Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $M_2(K)$?

share|cite|improve this question
Ha? What are $C(q)$ and $M_2(K)$? – Gunnar Þór Magnússon 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
up vote 0 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.