1
$\begingroup$

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,

enter image description here

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.

enter image description here

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?

$\endgroup$
  • 1
    $\begingroup$ 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 Sep 26 '16 at 15:58
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. Sep 26 '16 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 $A$. $\endgroup$ – Robert Israel Sep 26 '16 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.