Prove: $T^*=-T\Rightarrow \langle T(v),v\rangle=0$ over $\mathbb{F}$

Question

Prove: $T^*=-T\Rightarrow \langle T(v),v\rangle=0$ over $\mathbb{F}$

Can one prove this above any field? I see that $T$ is a anti-symmetric operator can it help?

Answer 1

Assuming that you're dealing with a symmetric bilinear form over the $\mathbb{F}$-vector space $V$ and that you have an operator $T\colon V\to V$ such that $$ \langle T(v),w\rangle=\langle v,-T(w)\rangle $$ for all $v,w\in V$, then, for $w=v$ we get $$ \langle T(v),v\rangle=\langle v,-T(v)\rangle=-\langle v,T(v)\rangle =-\langle T(v),v\rangle $$ and we can derive that $2\langle T(v),v\rangle=0$ so $$ \langle T(v),v\rangle=0 $$ provided the characteristic of $\mathbb{F}$ is different from $2$.

If the characteristic is $2$, then $-T=T$ and there's nothing you can say about $\langle T(v),v\rangle$ in general.