I'm solving problem 3.D in H. Georgi Lie Algebra etc for fun where one is to compute the matrix elements of the direct product $\sigma_2\otimes\eta_1$ where $[\sigma_2]_{ij}\text{ and }[\eta_1]_{xy}$ are two different Pauli matrices in two different two dimensional spaces.
Defining the basis in our four dimensional tensor product space
$$\tag{1}\left|1\right\rangle = \left|i=1\right\rangle\left|x=1\right\rangle\\
\left|2\right\rangle = \left|i=1\right\rangle\left|x=2\right\rangle\\
\left|3\right\rangle = \left|i=2\right\rangle\left|x=1\right\rangle\\
\left|4\right\rangle = \left|i=2\right\rangle\left|x=2\right\rangle$$
Now we know that when we multiply representations, the generators add in the sense of
$$\tag{2}[J_a^{1\otimes2}(g)]_{jyix} = [J_a^1]_{ji}\delta_{yx} +\delta_{ji}[J_a^2]_{yx}, $$ where the $J$s are the generators corresponding to the different representations $D_1$ and $D_2$ ($g$ stands for the group elements).
Using all of this I find that in the basis of $(1)$ the matrix representation of the tensor product is given by
$$\tag{3}\sigma_2\otimes\eta_1 = \begin{pmatrix}
0 & \mathbf{1} & -i & 0 \\
1 & 0 & 0 & -i \\
i & 0 & 0 & 1 \\
0 & i & 1 & 0 \end{pmatrix}$$
(The bold $\mathbf{1}$ is just notation, see below!)
I am not asking you to redo the calculations for me but does $(3)$ make sense?
Appendix.
My calculations were done in the following fashion [using equation $(2)$]:
$$\tag{4}\langle 1| \sigma_2\otimes \eta_1 |1\rangle =
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=1\rangle
\\
=
[\sigma_2]_{11}\delta_{11}+\delta_{11}[\eta_1]_{11}
\\
= 0.$$
Similarly for eg
$$\tag{5}
\langle 1| \sigma_2\otimes \eta_1 |2\rangle =
\\
\langle j=1,y=1| \sigma_2\otimes \eta_1 |i=1,x=2\rangle
\\
=
[\sigma_2]_{11}\delta_{12}+\delta_{11}[\eta_1]_{12}
\\
= 1.
$$
This is how the bold $\mathbf{1}$ was obtained.
So are my calculations $(4), (5)$ totally wrong?
The Pauli matrices
$$\begin{align}
\sigma_1 &=
\begin{pmatrix}
0&1\\
1&0
\end{pmatrix} \\
\sigma_2 &=
\begin{pmatrix}
0&-i\\
i&0
\end{pmatrix} \\
\sigma_3 &=
\begin{pmatrix}
1&0\\
0&-1
\end{pmatrix} \,.
\end{align}
$$
Best Answer
I think it is easier to compute direct products when you write the matrices in component form; basically, you just have to multiply each element of the first matrix by the whole second matrix: $$ \mathbf{A}\otimes\mathbf{B} = \begin{bmatrix} A_{11} \mathbf{B} & \cdots & A_{1n} \mathbf{B} \\ \vdots & \ddots & \vdots \\ A_{n1} \mathbf{B} & \cdots & A_{nn} \mathbf{B}\end{bmatrix} $$ In your case, using the Pauli matrices $\boldsymbol{\sigma}_2$ and $\boldsymbol{\eta}_1$, we get: $$ \boldsymbol{\sigma}_2 \otimes \boldsymbol{\eta}_1 = \begin{bmatrix} 0 & -i\boldsymbol{\eta}_1 \\ i\boldsymbol{\eta}_1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ i & 0 & 0 & 0\end{bmatrix} $$