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$A$ 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?

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1 Answer 1

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$$\lambda_1(w,z)=(\lambda_1w,z)=\underbrace{(Aw,z)=(w,Az)}_\text{due to $A=A^\ast$}=\lambda_2(w,z)$$

and since $\lambda_1 \neq \lambda_2$, the inner product must be zero.

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  • $\begingroup$ Thanks! I will accept the answer in 10min $\endgroup$ Apr 20, 2018 at 14:22
  • $\begingroup$ @RadekMartinez no problem. The "issue" with your solution is that you were not using the self adjoint hypothesis anywhere. $\endgroup$ Apr 20, 2018 at 14:22

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