Why is the following map an isomorphism between $Cl(V,\omega)$ and $\operatorname{End}(\Lambda(V))$? 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.
 A: I do not think that your question is all that bad, and in fact I think I have found a correct statement and a correct proof, see below. To summarize the proof, we have a basis indexed by subsets of $\lbrace 1,2, \ldots ,n \rbrace$ and to show the isomorphism property, we need to show we can go freely from any subset to any other. To achieve this, we use the
$x_i$ to delete elements and the $y_i$ to insert elements.
There are two things to be corrected in your original question :
1) The form $w$ is symmetric, not antisymmetric (nobody ever constructed Clifford algebras from antisymmetric forms)
2) The formula for $\sigma(y)(x)$ should be $2w(y,x)$ instead of $w(y,x)$.
To be more specific : $\lbrace x_1, \ldots ,x_n \rbrace$ is a basis for $W$. So $\Lambda (W)$ has a basis ${\cal B}=\big({\lambda_I}\big)_{I \subseteq E}$ indexed by the subsets $I$ of $E=\lbrace 1,2, \ldots ,n \rbrace$, where for $I=\lbrace i_1&lti_2&lt \ldots &lti_t \rbrace$ we put
$$
\lambda_I=w_{i_1} \wedge w_{i_2} \wedge \ldots w_{i_t} 
$$
(in particular, $\lambda_{\emptyset}=1$, the unity element in $\Lambda (W)$). Note that the wedge product by $x_i$ (let us denote that operation by $X_i$) acts on this basis by
$$
 X_i(\lambda_I)=\left\lbrace
\begin{array}{ll}
0 , & {\rm if } \ i \in I, \\
(-1)^s \lambda_{I \cup \lbrace i \rbrace}, & {\rm if } \  i \not\in I, {\rm where} \ s \
{\rm is \ the \ number \ of \ elements \ in \ } I {\rm \ below \ } i. 
\end{array}
\right.
$$
It is natural to reverse this process and define a new action $Y_i$ on $W$ by
$$
 Y_i(\lambda_I)=\left\lbrace
\begin{array}{ll}
0 , & {\rm if } \ i\not\in I, \\
(-1)^s \lambda_{I \setminus \lbrace i \rbrace}, & {\rm if } \  i \in I, {\rm where} \ s \
{\rm is \ the \ number \ of \ elements \ in \ } I {\rm \ below \ } i. 
\end{array}
\right.
$$
For each $I\subseteq E$, denote by $F_I$ and $G_I$, respectively, the subspace of ${\Lambda}(W)$ generated by the $\lambda_J$ such that $I\subseteq J$ ($ I \not\subseteq J$, respectively). Then $F_I$ and $G_I$ are complementary subspaces in ${\Lambda}(W)$. Let us also write $F_i$ instead of $F_{\lbrace i \rbrace}$ and $G_i$ instead of $G_{\lbrace i \rbrace}$ for short. Then by contruction, $Z_i=X_iY_i$ is the projector onto $F_i$ according to $G_i$, and $Y_iX_i$ is the projector onto $G_i$ according to $F_i$.
Similarly we have
$$
X_i^2=0, Y_i^2=0, X_iY_j+X_jY_i=2\delta_{ij}
$$
so that we have in fact defined an action of the Clifford algebra. Note also that if $a$ and $b$ are two elements of ${\sf End}(\Lambda(W))$, and $a$ is decomposable, say 
$a=\lambda_{I}$ for some $I \subseteq E$, then it is easily checked that
$$
Y_i(a \wedge b)=Y_i(a)\wedge b + (-1)^{|I|} a \wedge Y_i(b)
$$
(a property you mention in your original post).
Let us now show the isomorphism property. First note that the two spaces, ${\sf Cl}(V,w)$ and ${\sf End}({\lambda}(W))$ have the same dimensionality, namely $2^{2n}$. So it suffices to show that this homorphism is surjective, i.e. that its image $\cal I$ is the full space ${\sf End}({\Lambda}(W))$. Now the basis we have given for ${\Lambda}(W)$ provides a standard basis $(\gamma_{I,J})_{I,J \subseteq E}$ for ${\sf End}({\Lambda}(W))$,  where
$$
\gamma_{I,J} (\lambda_K)=\delta_{KJ}\lambda_I
$$
So all we need to show is that all the $\gamma_{I,J}$ are in $\cal I$.
Now $\cal I$ contains all the $Z_i$. So $\cal I$ contains $Z_I$ for any $I \subseteq E$, where we put $Z_{ \lbrace i_1 &lt i_2&lt \ldots &lt i_t \rbrace }=Z_{i_1}Z_{i_2} \ldots Z_{i_t}$. So $\cal I$ also contains all the $Z_{I,t}$ for $I \subseteq E$ and $0 \leq t \leq n$, where we put
$$
Z_{I,t}=\sum_{J \subseteq I, |J|=t } Z_J
$$
The action of all those elements on the basis $\cal B$ can be written as follows : for
any $K,I \subseteq E, i \in E, 0 \leq t \leq n$, we have
$$
Z_i(\lambda_{K})=|K \cap \lbrace i \rbrace| \lambda_K, \
Z_{I,t}(\lambda_{K})=\binom{|K \cap I|}{t} \lambda_K
$$
Note that the polynomials $1,x,\frac{x(x-1)}{2},\binom{x}{3}, \ldots , \binom{x}{n}$ form a basis of the space of polynomials of degree $\leq n$ in $x$. For $i\in E$, we know that there a polynomial $\Delta_i(x)$ such that $\Delta_i(x)=\delta_{ix}$ for all $x\in E$
(explicitly, $\Delta_i(x)=\prod_{j \neq i}{\frac{x-j}{i-j}}$). So for any $i\in E$ there are coefficients $a_{i0},a_{i1}, \ldots ,a_{in}$ such that 
$$
\delta_{ix}=\sum_{t=0}^{n} a_{it} \binom{x}{t} \ (x \in E)
$$
In fact, all the coefficients $a_{it}$ can be computed explicitly, but it is not necessary here and I am too lazy to do it. If we put
$$
{Z'}_I=\sum_{t=0}^{n} a_{|I|t} Z_{I,t}
$$
then by construction $Z'_I$ is in $\cal I$, and is the projector onto the line ${\sf span} (\lambda_I)$  according to the hyperplane ${\sf span} (\lambda_J)_{J \neq I}$.
Let $I,J \subseteq E$. We claim that there is a $f\in {\cal I}$ such that 
$f(\lambda_I)=\lambda_{J}$. Indeed, if we put $I \setminus J=\lbrace j_1,j_2, \ldots j_t \rbrace$ and $J \setminus I = \lbrace i_1,i_2, \ldots i_s \rbrace$ and $f=X_{i_1}X_{i_2} \ldots X_{i_t}Y_{j_1}Y_{j_2} \ldots Y_{j_t}$, then $f\in {\cal I}$ and $f(\lambda_I)=\lambda_{J}$. It follows that $\gamma_{I,J}=fZ'_I$ is also in $\cal I$, as wished.
A: 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 :(
