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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 ?

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  • $\begingroup$ $P$ might not be invertible. $\endgroup$
    – Ken Duna
    Commented Dec 18, 2016 at 17:17
  • $\begingroup$ The image of $P$ is invariant under $N$, so... $\endgroup$
    – Henricus V.
    Commented Dec 18, 2016 at 17:30

1 Answer 1

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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 $\text{rank}(P)$

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