Assume F is a field of characteristic different than 2. Let a, b be invertible elements in F, and let A(a,b) be the generalised quaternions. Using the Artin–Wedderburn theorem, there is a representation of A(a,b) over F. I found a representation as Q8 but it's not over F. So, how to find a representation as matrices over F?
closed as offtopic by Johannes Hahn, abx, Alex Degtyarev, Suvrit, Stefan Kohl Mar 3 '15 at 21:27This question appears to be offtopic. The users who voted to close gave this specific reason:



You could take the regular representation (left multiplication on $A$). So if $x^2 = a, y^2 = b$ then taking a basis $\{1,x,y,xy\}$ of $A$, $x$ would be represented by the matrix $$\left( \begin{array}{cccc} 0 & a & 0 & 0 \cr 1 & 0 & 0 & 0 \cr 0 & 0 & 0 & a \cr 0 & 0 & 1 & 0 \end{array} \right), $$ etc. 


You need a (necessarily noncentral) subfield $K$ of your quaternions $Q$, of degree $2$ over the center $k$ of the quaternions. Then there is a representation of $Q$ as $2$by$2$ matrices over $K$, by choosing any $K$basis of $Q$... 

