Let $A$ be a unital C*-algebra. Then we know that $M_n(A)$ is also a C*-algebra.
Let $x=[x_{ij}]\in M_n(A)$. I want to prove that if for every state $\phi$ on $A$ and for every $\{y_1,...,y_n\}\subset A$ we have $$\phi\big(\sum_{i,j}y_i^*x_{ij}y_j\big)\geq 0,$$ then this implies that $x=[x_{ij}]\geq 0$ in $M_n(A)$.
Since we know that $a\in A$ is positive if and only if $\phi(a)\geq 0$ for all state $\phi$, we may conclude that $$\sum_{i,j}y_i^*x_{ij}y_j$$ is a positive element in $A$ for all $\{y_1,...,y_n\}\subset A$. However after this I am unable to show that this implies that $x=[x_{ij}]\geq 0$ in $M_n(A)$. How can prove this?
Hints are most welcome.
Identifying $A$ as a $C^*$-subalgebra of $B(\mathcal H)$ for some Hilbert space $\mathcal H$, and hence identifying $M_n(A)$ as a $*$-subalgebra of $M_n(B(\mathcal H))$, observe that $\sum_{i,j}y_i^*x_{ij}y_j$ can be identified as an operator in $B(\mathcal H)$. So the question boils down to prove that if $\sum_{i,j}y_i^*x_{ij}y_j\in B(\mathcal H)$ is positive for every $\{y_1,...,y_n\}\subset B(\mathcal H)$ then $x=[x_{ij}]\geq 0$ in $M_n(B(\mathcal H)).$
To this end, we have for every $h\in \mathcal H$ and for every $\{y_1,...,y_n\}\subset B(\mathcal H)$ we have \begin{align*} 0&\leq \left\langle\sum_{i,j}y_i^*x_{ij}y_j h,h\right\rangle \\ &=\left \langle\begin{pmatrix}y_1^*&\cdots&y_n^*\end{pmatrix}[x_{ij}]\begin{pmatrix}y_1\\\vdots\\y_n\end{pmatrix}h,h\right\rangle\\ &=\left\langle\left( Y^*[x_{ij}]Y\right)h,h\right\rangle \\ &=\left\langle[x_{ij}]Yh,Yh\right\rangle. \end{align*} It follows then that for every $\hat h\in \mathcal H^n$ we have $$ \left\langle[x_{ij}]\hat h,\hat h\right\rangle\geq 0, $$ which implies that $x=[x_{ij}]\geq 0$ in $M_n(B(\mathcal H)).$