# Proof about eigenspace having $\text{rank}(P)$ as dimension

I'm stuck trying to prove the following.

Let $N,P \in \mathbb{R}^{n \times n}$ with $P \neq O$. Prove the following: if $P=NP$ then $N$ has an eigenspace $E$ with $dim(E) \geq \text{rank}(P)$.

I thought maybe doing something like: \begin{align*} &&P= NP \\ &\iff& P*P^-{1} = N\\ &\iff& N = P*\mathbb{I_n} * P^{-1} \end{align*} Now the $\mathbb{I_n}$ has the eigenvectors on its diagonal, or is that not correct?

The question does not state anything about the matrices being diagonalizable. I don't think my approach is any useful. Can someone point me in the right direction ?

• $P$ might not be invertible. Commented Dec 18, 2016 at 17:17
• The image of $P$ is invariant under $N$, so... Commented Dec 18, 2016 at 17:30

Notice that $P(x)=NP(x)=N(Px)$ for all $x\in \mathbb R^n$.
This means that $N(y)=y$ for all vector in the range of $P$.
So the $1$-eigenspace of $N$ contains the range of $P$, and therefore has dimension at least $\text{rank}(P)$