Let $\mathcal A$ be the algebra over the real numbers consisting of matrices of the form $$\begin{pmatrix} z&w\\ – \bar{w}& \bar{z} \end{pmatrix} \ (z, w \in \mathbb C). $$
$\mathcal A$ is in the bijection with the algebra of quaternions via $$a+bi+cj+dk \longleftrightarrow \begin{pmatrix} a+bi& c+di\\ – c+di & a-bi \end{pmatrix}. $$
I saw no other way of verifying the homomorphism condition then by calculation (which I only finished partially until I believed it).
Is there a more natural way of noting that the given bijection is a homomorphism, without the necessity of computing matrix- and quaternion products?
I got a few clues from @martinis answer:
Put $$ I_2 := \begin{pmatrix} 1&0\\0&1 \end{pmatrix} \ ,\ I := \begin{pmatrix} i&0 \\ 0&-i \end{pmatrix},\ J:= \begin{pmatrix} 0&1 \\ -1&0 \end{pmatrix},\ K := \begin{pmatrix} 0&i\\i&0 \end{pmatrix}.$$
Now notice that we have
$$a+bi+cj+dk \longleftrightarrow \begin{pmatrix} a+bi& c+di\\ – c+di & a-bi \end{pmatrix} = aI_2 +bI+cJ+dK. $$
Hence the bijection is $\mathbb R$-linear in $(a,b,c,d)$. For the homomorphism condition it now suffices to check whether the above defined matrices satisfy $$I^2=J^2=K^2=IJK=-I_2,$$ because then multiplication rules on both sides agree. It then follows that the bijection is multiplicative. And with that, the isomorphism is established.
For some reason the concept transport of structure pops into my mind.
Best Answer
Hint: Both sides are $\mathbb R$-linear in $(a,b,c,d)$, so you have to check multiplicativity only on the $\mathbb R$-basis $\{1,i,j,k\}$ of $\mathbb H$, that is one only has to check whether $$ I := \def\p#1#2#3#4{\begin{pmatrix}#1\\#3\end{pmatrix}}\p i00{-i}, J := \p 01{-1}0, K := \p 0ii0 $$ fulfill $I^2 = J^2 = K^2 = IJK = -\mathrm{Id}$.