# Subspace generated by three bivectors

I have the following exercise that I have not been able to solve:

Consider three simple bivectors $$B_i=\alpha_i\wedge\beta_i, i=1,2,3$$ in $$\mathbb{R}^{4}$$. The bivectors $$B_i$$ are linearly independent in the space $$\bigwedge^2\mathbb{R}^{4}$$. Let's define $$V_{i}=$$span$$\{\alpha_i,\beta_i\}$$. Also consider that dim$$(V_i\cap V_j)=1$$

a) Let $$x_1$$ be any nonzero vector in $$V_2\cap V_3$$ and cyclically (that is to say $$x_2\in V_1\cap V_3$$ and $$x_3\in V_1\cap V_2$$). Prove that the space span$$\{x_1,x_2,x_3\}$$ is three dimensional or one dimensional, but not two dimensional.

b)Now suppose that there exists $$n\in\mathbb{R}^4$$ such that $$n_{I}B_{i}^{IJ}=0$$ for $$i=1,2,3$$. Prove that span$$\{x_1,x_2,x_3\}$$ is actually three dimensional in this case.

I have tried to solve it by seeing $$\{x_1,x_2,x_3\}$$ as $$\{x_1\}\cup\{x_2\}\cup\{x_3\}$$, considering that span$$\{x_1,x_2,x_3\}=$$span$$\{x_1\}+$$span$$\{x_2\}$$+span$$\{x_3\}$$ and using the formula dim$$(W_1+W_2+W_3)=$$dim$$(V_1)$$+dim$$(V_2)$$+dim$$(V_3)$$-dim$$(V_1\cap(V_2+V_3))$$-dim$$(V_2\cap V_3)$$, but the formula doesn't help to prove that dim(span$$\{x_1,x_2,x_3\})\neq2$$

The inequality dim$$(W_1+W_2+W_3)\leq$$ dim$$(W_1)$$+ dim$$(W_2)$$+ dim$$(W_3)$$- dim$$(W_1\cap W_2)$$- dim$$(W_1\cap W_3)$$- dim$$(W_2\cap W_3)$$+ dim$$(W_1\cap W_2\cap W_3)$$ doesn't work either.

And I have found that trying to prove the second item is even more difficult to me.

Could anybody help me??

(a) It's straightforward to find examples for which $$\dim \operatorname{span}\{x_1, x_2, x_3\}$$ is $$1$$, $$3$$.
So, suppose $$\dim \operatorname{span} \{x_1, x_2, x_3\} < 3$$. Then (by relabeling if necessary) we may assume that $$x_3 \in S := \operatorname{span}\{x_1, x_2\}$$. What can we conclude about $$V_1, V_2, V_3$$ if $$\dim S = 2$$?
(b) Suppose $$\dim \operatorname{span} \{x_1, x_2, x_3\} = 1$$. Then, we may take $$\alpha_1 = \alpha_2 = \alpha_3 = x_1$$ and thus $$B_i = x_1 \wedge \beta_i$$ for $$i=1,2,3$$. What can we then conclude from the linear independence of $$\{B_1, B_2, B_3\}$$?
• In b) how do you use the assertion $n_I B^{IJ}=0$ ? Apr 15, 2021 at 23:51
• If $\{\beta_1, \beta_2, \beta_3\}$ is linearly dependent, then so is $\{B_1, B_2, B_3\}$, which contradicts our hypothesis, i.e., $\{\beta_1, \beta_2, \beta_3\}$ is linearly independent. Since we're proving the contrapositive of the original statement, it's enough to show that that the common kernel of the contraction maps $x_I \to x_I B_i^{IJ}$ is trivial, i.e., that the only $n_I$ satisfying the conditions $n_I B_i^{IJ}$ is the zero vector. Apr 16, 2021 at 4:01