# 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.

• $F\oplus F$ is the direct sum, not the tensor-product. Let $x=\begin{pmatrix} x_1\\ x_2\end{pmatrix}\in F\oplus F$, similar $x'=\begin{pmatrix} x_1'\\ x_2'\end{pmatrix}$. Then $$\langle x, x'\rangle_{F^2}=\langle x_1,x_1'\rangle_F + \langle x_2,x_2'\rangle_F.$$ Jan 14 '18 at 17:39
• Thank you. If $F\oplus F=F^2$? Jan 14 '18 at 17:41
• I mean what is the difference between direct sum and cartesian product ? Jan 14 '18 at 17:47
• @Thierry Strictly speaking, the Cartesian product of Hilbert spaces doesn't come with an inner product. However, cartesian products and direct sums of finite collections indeed result in the same underlying sets. Jan 14 '18 at 17:48
• @Thierry also, tensor products use $\otimes$ rather than $\oplus$. Jan 14 '18 at 17:49

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.