TITLE: Why is the following map an isomorphism between $Cl(V,\omega)$ and $\operatorname{End}(\Lambda(V))$?
QUESTION [8 upvotes]: Suppose you have a vector space $V$ of dimension $2n$. I know that there exists a basis $x_1,\dots,x_n,y_1,\dots,y_m$ such that $\omega(x_i,x_j)=\omega(y_i,y_j)=0$ and $\omega(x_i,y_j)=\delta_{ij}$, where $\omega$ is a bilinear symmetric nondegenerate form.
Now let $W=\operatorname{span}\{x_1,\dots,x_n\}$, and $W'=\operatorname{span}\{y_1,\dots,y_n\}$. If $Cl(V,\omega)$ denotes the corresponding Clifford algebra, then one can define a $Cl(V,\omega)$-module structure on the exterior algebra $\Lambda(W)$ as follows. For $x\in W$, let $\sigma(x)\in\operatorname{End}(\Lambda(W))$ as wedge multiplication by $x$ on the left. Also, for $y\in W'$, one can define $\sigma(y)x=\omega(y,x)$ for $x\in W$, which then extends to $\sigma(y)\in\operatorname{End}(\Lambda(W))$ by
$$
y(a\wedge b)=y(a)\wedge b+(-1)^{\deg a}a\wedge y(b).
$$
My question is, why does the $\sigma$ defined above in fact give an isomorphism between $Cl(V,\omega)$ and $\operatorname{End}(\Lambda(W))$? I'm viewing the endomorphisms as over the underlying field $k$, by the way.

REPLY [2 votes]: I would like to begin similiarly to a previous post with the comment that both algebras have the same dimension, but I find it easier to show that the map is injective by using properties of the Clifford algebra (because I find writing out the extension to be unhelpful). Let me also add that I'm working under the assumption the form is symmetric. (I have never heard of a Clifford algebra using anything but.) 
I think $^\dagger$
the suggested map is :$$
\sigma_{w+w'}(x):=w\wedge x+\omega(w',x)\wedge 1
$$
for all x in $W$ and $w+w'\in W+W'=V$, and then it is extended
 to have domain all of $\bigwedge W$. So, $\sigma$ is defined from $V$ into $End(\bigwedge W)$. 
If $\sigma_{w+w'}(x)=0$ for all $x\in V$, then both $w\wedge x=0$ and $\omega(w',x)=0$ for all $x$ in $V$, but then by the definition of $\omega$, both $w$ and $w'$ must be zero. Hence $\sigma$ is injective on V, and its image has dimension $2n$.
It is (or at least seemed) routine to verify that $\sigma_v\circ\sigma_v=\omega(v,v)$ for all $v\in V$, and so the universal mapping property would kick in and extend the map
$$
 v\mapsto \sigma_v
$$
to an algebra homomorphism from $C\ell(V,\omega)$ to $End(\bigwedge W)$.
Since $\sigma$ injected $V$ into $End(\bigwedge W)$, the extension is an injection of $C\ell(V,\omega)$ into $End(\bigwedge W)$ as well. Since these two algebras have the same finite dimension, this would be an isomorphism.
$^\dagger$ I'm not 100% confident though :(