How would you show any matrix in $F^{n\times n}$ is similar to a matrix of the form
$$\begin{pmatrix} R&0\\ 0&S \end{pmatrix}$$
where $R$ is nilpotent and $S$ is invertible?
Would you designate specific vector spaces and show linear transformations restricted to kernels?
Note that the problem is equivalent to finding a decomposition of the vector space $F^n$ into a direct sum $V\oplus W$ of $\phi$-stable subspaces ($\phi$ being the linear operator defined by your matrix) such that the restrictions of $\phi$ to $V$ and to $W$ are nilpotent respectively invertible. Once one has such a decomposition, choosing bases in $V$ and $W$ and expressing $\phi$ on the combined basis suffices.
Now let $P\in K[X]$ be any monic polynomial such that $P[\phi]=0$; you may take it to be the minimal polynomial of$~\phi$, or its characteristic polynomial if you prefer. Write $P=X^mQ$ with $m\geq0$ and $Q$ is not divisible by$~X$. Then since $X^m$ and $Q$ are relatively prime, $F^n=\ker(\phi^m)\oplus\ker(Q[\phi])$, by what many people call the Chinese remainder theorem*. The restriction of $\phi$ to $V=\ker(\phi^m)$ is obviously nilpotent and the restriction of $\phi$ to $W=\ker(Q[\phi])$ is invertible, as $W\cap\ker(\phi)\subseteq W\cap\ker(\phi^k)=W\cap V=\{0\}$.
Alternative ways to see that $\phi|_W$ is invertible are that it cannot have $0$ as eigenvalue, since this is no root of the polynomial$~Q$ annihilating $\phi|_W$, or by explictely computing the inverse as $S[\phi]$ where the polynomial$~S$ is the quotient $(1-c^{-1}Q)/X$ with $c=Q[0]$ the nonzero constant term of$~Q$ (use $XS\equiv1\pmod Q$).
Another possible approach is to consider the kernels of successive powers $\phi^k$. Since the dimensions of the kernels increase weakly with$~k$, it must happen that $\ker(\phi^{k+1})=\ker(\phi^k)$ for some $k\in\Bbb N$. At this point $\def\im{\operatorname{im}}\ker(\phi^k)\cap\im(\phi^k)\subseteq\ker(\phi^k)\cap\im(\phi)=\{0\}$, so the sum $\ker(\phi^k)+\im(\phi^k)$ is direct, and fills the whole space by rank-nullity. Then restriction of $\phi$ to $V=\ker(\phi^k)$ is obviously nilpotent (of order$~k$), and its restriction to $W=\im(\phi^k)$ is surjective since $\im(\phi^{k+1})=\im(\phi^k)$; we have our direct sum decomposition.
*A more appropriate name would be kernel decomposition theorem. The Chinese remainder theorem for (the PID) $F[X]$ just states that any system of congruences $Q\equiv R_i\pmod{P_i}$ ($i=1,\ldots,k$) with pairwise relatively prime polynomials $P_i$ has a unique solution$~Q$ modulo the product$~P=P_1P_2\ldots P_k$. The result needed here is that in this case for any endomorphism$~\phi$ one has $$ \ker(P[\phi]) = \ker(P_1[\phi])\oplus\cdots\oplus\ker(P_k[\phi]) $$ From the Chinese remainder theorem one gets the existence, for $1\leq j\leq n$, of $Q_i\in F[X]$ satisfying $Q_j\equiv \delta_{i,j}\pmod{P_i}$ (for $i=1,\ldots,k$). With $\pi_j$ the restriction to $\ker(P[\phi])$ of $Q_j[\phi]$, it acts by the scalar $\delta_{i,j}$ on the subspace $\ker(P_i[\phi])$ (since $Q_j\equiv \delta_{i,j}\pmod{P_i}$), its image is contained in $\ker(P_j[\phi])$ (since $P_jQ_j\equiv0\pmod P$) and $\pi_1+\cdots+\pi_k=1$ (since $Q_1+\cdots+Q_k\equiv1\pmod P$). This implies that the $\pi_j$ are the projections onto the factors of a direct sum decomposition as announced. See also this answer.
Do Jordan decomposition and put the block corresponding to the zero eigenvalue in the left corner.
