tensor of operators is Kronecker product?

Question

Say $A$ and $B$ are operators on Hilbert spaces $H_A,H_B$ respectively. If the Hilbert spaces are finite dimensional, then I know the tensor $A\otimes B$ can be represented by the Kronecker product $[a_{ij}B]$.

Question 1: Does the Kronecker product formula $[a_{ij}B]$ still work in infinite dimensions?

Question 2: If not, does it work when $H_A$ is finite dimension and $H_B$ is infinite dimensional (possibly an operator on a non-separable space)?

Answer 1

We can make the Kronecker-product formula work in the following way. If $\{u_j\}_{j \in \Bbb N}$ is a basis of $H_A$, then we have $$ H_{A} \otimes H_B \cong \bigoplus_{j \in \Bbb N} H_B, $$ with an isomorphism between the two spaces defined by $\phi : H_{A} \otimes H_B \to \bigoplus_{j \in \Bbb N} H_B$, $$ \phi(u_j \otimes v) = (0,\dots,0,v,0,0,\dots). $$ Now, if $a_{jk}$ are defined such that $Au_k = \sum_{j \in \Bbb N} a_{jk}u_j$, then we have $$ \phi(A \otimes B) \phi^{-1}(v_1,v_2,\dots) = \phi(A \otimes B)\sum_{k \in \Bbb N} u_k \otimes v_k = \phi\sum_{k \in \Bbb N} \left(\sum_{j \in \Bbb N} a_{jk}u_j \otimes (Bv_k) \right)\\ = \phi\sum_{j \in \Bbb N} u_j \otimes \sum_{k \in \Bbb N}(a_{jk}B)(v_k) \\ = \left(\sum_{k \in \Bbb N}(a_{1k}B)(v_k),\sum_{k \in \Bbb N}(a_{2k}B)(v_k),\dots \right), $$ which indeed corresponds to the operator matrix product $$ \pmatrix{a_{11}B & a_{12} B & \cdots\\ a_{21}B & a_{22}B & \cdots\\ \vdots & \vdots & \ddots} \pmatrix{v_1\\ v_2 \\ \vdots}. $$

Answer 2

$\DeclareMathOperator{\End}{End}$ Yes, the Kronecker product formula works, and not just for Hilbert spaces.

More precisely, in the non-topological case, if $V,W$ are $k$-vector spaces with (algebraic) bases $(v_i)_i$, $(w_k)_k$ and $A\in \End(V)$, $B\in \End(W)$, then you can express $A,B$ using matrix coefficients $a_{i,j},b_{k,l}$ as $A(v_{i})=\sum_{j}a_{i,j}v_j$, $B(w_k)=\sum_{l}b_{k,l}w_l$, and then $$(A\otimes B)(v_i\otimes w_k)=(Av_i)\otimes (Bw_k)=(\sum_{j}a_{i,j}v_j)\otimes(\sum_{l}b_{k,l}w_l)=\sum_{j,l} a_{i,j}b_{k,l} (v_j\otimes w_l),$$ i.e. the matrix coefficients of $A\otimes B$ are, indeed, given by the Kronecker product (with respect to the tensor product of the bases or $V,W$ with respect to which we compute the matrix coefficients of $A,B$).

Now, if $V,W$ are Hilbert spaces and $(v_i)_i$, $(w_k)_k$ are their orthonormal bases, then $(v_i\otimes w_k)_{i,k}$ is an orthonormal basis of $V\otimes W$ and exactly the same computation works. The only difference is that the summation is no longer essentially finite, but only absolutely convergent. I suppose this should work the same way in any context where "matrix coefficients" make any sense.