I am studying quantum information theory on my own, and I am stuck on a simple mathematical step. I can't understand where I am wrong.
Let $\mathcal{H} = \mathcal{H_1}\otimes \mathcal{H_2}$, then every operator $U$ in $\mathcal{H}$ could be decomposed as $U =\sum_{ij}c_{ij} (A_i\otimes B_j) $ where $A_i$ and $B_j$ are orthonormal operator bases in $\mathcal{H_1}$ and $\mathcal{H_2}$. This property is widely used in NC textbook, and also to prove the Schmidt decomposition Theorem for operators (https://arxiv.org/abs/quant-ph/0208077).
Altough this property is very simple and I have an "intuitive view" of why this is true, I cannot prove it.
This is what I have done:
Let $B_1 = B(\mathcal{H}_1,\mathcal{H}_1)$ and $B_2 = B(\mathcal{H}_2,\mathcal{H}_2)$ the space of all (continuous) linear operators on respectively $\mathcal{H}_1$ and $\mathcal{H}_2$. Then we could form $B=B_1\otimes B_2$ such that for all $X \in B$, given the basis $A_i\otimes B_j $, the previous decomposition is true by definition of tensor product space.
However, I think this is not a proof, because $U\in B(\mathcal{H}_1\otimes\mathcal{H}_2,\mathcal{H}_1\otimes\mathcal{H}_2)$, while $X\in B(\mathcal{H}_1,\mathcal{H}_1) \otimes B(\mathcal{H}_2,\mathcal{H}_2)$.
Am I wrong? Does this means that the two spaces are the same? How can I prove it?
Edit:
I think I've found a proof in the finite dimensional case.
Suppose that $X\in B(\mathcal{H}_1 \otimes \mathcal{H}_2,\mathcal{H}_1 \otimes \mathcal{H}_2)$. If $|i\rangle |j\rangle$ is a orthonormal basis in $\mathcal{H}$ then $\sum_{ij}|i\rangle |j\rangle\langle i|\langle j| = I$ thus: $$X = \sum_{ij}\sum_{i'j'}|i\rangle |j\rangle\langle i|\langle j| X |i'\rangle |j'\rangle\langle i'|\langle j'|$$ Where $\mathbf{X}_{iji'j'} = \langle i|\langle j| X |i'\rangle |j'\rangle$ is the "matrix" representation of $X$. Then: $$ X = \sum_{ij}\sum_{i'j'} \mathbf{X}_{iji'j'} (|i\rangle\langle i'|)\otimes (|j\rangle\langle j'|)$$
$X$ is then a linear combination of elements in $B(\mathcal{H}_1,\mathcal{H}_1) \otimes B(\mathcal{H}_2,\mathcal{H}_2)$. Thus $X\in B(\mathcal{H}_1,\mathcal{H}_1) \otimes B(\mathcal{H}_2,\mathcal{H}_2)$
I think that this argument could also be used to prove the result in the infinite-dimensional case, taking the limit of partial sums, but this requires more work, I think.
Observe that there is a natural inclusion $\pi$ of the algebraic tensor product $B(H_1)\odot B(H_2)$ inside $B(H_1\otimes H_2)$. Every vector of $H_1\otimes H_2$ is clearly cyclic for $\pi$, which shows that $\pi$ is an irreducible representation. From this we conclude that the commutant of $B(H_1)\odot B(H_2)$ is trivial, so that the weak closure of $B(H_1)\odot B(H_2)$ coincides with the whole $B(H_1\otimes H_2)$ by von Neumann's bicommutant theorem.