I know how to prove:

Given an $n \times n$ matrix $A$ with $\lambda$ and $\mu$ being distinct eigenvalues and $x$ and $y$ are their corresponding eigenvectors. Show that $y^{H}x=0$.

However, if I change $A$ to be a normal matrix, I would like to ask if we can get a similar result that $y^{*}x=0$? (I denote the $*$ to be conjugate transpose.)

Thank you your attention so much.

Update for more.

I do not know how to prove this statement:

If $A$ is not Hermitian and suppose $Ax=\lambda x$ and $y^{H}A=\mu y^{H}$. Show that $y^{H}x \neq 0$. (Algebraic multiplicity of $\lambda =1$.)

My Trial: Suppose $y^{H}x=0$.

Then, from $y^{H}A=\mu y^{H}$, we have $y^{H}Ax=\mu y^{H}x \Rightarrow $ $\lambda y^{H}x=\mu y^{H}x=0.$

But I fail to derive contradiction:(

  • $\begingroup$ See, e.g., here. $\endgroup$ – Algebraic Pavel May 5 '14 at 10:18
  • $\begingroup$ @Pavel Thank you for your links. $\endgroup$ – nam May 5 '14 at 14:32

About your update:

Let $\mu\neq 0$. Then $$y^Hx = \frac1\mu(\mu y^H x) = \frac1\mu y^HAx = \frac\lambda\mu y^Hx$$ meaning that if $\lambda \neq \mu$, $y^Hx=0$ must hold.

As for the main question, I first offer a hint:

Hint: First show that if $A$ is normal, then $Ax=\lambda x$ holds iff $A^*x\bar\lambda x$.

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  • $\begingroup$ thx but why your second equality holds? $\endgroup$ – nam May 5 '14 at 7:03
  • $\begingroup$ Because I replaced $\mu y^H$ with $y^H A$ which is true. $\endgroup$ – 5xum May 5 '14 at 7:07
  • $\begingroup$ Thanks for your help...but the question asks me to prove $y^{H}x \neq 0$. Do I misunderstand your words or the question is wrong indeed? $\endgroup$ – nam May 5 '14 at 7:15
  • $\begingroup$ $y^Hx\neq 0$ is false if $\lambda \neq \mu$ and $\mu\neq 0$, as I have shown. $\endgroup$ – 5xum May 5 '14 at 7:17
  • $\begingroup$ Thank you for your comment very much. Could you please relieve my worry about the question that if $A$ is normal, can we claim that $y^{H}x=0$ as well? $\endgroup$ – nam May 5 '14 at 7:36

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