doubt about symmetric matrices and vector norms Let $A$ be a symmetrical $N\times N$ matrix that is not a product of a scalar by an orthogonal matrix.
What must be imposed to two N-dimensional vectors $v_1$ and $v_2$ with $|v_1| = |v_2| = 1$,  so that we have $|Av_1| = |Av_2|$  (Euclidean norm)?
for example that $k_1$, $-k_1$, and $k_2$ other than $\pm k_1$ are eigenvalues of $A$, so $Ax_1 = k_1x_1$ and $Ax_2 = -k_1x_2$ and $Ax_3 = k_2x_3$, so
$| x_1| = | x_2 | = | x_3 | = 1$ like this
$| Ax_1 | = | Ax_2 |$ and $| Ax_3 |$ is different
$| Ax_1 | = | Ax_2 |$
Besides that
$$
|(x_1 + x_3) / \sqrt{2} | = | (x_1-x_3) / \sqrt{2} | \\
= 1 e | A (x_1 + x_3) / \sqrt{2} | \\
= | A (x_1-x_3) / \sqrt{2} | \\
= \sqrt{(k_1^2 + k_3^2) / 2},
$$
again it is different $| Ax_1 | , | Ax_2 |$ and $| Ax_3 |.$ What I want to know is under what conditions do two unit vectors have the same norm?
 A: Considering the suggestion @Ben Grossmann
I thought of if A:V->V $A^2$, and if $\lambda _i$ i = 1, .., k are the distinct self-values ​​of $ A^2 $ of $A^2$, and $\lambda _1> \lambda _2> ...>\lambda_k $, and $ P_i $ is a V projector in $ Kernel (A ^ 2- \lambda _i I)$ then $ |Ax| = |Ay| $ if and only if, $ |P_ix|^2 =|P_iy|^2 $ for i = 1, .. k. Of course if this is true then the rules are the same. On the other hand, |Ax| = |Ay|, then $ |x|^2 = |P_1x|^2 + |P_2x|^2 + .. +|P_kx|^2 =1 $ and the same goes for y, so $ \lambda_1 (|P_1y |^2- |P_1x|^2) =\lambda _2 |P_2x|^2-|P_2y |^2 + ... + \lambda _k|P_kx |^2-|P_ky |^2 $, just watch 1- | P_1x |^2 =|P_2x|^2 + .. + |P_kx |^2 $ and the same goes for y, so |P_1y |^ 2-| P_1x |^2 =|P_2x |^2 - |P_2y |^2 + .. +| P_kx |^2-|P_kx |^2-|P_ky|^2$. so we get to $\lambda_1 (|P_2x |^2-|P_2y|^2 + .. + |P_kx|^2-|P_kx|^2- |P_ky|^2) =\lambda _2 |P_2x|^2-|P_2y|^2 + ... +\lambda_k |P_kx |^2-|P_ky |^2 $. So we have $\lambda_1 = \frac {\lambda _2 |P_2x |^2-|P_2y | ^ 2 + ... + \lambda_k|P_kx |^2-|P_ky|^2} {(|P_2x |^2 -|P_2y |^2 + .. + |P_kx |^2+|P_kx|^2- |P_ky|^2)} $. Now notice, if B is matrix of A restricted to the complement of $P_1(V)$ that this is just the vector $ |\frac {B(x-y-P_1(x-y))} {| x-y-P_1 (x-y)|} $, norm $ |B| =\lambda_2 $, just like $ \lambda_1> \lambda_2 $ B would never assign this norm to a vector of norm 1. Thus, |P_1y|^2-|P_1x|^ 2=0 is left over, as well as for all i , and show what we wanted.
