Q:

wedge product - distributivity over addition

Wedge and tensor algebra are very new concepts to me and I want to understand how to prove the following property of the wedge product: 
$$\omega\wedge(\eta_{1}+\eta_{2})=\omega\wedge\eta_{1}+\omega\wedge\eta_{2}$$
where $$\omega,\eta_{1},\eta_{2}\in\Lambda^{1}(V):=V^{\star}\hspace{5mm}(\diamondsuit)$$ 
By definition of the wedge product we have
$$\omega\wedge\eta_{1}(v_{1},v_{2})=\frac{(1+1)!}{1!1!}Alt(\omega\otimes\eta_{1})(v_{1},v_{2})=2!\frac{1}{2!}\displaystyle\sum_{\sigma\in S_{2}}\text{sgn}\;\sigma\cdot\omega(v_{\sigma(1)})\cdot\eta_{1}(v_{\sigma(2)})=$$
$$=\omega(v_{1})\eta_{1}(v_{2})-\omega(v_{2})\eta_{1}(v_{1})$$
Then the rhs of $(\diamondsuit)$ should be something like this
\begin{align}
\omega\wedge\eta_{1}(v_{1},v_{2})+\omega\wedge\eta_{2}(v_{1},v_{2})= &\omega(v_{1})\eta_{1}(v_{2})-\omega(v_{2})\eta_{1}(v_{1})+\omega(v_{1})\eta_{2}(v_{2})-\omega(v_{2})\eta_{2}(v_{1})\\
=&\omega(v_{1})[\eta_{1}(v_{2})+\eta_{2}(v_{1})]-\omega(v_{2})[\eta_{1}(v_{1})+\eta_{2}(v_{1})]\\
=&\omega\wedge(\eta_{1}+\eta_{2}))(v_{1},v_{2})&
\end{align}
On the other hand 
\begin{align}
\omega\wedge(\eta_{1}+\eta_{2}))(v_{1},v_{2})&=\frac{(1+1)!}{1!1!}Alt(\omega\otimes(\eta_{1}+\eta_{2}))(v_{1},v_{2})\\
&=\displaystyle\sum_{\sigma\in S_{2}}\omega(v_{\sigma(1)})\cdot (\eta_{1}+\eta_{2})(v_{\sigma(2)})\\
&=\omega(v_{1})(\eta_{1}+\eta_{2})(v_{2})-\omega(v_{2})(\eta_{1}+\eta_{2})(v_{1})\\
&=\omega(v_{1})\eta_{1}(v_{2})+\omega(v_{1})\eta_{2}(v_{2})-\omega(v_{2})\eta_{1}(v_{1})-\omega(v_{2})\eta_{2}(v_{1})
\end{align}
Is this approach correct? Also should this work for tensors of higher degrees? 
Thank you

A:

It's correct except for three typos. First, in the first line of your last equations, it should be $\text{Alt}(\omega\otimes (\eta_1+\eta_2))$.  Second, in two places you've messed up on distributing a negative: line 2 of the second computation, last line of the last computation.

