Is there any geometric interpretation of the Kronecker product of two $2\times 2$ matrices?

Related: http://math.stackexchange.com/questions/18881/motivationfortensorproduct Also, there is a nice quantum mechanical interpretation in terms of a system of two qubits. – Alexei Averchenko Nov 06 '12 at 15:22
1 Answers
It pays to notice that the Kronecker product is really the tensor product acting on operators: if $f: A \to B$ and $g: X \to Y$ are linear mappings, then $f \otimes g: A \otimes X \to B \otimes Y$ is given by the Kronecker product in any bases of the form $a_i \otimes x_j$, $b_k \otimes y_l$, where $\{x_i\}$ is the basis of $X$ etc.
Now consider a pure vector $a \otimes x \in A \otimes X$. By definition we have $(f \otimes g)(a \otimes x) = f(a) \otimes g(x)$. E.g. if $B = Y$ and $g = 1$, we have $(f \otimes 1)(a \otimes x) = f(a) \otimes x$, so the "$X$ part" is untouched. Thus, at least when both $f$ and $g$ are endomorphisms you can picture $f \otimes g = (f \otimes 1) \circ (1 \otimes g) = (1 \otimes g) \circ (f \otimes 1)$ as acting independently on factors of its domain by $f$ and $g$ respectively. This is obviously a corollary of $(f \otimes g)(a \otimes x) = f(a) \otimes g(x)$, which remains true in general.
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