Let$^1$
- $U$, $V$ and $H$ be $\mathbb R$-Hilbert spaces
- $Q\in\mathfrak L(U)$ be nonnegative and symmetric
- $U_0:=Q^{1/2}U$ be equipped with $$\langle u,v\rangle_{U_0}:=\langle Q^{-1/2}u,Q^{-1/2}v\rangle_U\;\;\;\text{for }u,v\in U_0$$
- $\iota\in\operatorname{HS}(U_0,V)$ be an embedding
- $T:=\iota\iota^\ast$
- $V_0:=T^{1/2}V$ be equipped with $$\langle u,v\rangle_{V_0}:=\langle T^{-1/2}u,T^{-1/2}v\rangle_V\;\;\;\text{for }u,v\in V_0$$
We can show that
- $V_0=\iota U_0$
- $\iota$ is an isometry between $U_0$ and $V_0$
- $T$ is a nonnegative and symmetric element of $\mathfrak L(V)$ with finite trace
Now, let
- $L\in\mathfrak L(H)$
- $\Phi\in\operatorname{HS}(U_0,H)$
- $B:=\Phi T^{1/2}$
How can we calculate the trace $\operatorname{tr}LBB^\ast$ of $LBB^\ast$?
Let $(f_n)_{n\in\mathbb N}$ be an orthonormal basis of $V$. Then, we obtain $$\operatorname{tr}LBB^\ast=\operatorname{tr}B^\ast LB=\sum_{n\in\mathbb N}\langle LBf_n,Bf_n\rangle_V\tag 1\;.$$
The question is: Can we further simplify $(1)$ in order to find an expression for $\operatorname{tr}LBB^\ast$ which doesn't inolve $\iota$ (and hence $T$)?
$^1$ Let $\mathfrak L(A,B)$ and $\operatorname{HS}(A,B)$ denote the space of bounded, linear operators and Hilbert-Schmidt operators, respectively. Moreover, let $\mathfrak L(A):=\mathfrak L(A,A)$.
If $C\in\mathfrak L(A,B)$ is not injective, then let $$C^{-1}:=\left.C\right|_{\ker(C)^\perp}^{-1}:C\left(\ker(C)^\perp\right)=C(A)\to\ker(C)^\perp$$ denote the pseudo inverse of $C$.
Let the inner product of a Hilbert space $A$ be denoted by $\langle\;\cdot\;,\;\cdot\;\rangle_A$.