Quadratic forms and Clifford Algebra Part 2 So just to ask, if $q(x, y) = ax^2 + by^2$ is a quadratic form in two variables over a field $K$ ($a, b \in K$) with char $K \neq 2$, how is $C(q)$ isomorphic to $M_2(K)$?
 A: If the field is algebraically closed (or at least $a$ and $b$ have square roots in $K$), then it's easy to see that $A=(\frac{1}{\sqrt{a}},0)$ and $B=(0,\frac{1}{\sqrt{b}})$ form an orthonormal basis for the two dimensional space with respect to the quadratic form. Further, $q(A)=q(B)=1$. We have that $\{1,A,B,AB\}$ is a basis for the Clifford algebra, all elements squaring to 1.
So: can you think of two distinct anticommuting elements $m_1$ and $m_2$ in the matrix ring which square to the identity matrix, and whose product $m_1m_2$ squares to the identity matrix? 
Hint: you can do it with just the elements $\{0,1,-1\}$ in the matrices. (It's important that the characteristic is not 2 so that $1\neq -1$.) Keep it simple! After you have found these elements, the obvious map gives you an isomorphism between the matrix ring and the Clifford algebra.

The added condition that $ab=0$ keeps $a$ and $b$ from being zero, but in the general case, it may be that an orthonormal basis contains elements squaring to -1.
