Let $E$ be a finite dimensional complex vector-space and $A\in\Hom(E)$. Prove that $A$ is diagonalizable iff for all $\l\in\C$: $E=\ker(A-\l)+\im(A-\l)$.
Since $\dim\ker(A-\l)+\dim\im(A-\l)=\dim E$ the second statement is equivalent to
$\ker(A-\l)\cap\im(A-\l)=\{0\}$. Suppose $A$ is diagonalizable and $x\in\ker(A-\l)\cap\im(A-\l)$, i.e. $(A-\l)x=0$ and $x=(A-\l)y$. Hence $y\in\ker(A-\l)^2=\ker(A-\l)$ and thus $x=0$. Conversely, assume $\ker(A-\l)\cap\im(A-\l)=\{0\}$: if $x\in\ker(A-\l)^2$, then $(A-\l)x\in\ker(A-\l)$, i.e. $(A-\l)x\in\ker(A-\l)\cap\im(A-\l)=\{0\}$ and therefore $x\in\ker(A-\l)$.