Tensor product of the spaces of quaternions and complex numbers

The canonical isomorphism $\Phi:\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}\to \mathrm{M}_2(\mathbb{C})$ is constructed as follows:

1. $\mathbb{H}\simeq \mathbb{C}^2$ as vector spaces over $\mathbb{C}$. Explicitly, $\gamma: \mathbb{C}^2\to \mathbb{H}$ is $\gamma(z_1,z_2)=z_1 + z_2j$ [Here $i$ is the imaginary unit of $\mathbb{C}$, and $i,j,ij=k$ are the three imaginary units of $\mathbb{H}$]. Given $\mathbb{H}\ni u =z+wj$, we have $$\bar{u}=\bar{z}+\overline{wj}=\bar{z}+\bar{j}\bar{w}=\bar{z}-j\bar{w}=\bar{z}-wj$$

2. We define the $\mathbb{C}$-algebra homomorphism $\Phi:\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}\to \mathrm{M}_2(\mathbb{C})$ via $$\Phi(1\otimes u)V=\gamma^{-1}(\gamma(V)\bar{u}), \qquad (z\in \mathbb{C}, u\in \mathbb{H}, V\in \mathbb{C}^2)$$ Explicitly, if $u=z+w j$, then $$ \phi(u):=\Phi(1\otimes u)=\begin{pmatrix} \bar{z} & \bar{w}\\ -w& z \end{pmatrix} $$ To finish, note that $\dim_{\mathbb{C}}\mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}=\dim \mathrm{M}_2(\mathbb{C})=4$, so it suffices to check that $\Phi$ is surjective. Define $I=\phi(i)$, $J=\phi(j)$, $K=\phi(ij)$, then it is easy to check that $$\mathrm{M}_2(\mathbb{C})\ni \begin{pmatrix} a & b\\ c & d \end{pmatrix}= \frac{a+d}{2}+\frac{i(a-d)}{2}I+\frac{b-c}{2}J+\frac{i(b+c)}{2}K $$ I want to also mention that the isomorphism $\phi$ is the natural isomorphism $\mathrm{Sp}(1)\to \mathrm{SU}(2)$, where $\mathrm{Sp}(1)$ is the group of unit quaternions.