-1}$ of $V$ for which the matrix $\mat\Phi$ of $\Phi$ w.r.t. $\family{e_k}_{k=0}^{n-1}$ is upper triangular. Let $\family{\lambda_k}_{k=0}^{n-1}$ be the diagonal of this matrix. Then (invoking a known theorem) $\family{\lambda_k}_{k=0}^{n-1}$ is precisely the eigenvalues of $\Phi$. Let $\family{\mu_j}_{j=0}^{m-1}$ be the distinct values of $\family{\lambda_k}_{k=0}^{n-1}$. By hypothesis, $\ker(\Phi – \mu_0\id) \oplus \im(\Phi – \mu_0\id) = V$, so it we wish to show that (by induction) $\im(\Phi – \mu_0\id) = \bigoplus_{j=1}^{m-1} \ker(\Phi – \mu_j\id)$.
I am having trouble with the converse. I know that showing $V = \bigoplus_{j=0}^{m-1} \ker(\Phi – \mu_j\id)$ and invoking another known proposition would conclude the proof. The proof would also work if I can somehow show that $\dim V = n = \sum_{j=0}^{m-1} \dim\ker(\Phi – \mu_j\id)$ since eigenspaces $\dim\ker(\Phi – \mu_j\id)$ of different eigenvalues are linearly independent.
I don't think this proof requires Jordan normal form so a proof involving only triangular and diagonal matrices would be the best.
Best Answer
If $\Phi$ is not diagonalizable, then there is some $\lambda$ for which $$\mathrm{ker}(\Phi - \lambda \mathrm{id}) \subsetneq \mathrm{ker}(\Phi - \lambda \mathrm{id})^2,$$ i.e. it has a Jordan chain of length at least $2$. (If not, then its JNF is a diagonal matrix!) Choose an element $v$ such that $w := (\Phi - \lambda \mathrm{id})v \ne 0$ but $(\Phi - \lambda \mathrm{id})^2 v = 0;$ then $$w \in \mathrm{ker}(\Phi - \lambda \mathrm{id}) \cap \mathrm{im}(\Phi - \lambda \mathrm{id}),$$ so the sum is not direct.