$\phi:\mathbb{R}^m\times \mathbb{R}^m\to\mathbb{R}$ alternating form of deg $2$. If $m$ is odd, $\exists v\neq 0$, $\phi(v,w) = 0$

Question

Let $\phi:\mathbb{R}^m\times \mathbb{R}^m\to\mathbb{R}$ be an alternating form of degree $2$. If $m$ is odd, there exists a non zero vector $v\in \mathbb{R}^m$ such that $\phi(v,w) = 0$ for all $w\in \mathbb{R}^m$

There are few theorems I know that would help to answer this. $v$ and $w$ are linearly dependent, then $\phi(v,w)=0$. Is the answer based on linear dependence? Working with the definition of alternating form, I tried to do $\phi(v,w) = -\phi(w,v)$ but got nothing. Somehow I need to relate the dimension of the space with the existence of such vectors. Anybody can help?

Answer 1

Hint: Consider the matrix $A_{ij}=\phi(\vec{e}_i,\vec{e}_j)$. This matrix is skew symmetric and has an odd number of rows/columns. Therefore, its determinant is $0$ (the proof of this can be found wikipedia). Can you use this to show that there is a vector $\vec{v}$ so that $\phi(\vec{v},\vec{e}_i)=0$ for all $i$?

Additional hint $1$:

Show that $(A\vec{v})_i=\phi(\vec{e}_i,\vec{v})$.

Additional hint $2$:

Find a vector $\vec{v}\not=\vec{0}$ so that $A\vec{v}=\vec{0}$.

The $\vec{v}$ constructed in the two additional hints is exactly the $\vec{v}$ needed by the question.