Showing $A$ is diagonalizable if $A$ commutes with a projection.

Question

Let V be a finite dimensional real or complex vector space, B the base of V and A: V -> V an endomorphism. Prove:

i) $\forall b \in B$ exists a projection $P: V \rightarrow V$, such that $\operatorname{Im}(P) = \operatorname{span}({b})$ and $P(b) = b$.
ii) If $A$ commutes with all projections $P: V \rightarrow V$, then A is diagonalizable.

I'm especially having trouble with the second part and I would greatly appreciate if anyone could help me out.

Answer 1

Let $B=\{b_1, \ldots, b_n\}$ and let $P_i$ be the projection associated with basis element $b_i$ as given by (i).

Show that $$P_i A b_i = A P_i b_i = A b_i.$$ Conclude that this implies $A b_i = \lambda_i b_i$ for some scalar $\lambda_i$.