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Eigenvalues of a Hermitian matrix are all real. Eigenvectors $x_1,x_2$ of two eigenvalues $\lambda_1, \lambda_2$ of a Hermitian matrix are orthogonal with respect to the conjugate product. The conjugate product is defined as $x_1^Hx_2 = \bar{x_1}^Tx_2$. A proof is given as the following,

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However, I am wondering if $x_1,x_2$ are orthogonal with respect to any inner product. A complex inner product is defined as the following, where $C$ is a field of complex numbers, and $F$ is the field of real numbers.

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If $<x_1,x_2> = 0$ then we say $x_1,x_2$ are orthogonal with respect to $<>$. Again my question is that does the orthogonality of two eigenvectors $x_1,x_2$ holds with respect to any inner product? If not, can anyone help provide a counter example?

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    $\begingroup$ The notion of Hermitian is (or can be) defined in terms of the inner product, so when you use a different inner product, all bets are off. $\endgroup$
    – copper.hat
    Commented Sep 26, 2016 at 15:58

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No, if you choose an arbitrary inner product there's no reason for the eigenvectors to be orthogonal. Try almost any example.

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  • $\begingroup$ Thank you! Could you help with a counterexample for vectors of $C^n$? I find another inner product $\text{Re} {x^Hy}$ but any vectors orthogonal wrt. the conjugate product will also be orthogonal wrt. $\text{Re} {x^Hy}$. $\endgroup$
    – Ralph B.
    Commented Sep 26, 2016 at 18:42
  • $\begingroup$ That's not an inner product: the Re spoils linearity. Try inner product $\langle x,y \rangle = x^H Q y$ where $Q$ is a positive definite (Hermitian) matrix that does not commute with $A$. $\endgroup$ Commented Sep 26, 2016 at 20:30

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