Demonstration of Electromagnetic Tensor antisymmetry I've already made a post about this topic here, but I realized that I didn't understand the explanation on that post. in Chapter 7 of Rindler's book on relativity, in section about electromagnetic field tensor, he states that
and introducing a factor 1/c for later convenience, we can ‘guess’ the tensor equation, $$ F_\mu= \frac{q}{c} E_{\mu \nu} U^\nu$$
thereby introducing the electromagnetic field tensor$$E_{\mu \nu}$$
We would surely want the
force $F\mu$ to be rest-mass preserving, which, according to (6.44) and (7.15), requires
$$F_\mu U^\mu = 0$$. So we need
$$E_{\mu \nu} U^\mu U^\nu = 0$$
for all $ U^\mu$ , and hence the antisymmetry of the field tensor
$$E_{\mu \nu}= −E_{\nu \mu}$$\
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.
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I'm really confused about the correct way to show that the equation $E_{\mu \nu} U^\mu U^\nu = 0$ implies the fact that $E_{\mu\nu}$ is antisymmetric tensor. What is  the correct demonstration of this implication?
OBS: i've saw some posts answering this kind of question with bilinear maps notation, instead of component notation. If possible, please make some demonstration using the index notation as in the post.
 A: It's probably clearer to go backwards:
$$ E_{\mu\nu} = -E_{\nu \mu}\\
\Leftrightarrow  E_{\mu\nu}U^\mu V^\nu = -E_{\nu\mu}U^\mu V^\nu \hspace{1em} \forall U, V\\
\Leftrightarrow  E_{\mu\nu}U^\mu V^\nu = -E_{\mu\nu}U^\nu V^\mu \hspace{1em} \forall U, V \hspace{1em}\text{Relabel RHS} \mu \leftrightarrow \nu\\
\Leftrightarrow E_{\mu \nu} (U^\mu + V^\mu)(U^\nu + V^\nu) = 0 \hspace{1em} \forall U, V
$$
Now we recognise that the addition of $V$ in the final equation doesn't actually change the condition, and without losing any generality we may take it to be zero.
So, we establish $E_{\mu\nu}U^\mu U^\nu \Leftrightarrow E_{\mu\nu} = - E_{\nu\mu}$
The only property used here is linearity of each tensor.
A: First decompose $E$ as the sum of its symmetric and antisymmetric parts: $$E_{ab} = E_{(ab)} + E_{[ab]}\:.$$
Now the idea is to prove that $$E_{(ab)} =0\tag{0}$$ so that $E_{ab} = E_{[ab]}$ is antisymmetric.
To this end observe that, in our hypothesis,
$$0= E_{ab}U^aU^b = E_{(ab)}U^aU^b + E_{[ab]}U^aU^b\:,\tag{1}$$
where
$$E_{[ab]}U^aU^b= E_{[ba]}U^bU^a= -E_{[ab]}U^bU^a = -E_{[ab]}U^aU^b =0\:.$$
Here (1) implies
$$E_{(ab)}U^aU^b =0\:.\tag{2}$$
Writing $U=X+Y$, we have from (2)
$$E_{(ab)}X^aX^b + E_{(ab)}Y^aY^b + 2E_{(ab)}X^aY^b=0\:.\tag{3}$$
where we have used
$$E_{(ab)}X^aY^b= E_{(ab)}Y^aX^b$$
as a consequence of the symmetry of $E_{(ab)}$. Using again (2) in (3) for $U=X$ and $U=Y$, we  obtain, for every choice of $X$ and $Y$,
$$E_{(ab)}X^aY^b=0\:.$$
In other words, all matrix elements of the matrix of elements $E_{(ab)}$ vanish, so that $E_{(ab)}=0$ and (0) is true conluding the proof.
