3
$\begingroup$

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) $ ?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

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 $Q$.

$\endgroup$
2
  • $\begingroup$ Fine, nice answer! $\endgroup$
    – gloerchen
    Aug 18, 2011 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. $\endgroup$ Aug 18, 2011 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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