# Vectors in Clifford Algebra

I'm studying Clifford Algebra $\mathcal{Cl}_2$ and got stuck in an exercise:

Let $\mathbf{a}=e_2+e_{12},\quad \mathbf{b}=(1/2)(1+e_1).$ Compute $\mathbf{ab}$.

The answer is zero, but I can't get to it and I'm having trouble putting $\mathbf{b}$ in the form $\mathbf{b}=b_1e_1+b_2e_2$.

Important information:

• $(1, e_1, e_2, e_{12})$ form a basis for the Clifford algebra $\mathcal{Cl}_2$
• The Clifford product of two vectors $\mathbf{a}=a_1e_1+a_2e_2\text{ and }\mathbf{b}=b_1e_1+b_2e_2$ is defined as $\mathbf{ab}=a_1b_1+a_2b_2+(a_1b_2-a_2b_1)e_{12}$
• And the following multiplication table: $$\begin{array}{cccc} & \mathbf{e_1} & \mathbf{e_2} & \mathbf{e_{12}} \\\\ \mathbf{e_1}& 1 & e_{12} & e_2 \\\\ \mathbf{e_2}& -e_{12} & 1 & -e_1 \\\\ \mathbf{e_{12}}& -e_2 & e_1 & -1 \end{array}$$

$$\begin{eqnarray*} 2{\bf a}{\bf b} &=& (e_2+e_{12})(1+e_1) \\ &=& e_2+e_{12}+e_2 e_1+e_{12}e_1 \\ &=& e_2 + e_{12} +(-e_{12}) + (-e_2) \\ &=& 0 \end{eqnarray*}$$
• @FernandoH.M.Bastos: Not a problem. What makes you think you can write ${\bf b}$ in terms of $e_1$ and $e_2$? Commented Jul 26, 2012 at 0:43
• @FernandoH.M.Bastos: The space is spanned by the linearly independent set $\{1,e_1,e_2,e_{12}\}$. By linear independence, you can't write any of the basis elements as a linear combination of the others. Think of it as trying to write $\hat z$ as a linear combination of $\hat x$ and $\hat y$. Commented Jul 26, 2012 at 0:55