The question; show that the linear mapping for which
$$ 1 \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, i \rightarrow \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}, j \rightarrow \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} k \rightarrow \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} $$ is an isomorphism of the quaternions, $\mathbb H$, onto the following subring of $M_2(\mathbb C)$
$$ \{ \begin{bmatrix} \alpha & -\beta \\ \overline\beta & \overline \alpha \end{bmatrix} \vert \alpha, \beta \in \mathbb C \}. $$
My Working so far:
Firstly, is it reasonable to merge the linear mapping definition a little bit? Say we have a arbitrary quaternion; $\mathbf x = a +bi + cj + dk$ then $\varphi(\mathbf x) = \begin{bmatrix} a + bi & c + di \\ -c + di & a - bi \end{bmatrix} = \begin{bmatrix} a + bi & c + di \\ -(\overline{c + di}) & \overline{ a + bi} \end{bmatrix} $
Moreover, to recall; an isomorphism $\phi: V \rightarrow W$ (where $V,W$ are vector spaces), is indeed one if for $\mathbf u, \mathbf v \in V$ and any two scalars $\lambda, \psi \in F$ (the mutual field) we have $$(*)\text{ }\phi(\lambda \mathbf u + \psi \mathbf v) = \lambda\phi(\mathbf u) + \psi\phi(\mathbf v)$$
Without showing my entire working (given it's already becoming clear as to what I should do) should I take two arbitrary quaternions defined like $\mathbf x$ with two scalars in $\mathbb C$ and try to show $(*)$?
I'm double checking because that seems like an awful amount of work for something quite simple.
What you have defined is a linear map (a homomorphism) not an isomorphism. An isomorphism is a bijective linear transformation.
Hint: show the given matrices form a basis for $M_2(\mathbb{C})$