Annihilator of a Tensor This is a question I have trouble understanding, hope you can clarify this to me. 
Problem:
Find the annihilator of the tensor $e_1\wedge e_2+e_3\wedge e_4$ in $V=\left<e_1,\,e_2,\,e_3,\,e_4\right>$. 
 A: $\{e_i\}$ is a dual basis, i.e., $$e_i^\ast (e_j)=\delta_{ij}.$$ That is any linear map from $V$ to ${\bf R}$ can be written by $$ \sum_{i=1}^4 c_i e_i^\ast$$ 
Note that $V\otimes V\ (=M_4({\bf R}))$ is vector space whose basis is $\{ e_i\otimes e_j \}$ So $$(e=)\ e_1\wedge e_2+e_3\wedge e_4=e_1\otimes e_2 - e_2\otimes e_1+ e_3\otimes e_4 - e_4\otimes e_3$$ is an element of $V\otimes V$. 
Consider dual space $(V\otimes V)^\ast $ whose basis is $\{ e_i^\ast \otimes e_j^\ast \}$
Here annihilator of $(e)$ is a collection $$A=\{f\in (V\otimes V)^\ast | \ f(e)=0\}$$
So $$A=(e_1^\ast\otimes e_2^\ast - e_3^\ast\otimes e_4^\ast)\oplus(e_1^\ast\otimes e_2^\ast + e_2^\ast\otimes e_1^\ast)\oplus (e_3^\ast\otimes e_4^\ast + e_4^\ast\otimes e_3^\ast)\oplus \bigoplus_{(i,j)\in J} e_i^\ast \otimes e_j^\ast $$ where $$J=\{ (i,j)|\ (i,j)\neq (1,2),\ (2,1),\ (3,4),\ (4,3)\} $$ 
A: Here is how I interpret the question:
The set of annihilators is
$Ani(e_1 \wedge e_2 + e_3 \wedge e_4) = \{v \in V\ |\ (e_1 \wedge e_2 + e_3 \wedge e_4) \wedge v = 0\}$
Note that any element $v \in V$ can be written as:
$v = \alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 + \alpha_4 e_4$
So $v \in Ani(e_1 \wedge e_2 + e_3 \wedge e_4)$ implies that
$(e_1 \wedge e_2 + e_3 \wedge e_4) \wedge (\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 + \alpha_4 e_4) = 0$
Hence we have that 
$\alpha_1 (e_1 \wedge e_3 \wedge e_4) \\
+ \alpha_2 (e_2 \wedge e_3 \wedge e_4) \\
+ \alpha_3 (e_1 \wedge e_2 \wedge e_3) \\
+ \alpha_4 (e_1 \wedge e_2 \wedge e_4) = 0$
Note that $\{(e_1 \wedge e_3 \wedge e_4), (e_2 \wedge e_3 \wedge e_4), (e_1 \wedge e_2 \wedge e_3), (e_1 \wedge e_2 \wedge e_4)\}$ are linearly independent in $\Lambda^3 (V)$ so for the above equation to be $0$ we require that $\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0$ hence
$Ani(e_1 \wedge e_2 + e_3 \wedge e_4) = \{0\}$
