Condition for the projection on $S_n(\mathbb{R})$ to be orthogonal For a matrix $B \in M_n(\mathbb{R})$ we denote $B_s = \frac{B+^tB}{2}$.

Let $A \in M_n(\mathbb{R})$ such that all the eigenvalues of $A_s$ are in $[-1;1]$. Prove that there exist an orthogonal matrix $C \in O_n(\mathbb{R})$ such that $C_s = A_s$ if and only if the dimensions of the eigenspaces of the eigenvalues in $(-1,1)$ of $A_s$ are even.

For now I don't see how to solve this problem. Yet I've noticed the following : 
It seems that $A_s$ is actually the projection of $A$ on the space $S_n(\mathbb{R})$ of symmetric matrix. So the problem is giving a condition on a matrix to see if there exist an orthogonal matrix such that it projections on $S_n(\mathbb{R})$ is equal to this matrix.
Moreover I know that an orthogonal matrix have all it's real eigenvalues in $[-1, 1]$, but does it help ? 
An other way to see the problem is to say that the function : 
$$ f:O_n \to \mathbb{R} $$
$$\text{     } x \mapsto \| x - A_s \|_2$$
go to $0$ at some point. Since this function is continuous and that $O_n(\mathbb{R})$ is compact maybe it's a good idea  yet the problem is that it doesn't capture the eigenspaces of $A_s$ so maybe it's the wrong way of doing. Yet here I am using the Frobenius norm, maybe there is a norm on the space of matrix which capture well the dimensions of the eigenspaces and maybe in this case we can consider the function $f$ with this norm.
Thank you ! 
 A: $(\Rightarrow)$ We want to prove the existence of a matrix $O$ in $\mathcal{O}_n(\mathbb{R})$ such that : $\frac{1}{2} \left ( O + ^t O \right) = A_s$ where $A_s$ is a matrix with all it's eigenvalues in $[-1, 1]$ and such that $\forall \lambda \in (-1; 1), \dim \operatorname{Ker} \left ( A_s - \lambda \right) \equiv 0 \pmod 2$.
Since an orthogonal matrix have all it's eigenvalues on the unit circle of $\mathbb{C}$, you can then prove that every orthogonal matrix $O$ is of the form : 
$$O = ^tP\begin{bmatrix} 
     R_1 & &  \\
     & \ddots & \\
     &         & R_k &\\
     &      \   & 
     &         & \pm1
    \end{bmatrix} P = ^tP DP$$
where the $R_k$ are $2 \times 2$ rotations matrix and where $P$ is an orthogonal matrix.
Moreover note that $A_s$ is symmetric thus by the spectral theorem there is a diagonale matrix $\Delta$ such that : $A_s = M \Delta ^tM$
Hence the problem boils down to proving the existence of an orthogonal matrix of the form $D$ such that : 
$$D +D^{-1} = \Delta$$
Since $R_k + R_k^{-1} =   \begin{bmatrix}
    \cos(\theta) & \\
     & \cos(\theta) \\
  \end{bmatrix}$ the equation has indeed a solution since every eigenvalues in $(-1, 1)$ appears an even number of times in $\Delta$ you just need to solve : $\cos(\theta) = \lambda$ for some $\lambda \in (-1; 1)$. Doing this we easily get our orthogonal matrix $O$.$\square$
$(\Leftarrow)$ So now you juste need to prove that if $O$ is an orthogonal matrix then all the eigenvalues in $(-1;1)$ of the matrix $\frac{1}{2} \left ( O + ^tO \right)$ have eigenspace of dimension $\equiv 0 \pmod 2$. But this is evident thank's to the above form of orthogonal matrix.$\square$
