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 $g_2$.

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

1 Answer 1


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 $$ \sigma = \sum_{i=1}^{n-1}\sum_{j=i+1}^n g_i\wedge h_{ij} = \sum_{i=1}^{n-1}g_i\wedge\left(\sum_{j=i+1}^n h_{ij}\right) = \sum_{i=1}^{n-1}g_i\wedge h_i. $$


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