This is a follow-up to a question I asked here yesterday. It's coming from a (non-examinable) exercise sheet and I really can't get my heard around how this question is posed and to be approached.
For clarity I'll re-type relevant parts of the initial post.
Suppose we have some field $K$ and non-zero elements $a,b,$ in $K$. Define $H=H(a,b)$ to be the $K$-algebra with basis $\{1,x,y,z \}$ over $K$ satisfying $$x^2=a, \\ y^2=b, \\ z=xy=-yx$$
Additional Information Now we're asked to suppose that $\sqrt{a}$ is in fact an element in the the field $K$.
Question We need to show that there is an algebra homomorphism $\phi : H \rightarrow M_2(K)$ such that
$$ \phi(x)= \left( \begin{array}{ccc} \sqrt{a} & 0 \\ 0 & -\sqrt{a} \end{array} \right) , \\ \phi(y)= \left( \begin{array}{ccc} 0 & b \\ 1 & 0 \end{array} \right) $$
Initial Thoughts It is clear that $\{\phi(x)\}^2$ is $a.\text{Id}$ and $\{\phi(y)\}^2$ is $b.\text{Id}$, which will presumably tie in nicely with the fact that $x^2=a$ and $y^2=b$.
If anyone can shed light on what this homomorphism actually is, and why it's a homomorphism, I'd be very grateful.
$H$ has a basis $1, x,y,z$. To give a linear map $\phi \colon H \to M_2(K)$ it is enough to give the values on the basis. As we want $\phi$ to be an algebra homomorphism, we let $$ \phi(1) = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} $$ $\phi(x)$ and $\phi(y)$ are given by the exercise. Moreover $$ \phi(x) \phi(y) = \begin{pmatrix} 0 & b \sqrt a\\ - \sqrt{a} & 0 \end{pmatrix} $$ So we let $$ \phi(z) = \begin{pmatrix} 0 & b \sqrt a\\ - \sqrt{a} & 0 \end{pmatrix}. $$ Alltogether we have $$ \phi(\alpha + \beta x + \gamma y + \delta z) = \begin{pmatrix} \alpha + \beta \sqrt a & \gamma b + \delta b\sqrt{a}\\ \gamma - \delta \sqrt a & \alpha - \beta \sqrt a \end{pmatrix} $$ Now we have to check that $\phi$ is multiplicative, it suffices to check that $$ \phi(x)^2 = a\,{\rm Id}, \phi(y)^2 = b\,{\rm Id}, \phi(x)\phi(y) = -\phi(y)\phi(x) = \phi(z) $$ which are all true. So $\phi$ is a homomorphism of $K$-algebras.