Let $V$ be a 3-dimensional vector space with an ordered basis $e_{1}$, $e_{2}$, $e_{3}$, and $A: V → V$ be a linear operator given by its matrix relative to the ordered basis $e_{1}$, $e_{2}$, $e_{3}$ as follows:
$A = \left( \begin{array}{ccc} 2 & 1 & 4 \\ -1 & 3 & 2 \\ 1 & -2 & 4 \end{array} \right)$.
$V \otimes V$ has basis $e_{i} \otimes e_{j} \forall 1 \leq i, j \leq 3$.
I have to find $A \otimes A: V \otimes V → V \otimes V$ in matrix form relative to the basis $(e_{i} \otimes e_{j})_{i, j}$.
I basically do not know how to do this at all. Can you lay out a rather detailed explanation of how to do so and what this means as you go along, preferably with the concrete example being worked through?
Would it simply be a $6 \times 6$ matrix with integer coefficients such that it is block diagonal, with the diagonal blocks being the $3 \times 3$ matrix presented earlier and the off-diagonal blocks being the $3 \times 3$ matrix of only $0$'s?
$A\otimes A$ is a 9 by 9 matrix, which can be written as: $$\begin{pmatrix} 2A & A & 4A \\ -A & 3A & 2A \\ A & -2A & 4A \end{pmatrix},$$ that is, $e_i \otimes e_j \mapsto \left(\sum_{k} a_{ki} e_k\right) \otimes\left(\sum_{l} a_{lj} e_l\right)$.