Let $\mathcal{H}$ be some Hilbert space, let $B(\mathcal{H})$ denote the bounded linear operators acting on $\mathcal{H}$, and let $M_n(B(\mathcal{H}))$ denote the $n \times n$ matrices with operator-valued entries. Let $A = [a_{ij}]$ be one such matrix. My question is,
If $\sum_{i,j=1}^n u_i^* a_{ij} u_j$ is a positive semi-definite (PSD) operator in $B(\mathcal{H})$ for all $n$-tuples $(u_1,…,u_n)$ of unitary elements in $B(\mathcal{H})$, does this imply that $A$ is PSD in $M_n(B(\mathcal{H}))$?
Best Answer
Yes. Let $\xi\in \mathcal H^n$. Choose $r $ such that $\|\xi_r\|\geq\|\xi_j\|$ for all $j $.
For each $j$, let $x_j\in B(\mathcal H)$ be a contraction such that $x_j\xi_r=\xi_j $. By the argument in this answer, there exists a unitary with $u_j\xi_r=\xi_j$. Then $$ \langle A\xi,\xi\rangle=\sum_{k,j}\langle a_{kj}\xi_j,\xi_k\rangle=\sum_{k,j}\langle a_{kj}u_j\xi_r,u_k\xi_r\rangle=\left\langle\left(\sum_{k,j}u_k^*a_{kj}u_j\right)\xi_r,\xi_r\right\rangle\geq0. $$