# Eigenvector of A by $A^2$

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.

• If $A$ is nonsingular, yes. Commented Jun 9, 2021 at 11:36
• Let $A$ be the matrix $$A=\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}\ .$$ What is $A^2$? Commented Jun 9, 2021 at 11:39
• @dan_fulea But your $A$ is not symmetric Commented Jun 9, 2021 at 18:06
• If it is not whether $A$ is singular, then the answer is no. As a counterexample, take $$A = \pmatrix{1&0\\0&0}, \quad v = \pmatrix{1\\1}, \quad \lambda = 1.$$ Commented Jun 9, 2021 at 18:13

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.