Let $V$ be a finite dimensional vector space over $k$, and $\phi$ an endomorphism of $V$. I am trying to prove that there exist $\phi_s, \phi_n$ which commute with each other, such that $\phi_s$ is diagonalizable, $\phi_n$ is nilpotent, and $\phi = \phi_s + \phi_n$.
Let $ \prod\limits_i (X - \lambda_i)^{n_i}$ be the characteristic polynomial of $\phi$, and let $$ V_i = \{v \in V : (\phi - \lambda_i 1_V)^{n_i} v = 0 \} $$ By the Chinese remainder theorem, there exists a polynomial $P \in k[X]$ such that $P \equiv 0 \pmod{X}$ and $P \equiv \lambda_i \pmod{ (X - \lambda_i)^{n_i}}$ for all $i$. This is clear whether or not one of the $\lambda_i$ is $0$. We set $\phi_s = P(\phi)$.
We have that $P(\lambda_i) = \lambda_i$. Also, $P(\phi)$ commutes with $(\phi - \lambda_i 1_V)^{n_i}$, from which it follows that $\phi_s$ stabilizes each $V_i$. Finally, the restriction of $\phi_s$ to $V_i$ is just scalar multiplication by $\lambda_i$, because we can write $P(X) - \lambda_i = g(X) (X - \lambda_i)^{n_i}$, hence $P(\phi) - \lambda_i 1_V = g(\phi)(\phi - \lambda_i)^{n_i}$.
The proof I'm reading claims that this information is enough to conclude that $\lambda_i$ are all the eigenvalues of $\phi_s$, $V_i$ are all the eigenspaces, and their direct sum is $V$. I don't see why. I can see that each $\lambda_i$ is an eigenvalue of $\phi_s$, and that each $V_i$ is contained in an eigenspace of $\phi_s$.
In general, if $P = \prod P_i \in k[X]$ with the $P_i$ pairwise relatively prime, you can write $\ker P(u) = \bigoplus \ker P_i(u)$ for any endomorphism $u$ (either using a Bézout relation, or invoking the structure of finite modules over a PID). Apply that to $P = \prod (X-\lambda_i)^{n_i}$ using the Cayley-Hamilton theorem.