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


$ Q^* Q $ has eigenvalue list $ (s_1, \dots, s_n) $ ?


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 $Q$.

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  • $\begingroup$ Fine, nice answer! $\endgroup$ – gloerchen Aug 18 '11 at 14:01
  • 2
    $\begingroup$ One additional remark: the condition that $s_i s_{n+1-i}$ is constant is also a sufficient condition for the existence of such $Q$, in which case $Q$ can even be a matrix with real entries. This follows from the easy fact that, for any $s>0$, there is a real 2 by 2 matrix $Q_s$ such that $Q_s^2=1$ and $Q_s^* Q_s$ is diagonal with diagonal entries $s,1/s$. Indeed, if $s_i s_{n+1-i}=1$, the block diagonal matrix with blocks $Q_{s_1},\dots,Q_{s_{[n/2]}}$ (and $1$ if $n$ is odd) does the job. $\endgroup$ – Mikael de la Salle Aug 18 '11 at 14:46

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