The matrix of left/right multiplication in $k^{2\times 2}$ is a tensor product of matrices?

Question

Here is an easy linear algebra exercise: (related to 1, but the answer there diverges with mine.)

Fix a matrix $A=\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}\in k^{2\times 2}$, compute the matrices of the following linear operators w.r.t. the standard basis $(E_{11},E_{12},E_{21},E_{22})$: $$\begin{cases} f_L:X\mapsto AX\\ f_R:X\mapsto XA \end{cases} $$

After a direct computation, I find out: $$ M_{f_L}=\begin{pmatrix} a_{11} & 0 & a_{12} & 0\\ 0 & a_{11} & 0 & a_{12}\\ a_{21} & 0 & a_{22} & 0\\ 0 & a_{21} & 0 & a_{22} \end{pmatrix}=A\otimes I_2$$ and $$ M_{f_R}=\begin{pmatrix} a_{11} & a_{21} & 0 & 0\\ a_{12} & a_{22} & 0 & 0\\ 0 & 0 & a_{11} & a_{21}\\ 0 & 0 & a_{12} & a_{22} \end{pmatrix}=I_2\otimes A^T$$

(I hope $ I_2\otimes A^T$ is correct, instead of $I_2\otimes A$ given by 1)

Question:Why does the tensor product of matrices appear here? Is there a more natural or elegant explanation/derivation, instead of direct computation?

Thanks in advance!

Answer 1

There is an explanation, but it is too complicated for the level on which this problem is usually stated. In a nutshell, the idea is that the space of matrices is itself a tensor product, and multiplications on the left and on the right only act on one factor of this product.

If $k^2$ is the space of column vectors with the standard basis $e_1 = \begin{bmatrix}1\\0\end{bmatrix}, e_2 = \begin{bmatrix}0\\1\end{bmatrix}$, and $k^{2*}$ is the space of row vectors with the basis $e^*_1 = [1\ 0], e^*_2 = [0\ 1]$. The space of matrices $k^{2 \times 2}$ is isomorphic to the tensor product $k^2 \otimes k^{2*}$: to every matrix $M = \begin{bmatrix}m_{11} & m_{12} \\ m_{21} & m_{22}\end{bmatrix}$ corresponds a tensor $$m_{11}\, e_1 \otimes e^*_1 + m_{12}\, e_1 \otimes e^*_2 + m_{21}\, e_2 \otimes e^*_1 + m_{22}\, e_2 \otimes e^*_2.$$ In particular, standard basis matrices $E_{ij}$ correspond to products of basis vectors $e_i \otimes e^*_j$.

We can also express this isomorphism in terms of its rows or columns: the tensor corresponding to the matrix $M$ with columns $m_1$ and $m_2$ is $m_1 \otimes e^*_1 + m_2 \otimes e^*_2$. To multiply $A$ by $M$ is the same as to multiply $A$ by every column of $M$, so left multiplication by $A$ transfers to the linear map on tensors which maps $$m_1 \otimes e^*_1 + m_2 \otimes e^*_2 \mapsto A m_1 \otimes e^*_1 + A m_2 \otimes e^*_2.$$ Or, in other words, it acts as $A$ on the first factor and as identity $I_2$ on the second factor. The matrix of such a map in the accordingly chosen basis is exactly the Kronecker product matrix $A \otimes I_2$.

Similarly, a matrix $M$ with rows $m^*_1$ and $m^*_2$ corresponds to the tensor $e_1 \otimes m^*_1 + e_2 \otimes m^*_2$, and multiplication by $A$ on the right acts as $I_2$ on the first factor, and as a multiplication by $A$ on the right on the second: $$e_1 \otimes m^*_1 + e_2 \otimes m^*_2 \mapsto e_1 \otimes m^*_1 A + e_2 \otimes m^*_2 A.$$ Note that multiplying a row by $A$ on the right is the same as multiplying its transpose (a column) by $A^\top$ on the left, so the matrix of right multiplication will be $I_2 \otimes A^\top$.