Prove or disprove: If $x^T A x = 0 $ for all $x$, then $ A = 0 $. Let $A$ be a square matrix and $x$ be a vector. Now consider the statement:
If $x^T A x = 0 $ for any $x$, then $A = 0$.
Is the above statement true or false?
How would you prove it?
 A: If $x ^t Ax = 0$ for every $x \in \mathbb R^n $  then the diagonal entries of $A$ 
are all zero by taking $x$ to be the column vector with the $i$-th 
component as  $1$ and $ 0$ elsewhere, $\forall i = 1,2,\ldots, n$. 
then choose your $x$ as follows  
the column vector with the $i$-th and the $j$-th 
components as  $1$ and $0$ elsewhere. Then  try to ahow $A^t=-A$
A: It is not true, $A=\begin{pmatrix}0&1\\-1&0 \end{pmatrix}$ is a counterexample.
Take $x=e_i+e_j$, you can get $A_{ij}+A_{ji}=0$, which means $A^T+A=0$
A: furthermore to the disproves of the fellers up, I would like to mention that if the question was in the complex field, the statement was true.
in other words:
$x^*Ax=0 \, \forall x \in C^n  \Leftrightarrow A=0 $
proof:
$\Leftarrow$ this direction is immediate
$\Rightarrow$
since the left side is zero for all x, lets choose few x's to show how A must be zero as well

*

*$x_1=e_k \Rightarrow x^*Ax=e^T_kAe_k=A_{kk} \Rightarrow A_{kk}=0 \, \forall k$

*$x_2=e_k+e_m \Rightarrow x^*Ax= (e^T_k+e^T_m)A(e_k+e_m)=A_{kk}+A_{km}+A_{mk}+A_{mm} $
since $A_{kk}=A_{mm}=0$ we get $\Rightarrow A_{km}=-A_{mk} \, \forall k\ne m$

*

*$x_3=e_k+ie_m \Rightarrow x^*Ax= (e^T_k-ie^T_m)A(e_k+ie_m)=A_{kk}+iA_{km}-iA_{mk}+A_{mm} $
$\Rightarrow A_{km}=A_{mk} \, \forall k\ne m$
from $x_2,x_3$ we get that $A_{km}=-A_{km}$ where this can only valid if $A_{km}=0 \, \forall k\ne m$
therefore $A=0 \,\blacksquare$
A: Since $x^tAx=\langle x,Ax\rangle$ (the inner product), the question may be stated: 
Is there a transformation that maps every vector to something orthogonal to itself? 
What should come to mind is a rotation matrix that rotates $\mathbb{R}^2$ by 90 degrees, which Shu Xiao Li has presented.
