Suppose I have unitary operators $$A: \mathbb{C}^{2^k} \rightarrow \mathbb{C}^{2^k}$$ $$B: \mathbb{C}^{2^j} \rightarrow \mathbb{C}^{2^j}$$
For some $k,j \in \mathbb{Z}, j,k \ge 0$. How do show that
$$ A \otimes I_{2^j} \circ I_{2^k} \otimes B = I_{2^k} \otimes B \circ A \otimes I_{2^j} $$
Work So Far:
A "proof by thought experiment" is as follows, $A$, $B$ can be different quantum operators operating on different sets of Qubits. Then assuming a system of $k+j$ Qubits (Of which we apply $A$ to the first $k$ Qubits and $B$ to the other group simultaneously) the entire operator can be represented as either
$$ A \otimes I_{2^j} \circ I_{2^k} \otimes B$$
or
$$ I_{2^k} \otimes B \circ A \otimes I_{2^j} $$
Depending on which you evaluate first. But since we know those operators are on disjoint sets of Qubits, the order SHOULD not matter. So we conclude that:
$$ A \otimes I_{2^j} \circ I_{2^k} \otimes B = I_{2^k} \otimes B \circ A \otimes I_{2^j} $$
But this proof now rests on whether my intuition of what Quantum Mechanics "SHOULD" do is correct. And thats a pretty weak foundation in my opinion. I was hoping someone can give insight into a good ol' fashion linear algebraic proof of this.