Show linear mapping is isomorphism 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 $M_2(\mathbb{C})$
