$A$ is a $n\times n$ complex matrix such that $A = A^*$, and let $\lambda_1$ and $\lambda_2$ be a real eigenvalues of $A$ with corresponding eigenvector $z$ and $w$ respectively. Prove that if $\lambda_1$ does not equal $\lambda_2$, then $w$ and $z$ are orthogonal.
My general idea is the multiply $\lambda_1$ by $\langle z , z \rangle$ , and then trying to do something to show $\langle z , w \rangle = 0$. Though I am unable to find a way of showing this from that.... Any ideas?
$$\lambda_1(w,z)=(\lambda_1w,z)=\underbrace{(Aw,z)=(w,Az)}_\text{due to $A=A^\ast$}=\lambda_2(w,z)$$
and since $\lambda_1 \neq \lambda_2$, the inner product must be zero.