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If $x$ and $y$ are eigenvectors of a Hermitian matrix corresponding to distinct eigenvalues, then $x$ and $y$ are orthogonal with respect to the standard inner product on $\mathbb{C}^{n\times1}$.

What I know that $x$ and $y$ are hermitian if $x = x^*$ and $y = y^*$ and that the standard inner product of $\mathbb{C}^{n\times1}$ is $y^*x$. But I don't know how to relate the orthogonality with respect to the standard inner product $\mathbb{C}^{n\times1}$.

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  • $\begingroup$ math.stackexchange.com/questions/900540/… $\endgroup$ – Tsemo Aristide May 14 '18 at 14:41
  • $\begingroup$ Orthogonality means that $y^\star x=0$. Now use the fact, that $x$ and $y$ are eigenvectors of a Hermitian matrix. Hint: Try multiplying the product in two different ways with the matrix and show that the result should be the same. Then use the fact, that the eigenvalues are different. $\endgroup$ – ctst May 14 '18 at 14:41
  • $\begingroup$ You might want to name that Hermitian matrix and use that instead of $x=x^*, \ y=y^*$. $\endgroup$ – Berci May 14 '18 at 14:44
  • $\begingroup$ the product of x and y? $\endgroup$ – Migz May 14 '18 at 14:45
  • $\begingroup$ hi @Berci what do you mean name that Hermitian matrix? $\endgroup$ – Migz May 14 '18 at 14:46
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Here $x$ and $y$ are not "hermitian". That applies to square matrices, not to vectors. The situation you have is that there is a hermitian matrix $H$ with $$ Hx=\lambda x,\ \ Hy=\mu y,$$ with $\lambda\ne\mu$.

Now, as $\lambda,\mu$ are both real (from $H$ hermitian), $$ \lambda y^*x=y^*(\lambda x)=y^*Hx=(Hy)^*x=(\mu y)^*x={\mu}y^*x. $$ From $\lambda\ne\mu$, the above equality can only happen if $y^*x=0$.

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