Let $A$ be an $n\times n$ matrix over a field $K$. Show that there exists an invertible matrix $P$ such that $P^{-1}AP$ is a diagonal sum of an invertible matrix and a nilpotent matrix. (Hint: use Fitting's Lemma.)
I guess one could choose $P$ to be made up of basis vectors, so then $P^{-1}AP$ is a diagonal matrix with the eigenvalues of $A$ on the diagonal. But what's a diagonal sum of two matrices? A sum of two matrices that are diagonal? A diagonal matrix where one part of it is an invertible matrix, and the rest is a nilpotent matrix?
Furthermore, how would I use Fitting's Lemma? It says that if a module $M$ is indecomposable (non-zero, can't be written as a direct sum of simple modules) and of finite-length (the longest chain of submodules is finite), every $End(M)$ is bijective or nilpotent.
I'm thinking that I can use the fact that an $R$-module is the same as an abelian group $M$ together with a ring homomorphism $End(M)$.
The matrix $A$ corresponds to an endomorphism $\varphi_A$ of the $K$-vector space $K^n$. Then, by Fitting's Lemma, there is a decomposition $K^n=V_1\oplus V_2$ such that $\varphi_A$ restricted to $V_1$ is nilpotent, and $\varphi_A$ restricted to $V_2$ is invertible, that is, there is a $K$-basis of $K^n$ such that the matrix associated to $\varphi_A$ in this basis is block-diagonal, and it's made by two blocks: one of them is given by a nilpotent matrix while the other one is an invertible matrix.