If A is an nx n matrix and X is an nx1 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