Let $A$ be a real symmetric matrix. Assume $A^{2}v=\lambda Av$ , can we somehow deduce $v$ is an eigenvector of $A$ with $\lambda$ as its corresponding eigenvalue? i.e $Av=\lambda v$. Thanks
Edit: Also assume $\lambda \ne 0$ and we do not know if it is singular or nonsingular.
Take $A = \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}$ and $\lambda = 2$. Then $A^2 = \lambda A$ so $v$ is arbitrary and not necessarily an eigenvector (e.g., $v = (1,2)^T)$.
However, you are guaranteed that either $\lambda$ is an eigenvalue of $A$, or $v$ is an eigenvector of $A$. Proof: if $Av = 0$, $v$ is an eigenvector but $\lambda$ is arbitrary. If $Av \ne 0$, then $A(Av) = \lambda(Av)$ so $\lambda$ is an eigenvalue of $A$ with eigenvector $Av$, as with the above counterexample.
It is true if $A$ is invertible as noted in the comments.