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Given an Hilbert space $\mathcal{H}$ and $x,y,z\in\mathcal{H}$, let $x\odot y$ be the compact operator defined by $x\odot y(z)=\langle z,y\rangle x$. Assume $T$ is compact, is it the case that $$ T=\sum_{i,j}\lambda_{i,j}e_i\odot e_j$$ where $\{e_i:i\in I\}$ is an orthonormal base for $\mathcal{H}$? This is true for all $T$ in $B^2(\mathcal{H})$ i.e. for all $T$ such that $\sum_{j\in I}\langle T e_j,T e_j\rangle<\infty$ (see for example Proposition 3.4.14 of the GTM book Analysis NOW by Pedersen). But $B^2(\mathcal{H})$ is a proper subclass of the compact operators, what happens for the other compact operators? Is this decomposition fine as well?

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Yes, there is such a decomposion, it is (more or less) the singular value decomposition of a compact operator.

To be more precise, let $(e_k)$ be an eigenbasis of $|T|$ and $U$ the partial isometry in the polar decomposition $T=U|T|$. By the spectral theorem for self-adjoint compact operators we have $$ |T|=\sum_i s_i e_i\odot e_i. $$ Thus, $$ T=U|T|=\sum_i s_i (Ue_i)\odot e_i=\sum_{i,j}s_i \langle Ue_i,e_j\rangle e_j\odot e_i. $$ The numbers $s_i$ are the eigenvalues of $|T|$ (taking into account their multiplicity), also called singular values of $T$.

Notice that the above series converges in operator norm for compact operators, whereas it converges in Hilbert-Schmidt norm for $T\in\mathcal{B}^2(\mathcal{H})$.

Both $\mathcal{B}^2(\mathcal{H})$ and the space of compact operators on $\mathcal{H}$ are therefore completions of the algebraic tensor product $\mathcal{H}\odot\mathcal{H}$, but with respect to different tensor norms (see here).

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