a positive matrix of operators Let $F$ be a complex Hilbert space. Let $A,B,C,D\in \mathcal{B}(F)$. Consider the operator matrix $T$ such that
\begin{equation*}
T=\begin{pmatrix}A & B \\
C & D
\end{pmatrix}
\end{equation*}

I see in a paper that $T$ is a positive operator on $F\oplus F$ if and only if $\langle Tx,x \rangle\geq 0$ for all $x\in F\oplus F$. But I don't understand how to calculate $\langle Tx,x \rangle$. What is the inner product on $F\oplus F$? Is $F\oplus F$ the direct sum of $F$ and $F$ or the tensor product of of $F$ and $F$?

Thank you.
 A: The notation $F\oplus F$ refers to the direct product of the Hilbert spaces $F$ and $F$. This is not the same as the tensor product, which is written $F\otimes F$. The scalar product on $F\oplus F$ is defined as follows:
If $x=\begin{pmatrix} x_1\\ x_2\end{pmatrix}\in F\oplus F$ with $x_1,x_2\in F$, and $x'=\begin{pmatrix}x_1'\\ x_2'\end{pmatrix}$ similarly, then
$$\langle x,x'\rangle_{F\oplus F}:= \langle x_1,x_1'\rangle_F +\langle x_2,x_2'\rangle.$$
This will give you an inner product on the vectorspace $F\oplus F$. One can verify that $F\oplus F$ is complete with this inner product, this follows out of $\|x\|^2_{F\oplus F}=\|x_1\|_F^2+\|x_2\|^2_F$. Specifically if $x^{(n)}$ is a Cauchy sequence in $F\oplus F$, we find that the components $x_1^{(n)}$, $x_2^{(n)}$ must also be Cauchy in $F$. Since $F$ is a Hilbert space these two sequences admit limits $x_1,x_2$. Let $x$ be the vector having these as components. Then:
$$\|x^{(n)}-x\|^2_{F\oplus F}=\|x_1^{(n)}-x_1\|^2_F + \|x_2^{(n)}-x_2\|^2_F.$$
Both terms on the right hand side converge to zero so $x^{(n)}\to x$ in $F\oplus F$ and thus every Cauchy sequence converges.
