# Proving Two Hermitian Matrices have the same eigenvectors

I am currently stuck on the following proof.

Suppose that a (n by n) unitary matrix U can be written as U=M+iN where M and N are Hermitian matrices.

Now assuming that M and N have n distinct eigenvalues it can be shown that they have the same eigenvectors.

My attempts to show it have been unsuccessful. I was wondering whether any one could give me advice on how to do this because I am getting nowhere at the moment.

Many Thanks

## 1 Answer

Since $I = U^*U = UU^*$, the following two operators are both equal to the identity:

1. $M^2 - N^2 + iMN - iNM$
2. $M^2 - N^2 -iMN + iNM$

This leads to $iMN - iNM = -iMN +iNM$ which gives us $$MN - NM = -MN + NM\\ 2MN = 2NM\\ MN = NM$$ so the two matrices $M$ and $N$ commute and must therefore have the same eigenvectors.

Commuting $n\times n$ matrices $M$ and $N$, each with with $n$ distinct eigenvalues will share eigenvectors because if $v$ is an eigenvector of $N$ with eigenvalue $l$, then so is $Mv$: $$N(Mv) = (NM)v = (MN)v = M(Nv) = M(lv) = l(Mv)$$ and since only scalar multiples of $v$ has that property, $Mv$ must be a scalar multiple of $v$, and therefore $v$ is an eigenvector of $M$ as well.