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.

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

1 Answer 1


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.


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