Prove that is $A$ is skew-symmetric, then $X^TAX = 0$ for all $X = [x_1 x_2 \cdots x_n]^T$ Recall that a matrix $A$ is skew-symmetric if and only if $A^T = -A$. Prove that if $A$ is skew-symmetric, then $X^TAX = 0$ for all $X = [x_1 x_2 \cdots x_n]^T$
 A: Note that $X^TAX$ is a scalar, so $X^TAX=(X^TAX)^T=X^TA^TX$
but $A^T=-A$, so $X^TAX=X^T(-A)X\Rightarrow2X^TAX=0\Rightarrow X^TAX=0$
A: $$
X^T A X = \sum_{i,j}x_iA_{ij}x_j = \sum_{i,j}x_jA_{ji}x_i= - \sum_{i,j}x_jA_{ij}x_i = -\sum_{i,j}x_iA_{ij}x_j = -X^TAX=0
$$
A: Whoops!  In my haste I took $X$ to be a matrix!  But you can get the gist of it:
First, the way it should be done:
$X^TAX = (X^TAX)^T = X^TA^TX = -X^TAX \Rightarrow X^TAX = 0, \tag{0}$
assuming we are not over a field of characteristic $2$, since for vectors $X$, $X^TAX$ is a scalar quantity, whence the leftmost equality in the above binds. QED
As I said, I neglected the little $T$ in the upper right corner of $X = [x_1, x_2, . . ., x_n]^T$ and so took the $x_i$ to be the columns of $X$; I think I'd best get some rest!
But in unedited form, here's what I got:
This is not true, as taking $X = I$ and 
$A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \tag{1}$
illustrates.  In this case
$X^TAX = A \ne 0. \tag{2}$
What is true is that
$x_i^TAx_i = 0 \tag{3}$
for each column $x_i$ of $X$, since
$(X^TAX)^T = X^TA^TX = -X^TAX, \tag{4}$
which shows that $X^TAX$ is skew-symmetric; now note that for any columns of $X$, $x_i$ and $x_j$, $x_j^TAx_i$ is a scalar; indeed $(X^TAX)_{ji} = x_j^TAx_i$ but the skew-symmetry of $X^TAX$ forces its diagonal entries to vanish, whence (3).  And that's about as far as I can take it!
Whew!  Sleepy Time!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
