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?
 A: 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
$B = -B^T. \tag{11}$
A: Let $B=(b_{ij})_{1\leq i,j\leq n}$. Then $x'Bx=\sum_{1\leq i,j\leq n} b_ {ij}x_{i}x_{j}$ where $x=(x_1,x_2,\ldots,x_n)'$.
Suppose first that $B'=-B$, i.e., $b_{ij}=-b_{ji}$ for all $1\leq i,j\leq n$. Then the sum above must vanish, since $\sum_{1\leq i,j\leq n} b_ {ij}x_{i}x_{j}=\sum_{1\leq i\leq j\leq n} (b_ {ij}+b_{ji})x_{i}x_{j}$.
Now let us assume that $\sum_{1\leq i,j\leq n} b_ {ij}x_{i}x_{j}=0$ for all $x$. Using $x=(0,0,\ldots,1,\ldots,0)'=:e_i$, the $i$-th unit vector you geht $b_{ii}=0$ for all $i$. Using $x=e_i+e_j$ with $i\not=j$ results in $b_ {ij}+b_{ji}=0$, as desired.
