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Let $V$ be a vector space over a field $k$ , and $Q:V\times V\to k$ a symmetric bilinear form. Then the Clifford algebra $\Cliff(Q,V)$ is the quotient of the tensor algebra $\mc{T}(V)$ by the relations
$$v\otimes w+w\otimes v=-2Q(v,w)\qquad \forall v,w\in V.$$
Since the above relationship is not homogeneous in the usual $\Z$ -grading on $\mc{T}(V)$ , $\Cliff(Q,V)$ does not inherit a $\Z$ -grading. However, by reducing mod 2, we also have a $\Z_2$ -grading on $\mc{T}(V)$ , and the relations above are homogeneous with respect to this, so $\Cliff(Q,V)$ has a natural $\Z_2$ -grading, which makes it into a superalgebra.
In addition, we do have a filtration on $\Cliff(Q,V)$ (making it a filtered algebra), and the associated graded algebra $\Gr\Cliff(Q,V)$ is simply $\Lambda^*V$ , the exterior algebra of $V$ . In particular, $$\dim\Cliff(Q,V)=\dim\Lambda^*V=2^{\dim V}.$$
The most commonly used Clifford algebra is the case $V=\R^n$ , and $Q$ is the standard inner product with orthonormal basis $e_1,\ldots,e_n$ . In this case, the algebra is generated by $e_1,\ldots,e_n$ and the identity of the algebra $1$ , with the relations
Trivially, $\Cliff(\R^0)=\R$ , and it can be seen from the relations above that $\Cliff(\R)\cong\C$ , the complex numbers, and $\Cliff(\R^2)\cong\mathbb{H}$ , the quaternions.
On the other ha nd, for $V=\C^n$ we get the particularly simple answer of $$\Cliff(\C^{2k}) \cong \mathrm{M}_{2^k}(\C) \qquad \Cliff(\C^{2k+1}) = \mathrm{M}_{2^k}(\C) \oplus \mathbf{M}_{2^k}(\C).$$
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