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Given $C$ a hermitian complex non-singular square matrix, and $\Delta$ real and diagonal with distinct, non-zero diagonal elements, I need to find an orthogonal matrix $H$ such that $$C=H\Delta H^H.$$

Is there a way to do this? One thought was to find the eigendecomposition of $C$: $$C=GDG^H.$$ Since $D$ is also real and diagonal, and $G$ is orthogonal, it has the same structure as the first equation and I thought I might be able to calculate $H$ from $G$ and $D$. However, I haven't been able to make much progress.

Is there a way to determine (a possible, if not unique) $H$ from the assumptions?

(I've found similar questions but none exactly the same; for instance in Matrix equation with a constraint the matrix $H$ is not constrained to be orthogonal.)

MBaz
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  • Well, you must have conditions on $C$ such as is is Hermitian and the eigenvalues are the same as $\Delta$ otherwise no such $H$ exists. – copper.hat Apr 08 '16 at 16:44
  • I've added the condition that $C$ is hermitian, thanks for pointing out my omission. However, I don't think that $D$ and $\Delta$ are equal. I have a numerical example showing otherwise. I can post it if it would help. – MBaz Apr 08 '16 at 17:04
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    $D,\Delta$ are not necessarily equal, but the have the same set of values, so there is a permutation matrix $P$ that will rearrange them for you. Replace $G D G^$ by $G P \Delta P^T G^$. – copper.hat Apr 08 '16 at 17:37
  • @copper.hat Thanks! Most helpful. I also found a scale factor, $x$. So I can write $C=\sqrt{x}GP\Delta P^TG^H\sqrt{x}$. If $J=\sqrt{x}GP$, is it necessarily the case that each columnt of $H$ is a rotation of the corresponding column of $J$? – MBaz Apr 08 '16 at 21:19

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