$V\oplus V^*$ acting on $\Lambda^{\bullet}V^*$ Let $V$ be an $n$ dimensional real vector space and $V^*$ be the dual vector space.
The elements of $V\oplus V^*$ acts on $\Lambda ^{\bullet} V^*$ naturally.
That is if $v+\xi$ is in $V\oplus V^*$ and $\varphi$ is in $\Lambda ^{\bullet} V$ , then $(v+\xi)\varphi=\iota_{v}\varphi+\xi \wedge \varphi$.
My question is that for arbitary $\varphi \in \Lambda ^{\bullet} V^*$, is there any $0 \neq X \in V\oplus V^*$ such that $X \varphi=0$?
 A: Let $V$ be a $4$ dimensional vector space, with basis $\{e_i\}$. Fix a dual basis $e^i$ of $V^*$. Let $\varphi=e^1\wedge e^2+e^3\wedge e^4$. I claim no $X\in V\oplus V^*$ will kill $\varphi$. Indeed, consider $X=v+\xi$. The action of $v$ decreases degree and the action of $\xi$ increase it, so if $X\varphi=0$, we must have $\iota_v\varphi=\xi\wedge \varphi=0$. 
But now write out $\xi=\sum_{i}{a_i} e^i$ and compute $\xi\wedge \varphi$. It is easy to see that the summand $a_1e^1\wedge e^3\wedge e^4$ cannot cancel with anything. So $a_1=0$. Symmetrically $a_2=a_3=a_4=0$.  Hence $\xi=0$. 
Now suppose $v=\sum b_i e_i$. Then $$\iota_v\varphi(e_j)=\varphi(\sum_i b_ie_i,e_j)=\sum_ib_i\varphi( e_i,e_j).$$
Plugging in $i=1$, $0=\iota_v\varphi(e_1)=b_2$, so $b_2=0$ and similarly for the other $b_i$. Thus $v=0$, and hence $X=0$.
As was suggested in a now-deleted comment, a similar trick will work for even dimensional vector spaces of dimension $\geq 4$. Let $\varphi=e^1\wedge e^2+e^3\wedge e^4+ e^5\wedge e^6+\cdots$ and go through a similar argument.
