2 added 3 characters in body

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

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, $Q^Q^* Q$ and $QQ^$ 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$. 1 Suppose we are given$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$and$QQ^$have the same eigenvalues and the monotonicity conditions on$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\$.