# Wedge product decomposition of alternating bilinear form

I am trying to solve the following excercise:

Let E be a real vector space with dimension $$n$$, and let $$\sigma:E\times E \longrightarrow \mathbb{R}$$ be an alternating bilinear form. Show that there exist $$g_1,\ldots,g_{2r} \in E^*$$, linearly independent, such that

$$\sigma = g_1 \wedge g_2 + g_3 \wedge g_4 + \ldots + g_{2r-1} \wedge g_{2r}$$

I have been given the following clue: if $$\sigma \neq 0$$, there exist $$v_1,v_2 \in E$$ such that $$\sigma(v_1,v_2)\neq 0$$. Show that there exist $$g_1,g_2 \in E^*$$ such that $$g_1(v_1) = g_2(v_2) = 1, g_1(v_2) = g_2(v_1) = 0$$.

I have proven that such $$g_1$$ and $$g_2$$ exist. These two are furthermore linearly independent (l.i.). My idea is then to apply the same idea to $$\sigma_2 := \sigma - g_1 \wedge g_2$$. However, when I find $$g_3$$ and $$g_4$$, I know that $$\{g_3,g_4\}$$ is l.i. but I don't know how to prove that the whole set $$\{g_1,g_2,g_3,g_4\}$$ is l.i. I have gotten as far as to prove that at least one of $$g_3$$ or $$g_4$$ must be l.i. from $$g_1$$ and $$g_2$$.

• Is $g_1\wedge g_2$ defined as $(g_1\wedge g_2)(u,v) = g_1(u)g_2(v)-g_1(v)g_2(u)$? Oct 22, 2019 at 18:57
• @amsmath yes, it is. Oct 22, 2019 at 18:59

I assume you know that $$\sigma$$ can be written as a sum of wedge products. If not, have a look below under the bar. So, the task boils down to proving that whenever the linear forms in a sum of $$r$$ wedge products are linearly dependent, then the sum can be reduced to $$r-1$$ summands. And this is true. Consider the sum $$\sum_{i=1}^r\,g_i\wedge h_i.$$ Assume that $$g_1,\ldots,g_r,h_1,\ldots,h_r$$ are linearly dependent. WLOG, let $$h_r = \sum_{j=1}^ra_jg_j + \sum_{i=1}^{r-1}b_ih_i$$. Then \begin{align*} \sum_{i=1}^r\,g_i\wedge h_i &= \sum_{i=1}^{r-1}\,g_i\wedge h_i + g_r\wedge\left(\sum_{i=1}^{r-1}a_ig_i + \sum_{i=1}^{r-1}b_ih_i\right)\\ &= \sum_{i=1}^{r-1}\,\big(g_i\wedge h_i + a_i(g_r\wedge g_i) + b_i(g_r\wedge h_i)\big)\\ &= \sum_{i=1}^{r-1}\,\begin{cases}(g_i+b_ig_r)\wedge (h_i+\frac{a_i}{b_i}g_i) &\text{if }b_i\neq 0\\g_i\wedge (h_i-a_ig_r) &\text{if }b_i=0\end{cases}, \end{align*} which proves the claim.
I'll assume that $$E = \mathbb R^n$$. First note that any linear form $$g$$ has the form $$g(u) = a^Tu$$ with some $$a\in E$$. Now, if $$h(u) = b^Tu$$, then $$(g\wedge h)(u,v) = g(u)h(v)-h(u)g(v) = u^Tab^Tv-u^Tba^Tv = u^T(ab^T-ba^T)v.$$ The matrix $$B = ab^T-ba^T$$ is skew-symmetric, i.e., $$B^T=-B$$. Moreover, $$\sigma(u,v) = u^TAv$$ with a skew-symmetric matrix $$A$$, namely $$A = (\sigma(e_i,e_j))_{i,j=1}^n$$.
Consider the matrices $$E_{ij} = e_ie_j^T-e_je_i^T$$. Then $$\{E_{ij} : i=1,\ldots,n-1,\,j=i+1,\ldots,n\}$$ is a basis of the space of skew-symmetric matrices. Hence, $$A = \sum_{i,j}c_{ij}E_{ij}$$ and so $$\sigma(u,v) = u^TAv = \sum_{i,j}c_{ij}u^TE_{ij}v = \sum_{i,j}c_{ij}u^T(e_ie_j^T-e_je_i^T)v = \sum_{i,j}u^Tc_{ij}(g_i\wedge g_j)(u,v),$$ where $$g_i(u) = e_i^Tu$$. Now, we want to have them linearly independent. Set $$h_{ij} := c_{ij}g_j$$ and $$h_i := \sum_{j=i+1}^nh_{ij}$$. Then $$\sigma = \sum_{i=1}^{n-1}\sum_{j=i+1}^n g_i\wedge h_{ij} = \sum_{i=1}^{n-1}g_i\wedge\left(\sum_{j=i+1}^n h_{ij}\right) = \sum_{i=1}^{n-1}g_i\wedge h_i.$$