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

Show that since A is an orthogonal matrix:

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

$\displaystyle AA^T = I$

then i went ahead and did

$\displaystyle AX = A^TX$

multiplying both sides by A

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

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

$\displaystyle A^2X - X = O$

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

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

Any help appreciated! =D