Positivity of an $n \times n$ matrix in $M_n(B(\mathcal H))$

Question

Let $\mathcal H$ be a Hilbert space and $B(\mathcal H)$ be the set of all bounded operators on $\mathcal H$. Let $B \subset B(\mathcal H)$ be a $C^*$-subalgebra of $B(\mathcal H)$. Let $\xi \in H$ be such that $\overline {B\xi}=\mathcal H$. Now consider $M_n(B)$, the set of all $n \times n$ matrices with entries from $B$. Let $a=[a_{ij}] \in M_n(B)$ satisfies the following condition,
For any $b_1,b_2,\cdots, b_n$ in $B$ we have $~~\displaystyle\sum_{i=1}^n\sum_{j=1}^n\left\langle a_{ij}b_j\xi,b_i\xi \right\rangle \ge 0.$ Now I want to show that the matrix $a \ge 0$ in $M_n(B)$.
My approach: To show that $a \ge 0$ in $M_n(B)$, I want to prove $b^*ab \ge 0$ in $M_n(B)$ for all column matrices $0\ne b$ with $n$ entries from $B$. So let $b=[b_1,b_2,\cdots, b_n]^T$ where $b_i \in B$. Now notice that $$b^*ab = [b_1^*,b_2^*,\cdots, b_n^*] [a_{ij}] [b_1,b_2,\cdots, b_n]^T =\sum_{i=1}^n\sum_{j=1}^n b_i^*a_{ij}b_j.$$ Now I want to show that, $\displaystyle \sum_{i=1}^n\sum_{j=1}^n b_i^*a_{ij}b_j \ge 0$ for any $b_1,b_2,\cdots,b_n \in B$. But from the given condition we have $$\sum_{i=1}^n\sum_{j=1}^n\left\langle a_{ij}b_j\xi,b_i\xi \right\rangle \ge 0 \implies \left\langle \left(\sum_{i=1}^n\sum_{j=1}^nb_i^*a_{ij}b_j\right)\xi,\xi \right\rangle \ge 0.$$ Now I want to utilize the condition of $\xi$ that is $\overline {B\xi}=\mathcal H$, basically I want to show $\displaystyle \left\langle \left(\sum_{i=1}^n\sum_{j=1}^nb_i^*a_{ij}b_j\right)\eta,\eta\right\rangle \ge 0$ for all $\eta \in \mathcal H$.
Please help me to solve this. Thank you for your time and contribution.

Answer 1

The trick is of course to apply your assumption not to the column vector $[b_j]_j$, but to the column vector $[b_j x]_j$ with $x\in B$ arbitrary. More precisely, if $\eta\in \mathcal H$, then there exists a sequence $(x_k)$ in $B$ such that $x_k \xi\to \eta$ (that's were you use the assumption that $\xi$ is a cyclic vector for $\mathcal H$).

What you showed in the question (but applied to $[b_j x_k]_j$ instead of $[b_j]_j$) is that $$ 0\leq \sum_{i,j=1}^n \langle x_k^\ast b_i^\ast a_{ij}b_jx_k \xi,\xi\rangle=\sum_{i,j=1}^n\langle b_i^\ast a_{ij}b_jx_k\xi,x_k\xi\rangle. $$ As $x_k\xi\to \eta$, the right side converges to $\sum_{i,j}\langle b_i^\ast a_{ij}b_j \eta,\eta\rangle$.