Subalgebra of Clifford algebra Let $Cl(V\oplus V^{*})$ the Clifford algebra of $V\oplus V^{*}$ with natural pairing. 
How can check a subset $ Λ^{•} (V^{*}) $ is a subalgebra of Clifford algebra $Cl(V\oplus V^{*})$?
 A: The natural quadratic form on $V\oplus V^*$ is
$$Q(v, \phi) = \phi(v)$$
where $v\in V$ and $\phi \in V^*$. It is an example of hyperbolic/split/metabolic quadratic space.
Elements of $V^*$ are naturally embedded into $V\oplus V^*$ via
$$\phi \mapsto (0, \phi)$$
and the value of the quadratic form on those is
$$Q(0,\phi) = \phi(0) = 0$$
Thus, the embedded copy of $V^*$ is a subspace with zero quadratic form. The Clifford algebra of such a space is, by definition, the exterior algebra. Since Clifford algebra is functorial, embedding of quadratic spaces gives algebra embeddings; thus, the Clifford algebra of $V^*$ with a zero quadratic form, which is $\Lambda(V^*)$, is embedded into $\operatorname{Cl}(V \oplus V^*)$. This is exactly the subalgebra generated by the embedded copy of $V^*$ via the diagram of embeddings
$$
\begin{matrix}
V^* & \rightarrow & V \oplus V^* \\
\downarrow & & \downarrow \\
\Lambda(V^*) & \rightarrow & \operatorname{Cl}(V \oplus V^*)
\end{matrix}
$$
The same reasoning applies to $V$ embedded into $V \oplus V^*$ and $\Lambda(V)$ embedded into $\operatorname{Cl}(V \oplus V^*)$.
