Let $(\tau,\mathbb C^2)$ be the irreducible representation of $D_4$ by matrix multiplication, namely for every $v\in\mathbb{C}^2$: $$\begin{bmatrix}\tau(s)\end{bmatrix}v=\begin{bmatrix}-1 & 0\\ 0 & 1 \end{bmatrix}v$$ and $$\begin{bmatrix}\tau(rs)\end{bmatrix}v=\begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix}v$$ Where $s,rs$ are two reflections that generate $D_4$. Moreover, consider the representation $(\rho,\mathbb{C})$ of $D_4$ - for every $\lambda\in\mathbb{C}$, $\rho(s)(\lambda)=-\lambda$ and $\rho(rs)(\lambda)=\lambda$. Find $\rho\otimes\tau$.
My attempt - since $s,rs$ are generators it's enough to determine $(\rho\otimes\tau)(s),(\rho\otimes\tau)(rs)$. Moreover, it's enough to determine their action on the basis of $\mathbb{C}\otimes\mathbb{C}^2$, $\{1\otimes (1,0),1\otimes (0,1)\}$. So we have:$$(\rho\otimes\tau)(s)(1\otimes(1,0))=\rho(s)(1)\otimes\tau(s)(1,0)=-1\otimes(-1,0)=1\otimes(1,0)$$ $$(\rho\otimes\tau)(s)(1\otimes(0,1))=\rho(s)(1)\otimes\tau(s)(0,1)=-1\otimes(0,1)=1\otimes(0,-1)$$ and by the identification $1\otimes(a,b)\to(a,b)$, we have:$$\begin{bmatrix}(\rho\otimes\tau)(s)\end{bmatrix}=\begin{bmatrix}1 & 0\\ 0 & -1 \end{bmatrix}$$ Similarly: $$(\rho\otimes\tau)(rs)(1\otimes(1,0))=\rho(rs)(1)\otimes\tau(rs)(1,0)=1\otimes(0,1)=1\otimes(0,1)$$$$(\rho\otimes\tau)(rs)(1\otimes(1,0))=\rho(rs)(1)\otimes\tau(rs)(0,1)=1\otimes(1,0)=1\otimes(1,0)$$ and by the identification $1\otimes(a,b)\to(a,b)$, we have: $$\begin{bmatrix}(\rho\otimes\tau)(rs)\end{bmatrix}=\begin{bmatrix}0 & 1\\ 1 & 0 \end{bmatrix}$$
However, I cannot tell if this is correct. Moreover, I think this representation can't be reducible since I'm familiar with all irreducible representations of dimension $1$ and this representation is not a direct sum of any two of those. However, I can't see why it would be isomorphic to $\tau$. Any help would be appreciated.