For a matrix $ Q = (q\_{ij}) \in GL\_n(\mathbb{C}) $ let 
$ \overline{Q} = (\overline{q\_{ij}}) $ be the matrix obtained by entry-wise complex 
conjugation (equivalently, $ \overline{Q} $ is the transpose of the adjoint $ Q^\* $). 


The question is: Assume there are given positive real numbers 
$ s\_1 \geq s\_2 \cdots \geq s\_n > 0 $, does there exist $ Q \in GL\_n(\mathbb{C}) $ 
such that 


$ Q \overline{Q} $ is a multiple of the identity matrix 


and 


$ Q^\* Q $ has eigenvalue list $ (s\_1, \dots, s\_n) $ ? 



Eigenvalues of certain positive matrices
Suppose we are given $s\_1\geq s\_2\geq \ldots \geq s\_n>0$ and let $Q\in GL\_n(\mathbb{C})$ satisfying the two conditions. Then 


$Q\overline{Q}=\lambda I\_n$ 


for some $\lambda\in\mathbb{C}$ and hence, by transposing, $Q^\*Q^T=\lambda I\_n$. Pick $v\_k\in\mathbb{C}^n$ such that 


$Q^\*Qv\_k = s\_kv\_k$


Then, $Q^\*Q^T(Q^{-T}Qv\_k) = Q^\*Qv\_k = s\_kv\_k = \lambda(Q^{-T}Qv\_k)$ that is 


$\lambda Qv\_k=s\_k Q^Tv\_k$


Multiplying by $\overline{Q}$ on both sides and conjugate we obtain,


$\overline{\lambda} Q\overline{Q}\overline{v}\_k = s\_k Q Q^\*\overline{v}\_k$


Since, $Q\overline{Q}=\lambda I\_n$ and $s\_k\neq0$ we have, 


$Q Q^\*\overline{v}\_k = (|\lambda|^2/s\_k)\overline{v}\_k$


Moreover, $Q^\* Q$ and $Q Q^\*$ have the same eigenvalues and the monotonicity conditions on $s\_1,\ldots ,s\_n$ ensure we have, 


$\frac{|\lambda|^2}{s\_n}=s\_1,\text{ } \frac{|\lambda|^2}{s\_{n-1}}=s\_2,\text{ }\ldots , \text{ } \frac{|\lambda|^2}{s\_1}=s\_n$


This shows the choice $s\_n=s\_{n-1}$, $s\_1\neq s\_2$ allows no such $Q$.


