linear form $\phi : \mathbb{R^3} \times \mathbb{R^3} \to \mathbb{R}$ is alternate if and only if $\phi(v,v)=0$,for all $v \in \mathbb{R^3}$

Question

Prove that a bilinear form $\phi : \mathbb{R^3} \times \mathbb{R^3} \to \mathbb{R}$ is alternate if and only if $\phi(v,v)=0$,for all $v \in \mathbb{R^3}$.

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My thought:-

If $\phi : \mathbb{R^3} \times \mathbb{R^3} \to \mathbb{R}$ is alternate then we have
$\phi(v_1,v_2)= - \phi(v_2,v_1)$
by taking $v_1=v_2=v$ we have $\phi(v,v)= - \phi(v,v)$.
so $\phi(v,v)= 0$.
But I am not sure about the converse part. here $\phi$ is a bilinear form.Can I write $\phi(v,w)+\phi(w,v)=\phi(v+w,v+w)$ ? I yes then the problem is solved.But I am not sure about the step.

Can I get some help?

Answer 1

No, you cannot write $\phi(v,w)+\phi(w,v)=\phi(v+w,v+w)$, or at least not immediately. We need to use additivity in each argument independently, as follows:

$$\begin{array}{ll} \phi(v+w,v+w) & =\phi(v,v+w)+\phi(w,v+w) \\ & =\phi(v,v)+\phi(v,w)+\phi(w,v)+\phi(w,w).\end{array}$$

If $\phi(x,x)=0$ for all $x\in V$, which terms above can you delete - and if you do, what's left?