A problem about diagonalize invariant subspaces

Question

Let $V$ be a non-zero finite-dimensional vector space, A belongs to End($V$). Also, for any invariant subspace $M$ of A, there exists an invariant subspace $N$ of A such that $V=M\oplus N$.

Prove: A is diagonalizable

Answer 1

Assuming you can use the Jordan canonical form of $A$, we have the following proof by contradiction:

Suppose not. Then there is some eigenvalue $\lambda$ and some generalized eigenvector $v$ such that $$(A-\lambda I)v \neq 0 \\ (A-\lambda I)^2v = 0.$$ In other words, $A (A-\lambda I)v = \lambda (A-\lambda I)v$ so that $\mathbb{C}\cdot (A-\lambda I)v$ is an $A$-invariant subspace.

By assumption, there is some $A$-invariant subspace $N$ such that $V = \left(\mathbb{C}\cdot (A-\lambda I)v\right) \oplus N$, so in particular, $v = c(A-\lambda I)v + n$ where $c \in \mathbb{C}$ and $n\in N$. Apply $A-\lambda I$ to both sides to show $$(A-\lambda I)v = c(A-\lambda I)^2v + (A-\lambda I)n = 0 + An - \lambda n \in N;$$ however, this shows $$(A-\lambda I)v \in (\mathbb{C}\cdot (A-\lambda I)v) \cap N = \{0\},$$ so $(A-\lambda I)v = 0$, a contradiction.