<s>[INST] How to prove that $x'Bx=0$ for all $n\times 1$ vectors $x$ if and only if $B'=-B$? 
Theorem. $x^TBx=0$ for all $n \times 1$ vectors $x$ if and only if $B^T=-B$.

Then how to prove this?
 [/INST] Suppose 
$B^T = -B; \tag 1$
then
$(x^TBx)^T = x^TB^Tx = -x^TBx; \tag 2$
since $x^TBx$ is a scalar, that is, a $1 \times 1$ matrix, we have
$x^TBx = (x^TBx)^T, \tag 3$
whence (2) yields
$x^TBx = -x^TBx \Longrightarrow x^TBx = 0; \tag 4$
likewise, if
$x^TBx = 0 \tag 5$
for all $x$, then
$y^TBx + x^TBy = x^TBx + y^TBx + x^TBy + y^TBy = (x + y)^TB(x + y) = 0, \tag 6$
so
$y^TBx = -x^TBy; \tag 7$
now again, the quantities on each side of this equation are scalars, so in particular
$y^TB^Tx = (x^TBy)^T = x^TBy, \tag 8$
so (7) may be written
$y^TBx = -y^TB^Tx, \tag 9$
for all $x$ and $y$; thus
$Bx = -B^Tx, \tag{10}$
implying
$B = -B^T. \tag{11}$
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