Let $U$ be a $\mathbb R$-Hilbert space, $Q\in\mathfrak L(U)$ be nonnegative and self-adjoint with finite trace, $U_0:=Q^{1/2}U$ be equipped with $$\langle u_0,v_0\rangle_{U_0}:=\langle Q^{-1/2}u_0,Q^{-1/2}v_0\rangle_U\;\;\;\text{for }u_0,v_0\in U_0$$ (where $Q^{-1/2}$ denotes the pseudo-inverse, if $Q$ is not injective) and $(\tilde e_n)_{n\in\mathbb N}$ be an orthonormal basis of $(\ker Q^{1/2})^\perp$. It's easy to see that $(Q^{1/2}\tilde e_n)_{n\in\mathbb N}$ is an orthonormal basis of $U_0U$. Moreover, $(\tilde e_n)_{n\in\mathbb N}$ can be supplemented to a basis of $U$.
I want to show that $$\left\|L\right\|_{\operatorname{HS}(U_0,\:H)}^2=\sum_{n\in\mathbb N}\left\|LQ^{1/2}\tilde e_n\right\|_H^2$$ and $$\left\|LQ^{1/2}\right\|_{\operatorname{HS}(U,\:H)}^2=\sum_{n\in\mathbb N}\left\|LQ^{1/2}e_n\right\|_H^2$$ are equal, for all $L\in\operatorname{HS}(U_0,H)$ (where $\operatorname{HS}(U_0,H)$ denotes the space of Hilbert-Schmidt operators from $U_0$ to $H$.
It seems like we need to show that $Q^{1/2}e_k=0$ (i.e. $e_k\in\ker Q^{1/2}$) for all $e_k$ which aren't in $(\tilde e_n)_{n\in\mathbb N}$. But I've no idea why this should be the case. I guess we somehow need to use that (since $(\ker Q^{1/2})^\perp$ is closed and $(\ker Q^{1/2})^{\perp\perp}=\overline{\ker Q^{1/2}}=\ker Q^{1/2}$) $$U=(\ker Q^{1/2})^\perp\oplus\ker Q^{1/2}\;,$$ but I don't know how we could do that. Maybe $(e_n)_{n\in\mathbb N}$ must be constructed by choosing elements from $\ker Q^{1/2}$.