Problem concerning eigenspace and rank of some matrix Problem: Let $N,P \in \mathbb{R}^{n \times n}$ be matrices, and let $P \neq 0$. Suppose that $P = NP$ and that $P$ is diagonalizable. Prove then that $N$ has an eigenspace with dimension greater than or equal to the rank of $P$.
Attempt: Let $E_{\lambda_i}$ be an eigenspace of $N$ corresponding to the eigenvalue $\lambda_i$. Then we have to prove that $\text{rank}(P) \leq \dim(E_{\lambda_i})$. Since $P$ is diagonalizable, there exists an invertible matrix $B$ such that $B^{-1} P B = D$ is a diagonal matrix. We know that the equation $n = \text{rank}(P) + \text{nullity}(P)$ always holds. Now, I tried to relate the rank of $P$ to the rank of $NP$, but I don't know how.

$N$ always has an eigenspace of dimension$>=$ rank $P$ as:
$N(Pe_i)=Pe_i$     ;  $ i=1,2,...,n $
$Pe_i$ is $i^{th}$ column of P, so, all columns of are there in eigen space of $N$, corresponding to e.v $1$. This proves the result. 
I don't think diagonalability of $P$ is required to prove this.
