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Let $g_i$ denote the $i$'th orthonormal basis of a Clifford algebra $G_{p,q}$, then any number in Clifford domain can be represented as follows: $$a=\sum_{i=1}^{2^{p+q}}(a_ig_{i})$$ where $a_i$ is some real coefficient.

Given Clifford numbers $a,b\in G_{p,q}$ how can I in general solve the following equation for the Clifford number $x=\sum_{i=1}^{2^{p+q}}(x_ig_{i})$, $x \in G_{p,q}$,?

$$(\sum_{i=1}^{2^{p+q}}(x_i(g_{i})^3))(-b)x = (\sum_{i=1}^{2^{p+q}}(a_i(g_{i})^3))ba$$

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  • $\begingroup$ What's relation between $x$ and $x_i$? $\endgroup$ – Shuchang Nov 25 '13 at 15:16
  • $\begingroup$ Source? Context? $\endgroup$ – rschwieb Nov 25 '13 at 23:13
  • $\begingroup$ @ShuchangZhang $$x=\sum_{i=1}^{2^{p+q}}(x_ig_{i})$$ $\endgroup$ – Sunny88 Nov 26 '13 at 6:51
  • $\begingroup$ @ShuchangZhang Consider the case of $G_{0,1}$, then this clifford algebra is isomorphic to complex numbers, so $g_1=1$ and $g_2=i$. Then $g_1^3=1$ and $g_2^3=i*i*i=-i$. $\endgroup$ – Sunny88 Nov 26 '13 at 7:00
  • $\begingroup$ @rschwieb The actual equation uses the operator from my previous question: math.stackexchange.com/questions/269335/… With this operator it looks like $x^*(-b)x=a^*ba$. $\endgroup$ – Sunny88 Nov 26 '13 at 7:04

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