## Need help proving something...seems simple

If A is an $n$x $n$ matrix and X is an $n$x1 column matrix and U is a subspace such that $U = \begin{Bmatrix}X\in{\mathbb{R}}^{n}| AX={A}^{T}X\end{Bmatrix}$

Show that since A is an orthogonal matrix:

$dim U = dim {E}_{1}({A}^{2})$ << for those who don't know the notation. ${E}_{1}({A}^{2})$ simply means the corresponding eigenvector of $A^2$ with respect to the eigenvalue, 1.

Someone wanna give me a hand? Here's what I did. I stated that since A is orthogonal, we know that

$AA^T = I$

then i went ahead and did

$AX = A^TX$
multiplying both sides by A

$A[AX] = A[A^TX]$

$A^2X = X$<<< since $AA^T = I$ since A is orthogonal

$A^2X - X = O$

$(A^2 - I)X = O$

and now i'm stuck.. don't know where to go next..

Any help appreciated! =D