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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 $M_2(K)$?

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Ha? What are $C(q)$ and $M_2(K)$? –  Gunnar Magnusson Jan 15 '13 at 13:42
    
Oh, $C(q)$ is the Clifford Algebra of $q$ and $M_2(K)$ is the 2x2 matrices over $K$. Sorry about that! –  Eric Jan 15 '13 at 13:44
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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.

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I forgot to mention the condition that $ab \neq 0$. Being algebraically closed of $K$ is not mentioned. =| However, with your question, is it: $m_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and $m_2 = - m_1$? –  Eric Jan 15 '13 at 15:15
    
@Eric Unfortunately, that choice of $m_1$ and $m_2$ commute rather than anticommute... but you are close! You can't miss it if you keep going. –  rschwieb Jan 15 '13 at 15:23
    
Ohhhh! Ok! Let me find $m_1$ and $m_2$. You Sir are great! –  Eric Jan 15 '13 at 15:50
    
Is it $m_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and $m_2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$? –  Eric Jan 15 '13 at 16:09
    
Just to clarify, without the condition on square roots, what happens to $A$ and $B$? –  Eric Jan 15 '13 at 16:14
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