Understanding matrix representation of tensor products

Question

My course material for linear algebra has this question, and I'm not sure on how to do it: For a linear operator $L$ on $V$, we can create a linear operator on $V\otimes V$ by $I\otimes L - L\otimes I$, where $I$ is the identity on $V$. If $L$ is given by the matrix \begin{align*} \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \end{align*} What is the matrix for $I\otimes L - L\otimes V$? I don't understand how to start or how to finish. I know about the Kronecker product, but it does not give me any meaningful insight into what is really happening and how these tensor products work. Should i look at $(I\otimes L)(e_i\otimes e_j)$ where $\{ e_1,e_2\}$ is a basis for say $\mathbb{C}^2$, and then try to represent it in a basis for $V\otimes V$ like $\{e_i \otimes e_j\}_{i\in I,j\in J}$? My problem is that I don't really understand how to represent tensor products in a basis or how to "work" with them.

Answer 1

$\def\kp{\color{red}{\otimes}} \def\cM{\cal M}$ $L \begin{bmatrix} e_1 \\ e_2 \end{bmatrix} =\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} e_1 \\ e_2 \end{bmatrix} $. The basis of $V \otimes V$ is $\{e_i \otimes e_j\}_{i,j=1,2}$. $(I \otimes L)( v_1 \otimes v_2) := Iv_1 \otimes L v_2$. $$ (I \otimes L - L \otimes I) \begin{bmatrix} e_1 \otimes e_1 \\ e_1 \otimes e_2 \\ e_2 \otimes e_1 \\ e_2 \otimes e_2 \end{bmatrix}=\begin{bmatrix} Ie_1 \otimes Le_1-Le_1 \otimes I e_1 \\ Ie_1 \otimes Le_2 - Le_1 \otimes Ie_2 \\ Ie_2 \otimes Le_1 - Le_2 \otimes Ie_1 \\ Ie_2 \otimes Le_2 - Le_2 \otimes I e_2 \end{bmatrix}=\begin{bmatrix} e_1 \otimes (e_1+2e_2)-(e_1+2e_2) \otimes e_1 \\ e_1 \otimes (4e_1 + 3e_2) - (e_1+2e_2) \otimes e_2 \\ e_2 \otimes (e_1+2e_2) - (4e_1+3e_2) \otimes e_1 \\ e_2 \otimes (4e_1+3e_2) - (4e_1+3e_2) \otimes e_2 \end{bmatrix}= \begin{bmatrix} 2e_1 \otimes e_2 -2e_2 \otimes e_1 \\ 4e_1 \otimes e_1 + 2 e_1 \otimes e_2- 2 e_2 \otimes e_2 \\ - 4 e_1 \otimes e_1-2e_2 \otimes e_1+ 2 e_2 \otimes e_2 \\ - 4 e_1 \otimes e_2 + 4e_2\otimes e_1 \end{bmatrix}= \begin{bmatrix} 0 & 2 & -2 & 0 \\ 4 & 2 & 0 & -2 \\ -4 & 0 & -2 & 2 \\ 0 & -4 & 4 & 0 \end{bmatrix} \begin{bmatrix} e_1 \otimes e_1 \\ e_1 \otimes e_2 \\ e_2 \otimes e_1 \\ e_2 \otimes e_2 \end{bmatrix} $$ We could have saved a lot of work by noting that the Kronecker product $\kp$ (which generates a matrix and not a tensor) of the matrix representations $[I],[L]$ of respectively $I,L$ $$ [I] \kp [L] - [L] \kp [I] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} \kp \begin{bmatrix} 1 & 2 \\ 4 & 3 \\ \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 4 & 3 \\ \end{bmatrix} \kp \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} =\left[\begin{matrix}1 & 2 & 0 & 0\\4 & 3 & 0 & 0\\0 & 0 & 1 & 2\\0 & 0 & 4 & 3\end{matrix}\right] -\left[\begin{matrix}1 & 0 & 2 & 0\\0 & 1 & 0 & 2\\4 & 0 & 3 & 0\\0 & 4 & 0 & 3\end{matrix}\right] = \begin{bmatrix} 0 & 2 & -2 & 0 \\ 4 & 2 & 0 & -2 \\ -4 & 0 & -2 & 2 \\ 0 & -4 & 4 & 0 \end{bmatrix}. $$