($X^{T}AX=0,\;\;\forall X$) iff $A$ is a skew symmetric matrix $X$ is a column vector and $A$ is a square Matrix. 
$\leftarrow$ 
$X^{T}AX=X^{T}A^{T}X=X^{T}(-A)X \implies X^{T}AX =0$
$\rightarrow$ 
I'm stuck here I tried doing the above backwards but I don't think I can say that 
$X^{T}A^{T}X=X^{T}(-A)X \implies A^{T}=-A$, can I ?
any help or hints will be appreciated. thanks !
 A: In a first step we prove this for the case $n=2$ where $X\in \mathbb{R}^n$.
Let $A\in \mathbb{R}^{2\times 2}$ be given such that $X^TAX=0$ for all $X$.
Then chosing $X=(1,0)$ and $X=(0,1)$ yields that the diagonal of $A$ has to be zero.
So, we assume
$$
 A= 
 \begin{bmatrix}
  0 & a \\ b & 0
 \end{bmatrix}
$$
Chosing $X=(1,1)$ yields $X^T AX = a+b$, thus $a=-b$, thus $A=-A^T$.
This easily extends to higher dimensions, by proving the same for all $2\times 2$ sub-matrices
$$
\begin{bmatrix}
 a_{ii} & a_{ij} \\ a_{ji} & a_{jj}
\end{bmatrix}
$$
with $1\leq i< j\leq n$.
A: Suppose that $x^TAx=0$ for all $x$, and define $B=A+A^T$. Then it is clear that:


*

*$B$ is symmetric

*$x^TBx=0$ for all $x$


As a symmetric matrix, $B$ has a complete set of orthonormal eigenvectors $v_i$ with eigenvalues $\lambda_i$, such that $Bv_i = \lambda_i v_i$, and any $x$ can be written as $\sum_{i=1}^n \langle x,v_i \rangle v_i$. 
Then:
$$0 = x^TBx = x^TB\sum_{i=1}^n \langle x,v_i \rangle v_i = x^T\sum_{i=1}^n \langle x,v_i \rangle \lambda_i v_i = \sum_{i=1}^n \langle x,v_i \rangle^2 \lambda_i $$
Taking $x=v_j$ shows that $\lambda_j=0$ for all $j$, i.e. $B \equiv 0$. 
It thus follows that $A+A^T=0$, and hence that $A$ is skew-symmetric.
A: HINT: What does $X^TAX=0$ mean when $X$ is a standard basis vector? And if $X$ is the sum of two distinct standard basis vectors?
A: The matrix $A$ can be presented as the sum of its symmetric and skew-symmetric part $A=A_{sym}+A_{sk}$.  
From this and $X^TA X=0 $ we have $X^TA_{sym}X=-X^TA_{sk}x$.  
It was proved previously (in OP)  that for  $A_{sk}$ for all vectors $X$ we have $x^TA_{sk}x=0$,
so we should have  for all vectors $X$ also $X^TA_{sym}X=0$.   
But this is not true for all vectors $X$ if $A_{sym}$ is non-zero matrix.  
At least it is not true for eigenvectors of $A_{sym}$ with non-zero eigenvalue - in this case $X^TA_{sym}X=X^T\lambda{X}=\lambda(X^TX)$ is not equal zero because $(X^TX= \Vert X \Vert ^2)$ is not zero  and $\lambda$ is not zero.  
So we can infer that symmetrical part of $A$ in these conditions must be zero matrix.
