I'm studying Chapter 16 on Multilinear Algebra of MacLane and Birkhoff's Algebra and I'm a little confused about using decomposable vectors. The question I haven't been able to solve is this:
Given $V$ a finite dimensional vector space, $s, t \in \Lambda_p(V)$ nonzero decomposable p-vectors, and $S = \{w \in V \mid w \wedge s = 0\}$, $T = \{w \in V \mid w \wedge t = 0\}$, show that $s+t$ is decomposable if and only if $S \cap T$ has dimension at least $p-1$.
One direction seems pretty obvious, since if we have $p-1$ linearly independent vectors $b_1, \dots, b_{p-1}$ in $S \cap T$, then we can expand that to bases $b_1, \dots, b_{p-1}, u$ of $S$ and $b_1, \dots, b_{p-1}, u'$ of $T$ where $u, u' \in V$, so that $s = b_1 \wedge \cdots \wedge b_{p-1} \wedge u\kappa$ and $t = b_1 \wedge \cdots \wedge b_{p-1} \wedge u'\kappa'$, and then $s+t = b_1 \wedge \cdots \wedge b_{p-1} \wedge (u\kappa + u'\kappa')$ is decomposable.
For the other direction I'm not really sure what to do, I've tried everything I can think of but I haven't gotten it to work.
For any multivector $x$ define $$ [x] = \{w \in V \;:\; w\wedge x = 0\} $$ so that e.g. $S = [s]$ and $T = [t]$.
If $[s+t] = S\cap T$ then $S = T$ and we're done, so suppose not. Then there is $v \in [s+t]\setminus S\cap T$ and $$ 0 = v\wedge(s+t) \implies v\wedge s = -v\wedge t. $$ This shows that if $v \in S$ or $v \in T$ then $v \in S\cap T$, so in fact we have $$ v\wedge s = -v\wedge t \ne 0. $$ Thus $$ [v]\oplus S = [v]\oplus T $$ and there is $\alpha \in V^*$ such that $av + v_s = \alpha(av + v_s)v + v_t$ for all $v_s \in S$, $v_t \in T$, and scalars $a$. We can now say that $$ S\cap T = \{w \in S \;:\; \exists a.\;\alpha(av+w) = a\}. $$ But now notice: $$ \alpha(av + w) = a \iff a(\alpha(v) - 1) + \alpha(w) = 0, $$ so if $\alpha(v) \ne 1$ we can always choose $a = \alpha(w)/(1-\alpha(v))$ and then $S = S\cap T = T$. So it must be that $\alpha(v) = 1$, and now $$ S\cap T = \{w \in S \;:\; \alpha(w) = 0\} = S\cap\ker\alpha. $$ If $S \subseteq \ker\alpha$ then again $S = S\cap T = T$ so this is not the case; now because $\ker\alpha$ is a hyperplane and $\dim S = p$ we finally see $$ \dim S\cap T = \dim(S\cap\ker\alpha) = p - 1. $$