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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 $\mathbb{C}^{n\times1}$.

Migz
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  • https://math.stackexchange.com/questions/900540/complex-hermitian-matrix-real-eigenvalues-and-orthogonal-eigenvectors – Tsemo Aristide May 14 '18 at 14:41
  • 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. – ctst May 14 '18 at 14:41
  • You might want to name that Hermitian matrix and use that instead of $x=x^, \ y=y^$. – Berci May 14 '18 at 14:44
  • the product of x and y? – Migz May 14 '18 at 14:45
  • hi @Berci what do you mean name that Hermitian matrix? – Migz May 14 '18 at 14:46
  • $x$ and $y$ are eigenvectors of a matrix, aren't they? It's that matrix that satisfy the self adjointness property, not the vectors. – Berci May 14 '18 at 14:48
  • @Migz the standard inner product, i.e. sum over the "pointwise" product of the entries (one complex conjugated). By name the matrix one means give it a letter/name, i.e. let $A$ be that Hermitian matrix and $x,y$ two eigenvectors (now we have them names). – ctst May 14 '18 at 14:53
  • I'm sorry, don't get it. Can you please enlighten me further? – Migz May 14 '18 at 14:54
  • I see, but how do I do your first comment to try multiplying the product in two different ways?@ctst – Migz May 14 '18 at 14:55
  • Calculate $<Ax,y>$ and use that $A$ is selfadjoint. – ctst May 14 '18 at 15:40

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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 $y^*x=0$.

Martin Argerami
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