tensor product of linear maps and Kronecker product

Question

I have linear map with matrix $A$. How can I prove, that the Kronecker product ($ A \otimes A$) satisfies the condition, that $ A \otimes A (u \otimes v) = A(u) \otimes A(v)$? To simplify this problem let be $A : \mathbb{R}^3 \rightarrow \mathbb{R}^3$.

Answer 1

A direct proof:

First, note that for the canonical basis vectors $e_i,e_j$, we have $$ (A \otimes A)(e_i \otimes e_j) = (A e_i) \otimes (Ae_j). $$ It follows that for $u = \sum u_i e_i$ and $v = \sum v_j e_j$, we have $$ (A \otimes A)(u \otimes v) = (A \otimes A)\left( \left(\sum_i u_i e_i \right) \otimes \left( \sum_j v_j e_j \right)\right) = (A \otimes A)\left( \sum_{i,j}u_i v_j e_i \otimes e_j\right) \\ = \sum_{i,j}u_i v_j [(A \otimes A)(e_i \otimes e_j)] = \sum_{i,j}u_i v_j (A e_i) \otimes (A e_j) \\ = \left( A \sum_i u_i e_i\right) \otimes \left( A \sum_j v_j e_j\right) = (Au) \otimes (Av). $$

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What I find to be a more enlightening proof is as follows. First, note that there exists a linear map on $\Bbb R^n \otimes \Bbb R^n = \Bbb R^{n^2}$ such that $T(u \otimes v) = (Au) \otimes (Av)$ for all $u,v \in \Bbb R^n$ (this comes down to the definition of the tensor product). Now, we compute the matrix of $T$ relative to the basis $\mathcal B = \{e_i \otimes e_j : 1 \leq i,j \leq n\}$ where the pairs $(i,j)$ are taken in lexicographical order.

For instance, the first column of the matrix is $$ T(e_1 \otimes e_1) = (Ae_1) \otimes (Ae_1) = \sum_{i,j = 1}^n a_{i1}a_{j1} \ e_i \otimes e_j =\\ a_{11}(a_{11}e_1 \otimes e_1 + \cdots + a_{1n}e_1 \otimes e_n) + \cdots + a_{1n}(a_{11}e_n \otimes e_1 + \cdots + a_{1n}e_n \otimes e_n) = \\ \pmatrix{a_{11} Ae_1 \\ \vdots \\ a_{n1} Ae_1}_{\mathcal B} $$ and similarly, the second column is $$ T(e_1 \otimes e_2) = \pmatrix{a_{11} Ae_2 \\ \vdots \\ a_{n1} Ae_2}_{\mathcal B}. $$ In general, we have $$ T(e_i \otimes e_j) = \pmatrix{a_{1i} Ae_j \\ \vdots \\ a_{ni} Ae_j}_{\mathcal B}. $$ Putting all these columns together tells us that the matrix of $T$ relative to the basis $\mathcal B$ is the usual Kronecker product, i.e. $$ [T]_{\mathcal B} = A \otimes A = \pmatrix{a_{11} A & \cdots & a_{1n}A\\ \vdots & \ddots & \vdots \\ a_{n1}A & \cdots & a_{nn}A}. $$ The conclusion follows.