Eigenvalues of certain positive matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:30:18Z http://mathoverflow.net/feeds/question/73140 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73140/eigenvalues-of-certain-positive-matrices Eigenvalues of certain positive matrices gloerchen 2011-08-18T11:52:49Z 2011-08-18T13:38:28Z <p>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 $Q^*$). </p> <p>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 </p> <p>$Q \overline{Q}$ is a multiple of the identity matrix </p> <p>and </p> <p>$Q^* Q$ has eigenvalue list $(s_1, \dots, s_n)$ ? </p> http://mathoverflow.net/questions/73140/eigenvalues-of-certain-positive-matrices/73149#73149 Answer by alext87 for Eigenvalues of certain positive matrices alext87 2011-08-18T13:33:11Z 2011-08-18T13:38:28Z <p>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 </p> <p>$Q\overline{Q}=\lambda I_n$ </p> <p>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 </p> <p>$Q^*Qv_k = s_kv_k$</p> <p>Then, $Q^*Q^T(Q^{-T}Qv_k) = Q^*Qv_k = s_kv_k = \lambda(Q^{-T}Qv_k)$ that is </p> <p>$\lambda Qv_k=s_k Q^Tv_k$</p> <p>Multiplying by $\overline{Q}$ on both sides and conjugate we obtain,</p> <p>$\overline{\lambda} Q\overline{Q}\overline{v}_k = s_k Q Q^*\overline{v}_k$</p> <p>Since, $Q\overline{Q}=\lambda I_n$ and $s_k\neq0$ we have, </p> <p>$Q Q^*\overline{v}_k = (|\lambda|^2/s_k)\overline{v}_k$</p> <p>Moreover, $Q^* Q$ and $Q Q^*$ have the same eigenvalues and the monotonicity conditions on $s_1,\ldots ,s_n$ ensure we have, </p> <p>$\frac{|\lambda|^2}{s_n}=s_1,\text{ } \frac{|\lambda|^2}{s_{n-1}}=s_2,\text{ }\ldots , \text{ } \frac{|\lambda|^2}{s_1}=s_n$</p> <p>This shows the choice $s_n=s_{n-1}$, $s_1\neq s_2$ allows no such $Q$.</p>