Let $V$ be a $4$-dimensional vector space with an ordered basis $e_{1}$, $e_{2}$, $e_{3}$, $e_{4}$, and $A: V → V$ be a linear operator given by its matrix relative to the ordered basis $(e_{1}, e_{2}, e_{3}, e_{4})$ as follows:
$A \doteq \left( \begin{array}{ccc} 17 & \mathbb{e} & 12 & 1 \\ a & \frac{137}{\gamma}+\frac{1+\sqrt{5}}{2} & \frac{\pi^{2}}{6} & 2 \\ \mathbb{i} & 0 & 23 & 3 \\ \frac{1}{4} & \frac{3}{2} & \frac{11}{4} & 4 \end{array} \right) \in \mathcal{M}_{4,4}(\mathbb{C})$, so that $a \in \mathbb{C}$.
$V \otimes V$ has basis $e_{i} \otimes e_{j} \forall 1 \leq i, j \leq 4$.
I have to find $A \otimes A: V \otimes V → V \otimes V$ in matrix form relative to the basis $(e_{i} \otimes e_{j})_{i, j}$.
Based on another question, the answer that I am getting would be:
$A \otimes A \doteq \left( \begin{array}{ccc} 17 A & \mathbb{e} A & 12 A & A \\ a A & (\frac{137}{\gamma}+\frac{1+\sqrt{5}}{2})A & \frac{\pi^{2}}{6} A & 2A \\ \mathbb{i}A & 0A & 23A & 3A \\ \frac{1}{4}A & \frac{3}{2}A & \frac{11}{4}A & 4A \end{array} \right) \overset{\underset{\mathrm{?}}{}}{\in} \mathcal{M}_{4^{2},4^{2}}(\mathbb{C})$.
Is this correct?
And am I to interpret this as a single matrix (where blocks of entries have been grouped together as multiples of $A$) or as a matrix of matrices? Is there a difference?
Also, even if I got this problem correct, can you explain to me why it is correct? I still do not fully understand how this all works (I just am working by analogy). More specifically, what does each block of this second matrix as presented mean and how do I understand what each row and column of it means once we expand the blocks out (which basis vectors are being acted upon each time?)? That is, if I were to label each entry of the second matrix with subscripts, how would I do so and what do the subscripts mean (to what do they correspond)?
The way that I currently understand it is that $A$ maps $e_{1}$ in a particular way so as to result in a particular linear combination of the basis elements $e_{i} \forall i \in \mathbb{N}: i \leq 4$, namely $A e_{1} = 17 e_{1} + a e_{2} + \mathbb{i} e_{3} + \frac{1}{4} e_{4}$. Then, whenever we have $e_{1} \otimes e_{j} \forall j \in \mathbb{N}: i \leq 4$, a first application of $A$ will produce this same linear combination of basis elements $e_{i} \otimes e_{j} \forall i \in \mathbb{N}: i \leq 4$, where $j$ is fixed accordingly. This "sub-multiplication" produces the coefficients of the first block-column of $A \otimes A$. The second application of $A$ would then act on each $e_{j}$ as it would have originally, which is why each entry in the first block-column comes out to have a factor of $A$ in it. The same holds accordingly true for every other allowed value of $i$. So, the $i$th block-column corresponds to any vector of form $e_{i} \otimes e_{j}$ for the same fixed $i$. Therefore, in this block-column (once we expand the notation), the $j$th column (so, the $i+j$th column overall) is how $A \otimes A$ maps $e_{i} \otimes e_{j}$. If I were to label the block-columns, I would just go from left to right and labelling them $i$, since $j$ does not matter for the blocks; then, when I expand this matrix into individual entries, for every fixed $i$, I would go through the columns labelling the possible $j$'s. There are only four $i$ values and for each there are only four $j$ values, producing sixteen total combinations (columns). Likewise, the block-rows correspond to the outputs for every fixed $i$ and the actual rows organize these for each fixed $j$; there are for $i$ values and for each of them there are four $j$ values (making a total of sixteen rows). The matrix is a $16 \times 16$ matrix. In each case, the $j$ values are interleaved throughout the $i$ values, rather than one block corresponding only to, say, $i=1=j$ and then another block only corresponding to $i=3=j$, and yet another being $i=2, j=4$. The blocks only represent $i$ values (for input in the block-columns and output in the block-rows), and the $j$ values "live in rooms within" each these blocks (columns corresponding to inputs and rows corresponding to outputs, dependent on the reigning $i$). Incidentally, the set membership that I questioned would be true and we are to understand the second matrix as just being composed of many blocks when we expand it, but it can be helpful to understand it as a matrix of matrices where each column acts on the vector for the corresponding fixed $i$. I figure that in another basis (perhaps $e_{j} \otimes e_{i}$?), the blocks would correspond to $j$ values and the $i$ values would be interleaved throughout those blocks, but then the order of the block-entries would be scrambled somewhat?
(The answer, it seems, is provided in the original post)
$A \otimes A \doteq \left( \begin{array}{ccc} 17 A & \mathbb{e} A & 12 A & A \\ a A & (\frac{137}{\gamma}+\frac{1+\sqrt{5}}{2})A & \frac{\pi^{2}}{6} A & 2A \\ \mathbb{i}A & 0A & 23A & 3A \\ \frac{1}{4}A & \frac{3}{2}A & \frac{11}{4}A & 4A \end{array} \right) \in \mathcal{M}_{4^{2},4^{2}}(\mathbb{C})$.
The explanation (as I understand it) can also be found in the original post. Essentially, each (in the example, the $i$th) block-column corresponds with the transformation of the appropriate basis vector $e_{i} \otimes e_{j} \forall j \in \{ 1,2,3,4 \}$, where $i$ is fixed.