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First, a quick definition: I say that a matrix $M$ with complex entries is complex-orthogonal if $MM^T = I$. Note that $T$ here refers to the transpose, not the conjugate transpose. That is, we are taking the adjoint with respect to a bilinear (as opposed to sesquilinear) dot-product over $\Bbb C$. A complex-orthogonal matrix need not be unitary, and vice versa (unless of its entries are in $\Bbb R$).

Consider the following claim:

Claim: For any $A \in \Bbb C^{n \times n}$, there exists a complex-orthogonal $U$ such that $UAU^T$ has equal diagonal entries.

My Question: Does this claim hold?

It may be helpful to take a look at this paper which proves that for any $A$, there is an $S$ such that $SAS^{-1}$ has equal diagonal entries. It goes on to show that such an $S$ may be taken to be a unitary matrix. Another proof of the same fact is presented in p. 77 of R.A. Horn and C.R. Johnson’s Matrix Analysis (1985/7, Cambridge Univ. Press).

Notably, both of these proofs exploit facts for which an analog fails to exist in my case. For Horn and Johnson's proof, we need the fact that the unitary matrices form a compact subset of $\Bbb C^{n \times n}$. For the proof linked, we need the numerical range to have certain "nice properties", which I believe fail in my case.

Any insight here is appreciated.

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  • $\begingroup$ (I found my mistake) but can you give an example of $A,U$ ? $\endgroup$ – reuns Sep 18 '16 at 3:55
  • $\begingroup$ For example, we can take $$ U = \frac 14 \pmatrix{5&3i\\3i&-5}, \qquad UAU^T = \pmatrix{0&1\\1&0} $$ and compute $A = U^T[UAU^T]U$ to find $$ A = \frac 1{16}\pmatrix{30i & -34\\-34 & -30i} $$ $\endgroup$ – Omnomnomnom Sep 18 '16 at 4:06

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