Values of $n$ so that exist a matrix $A\neq 0$ so that $Ax$ is orthogonal a $x$ for all $x\in \mathbb{R}^n$. Find the values of $n$ so that exist a matrix $A\neq 0$ with reals entries so that $Ax$ is orthogonal a $x$ for all $x\in \mathbb{R}^n$.
I try solved this exercise using the theory of orthogonal matrix of my course but I not can't prove this exercise.
Thanks for your help!

One needs to restrict $A$ to be nonsingular, lest we have $Ax = 0$ for some $0 \ne x \in \Bbb R^n$ and the issue of the orthogonality of $x$ and $Ax$ becomes meaningless (I assume orthogonality is technically undefined for zero vectors).  If we so assume, then:
There exists a real $n \times n$ nonsingular matrix with
$\langle x, Ax \rangle = 0, \; \forall 0 \ne x \in \Bbb R^n \tag 1$
if and only if $n$ is even, to wit:
Let $J$ be the $2 \times 2$ matrix
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 2$
for $y \in \Bbb R^2$, we have
$y = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}; \tag 3$
then
$Jy = \begin{pmatrix} -y_2 \\ y_1 \end{pmatrix}; \tag 4$
thus
$\langle y, Jy \rangle = y^TJy = -y_1y_2 + y_2y_1 = 0; \tag 5$
if $n = 2m$ is even, we can choose $A$ to be the block diagonal matrix consisting of $m$ blocks, each equal to $J$, thus:
$A = \begin{bmatrix} J & 0 & \ldots & 0 & 0 \\ 0 & J & 0 & \ldots & 0 \\ \vdots \\ 0 & 0 & \ldots & 0 & J \end{bmatrix}; \tag 6$
each block of $J$ then operates on $(x_{2i + 1}, x_{2i + 2})^T$, $0 \le i \le m - 1$, exactly as $J$ operates on $y$; thus 
$\langle x, Ax \rangle = 0, \; \forall x \in \Bbb R^n; \tag 7$
this shows the existence of the requisite $A$ when $n = 2m$.  If, on the other hand $n$ is odd, then since the characteristic polynomial $A -\lambda I$ is of odd degree, $A$ has at least one real eigenvalue $\rho$; then 
$Ax = \rho x \tag 8$
for some $x \ne 0$; thus,
$\langle x, Ax \rangle = \langle x, \rho x \rangle = \rho \langle x, x, \rangle \ne 0, \tag 9$
since the nonsingularity $A$ implies $\rho \ne 0$; every nonsingular $A$ of odd size thus fails to sastisfy $\langle x, Ax \rangle = 0$ for some $x$.
