Suppose $A$ is a real symmetric matrix with eigenvalue $\lambda$ and $A^2$ has eigenvalue $\lambda^2$. Then can $A$ also have eigenvalue $-\lambda$?

Say that $A$ is $n \times n$. Since $A$ is real and symmetric, it has $n$ orthogonal eigenvectors $v_1, \ldots, v_2$ with corresponding real eigenvalues, $\lambda_1, \ldots, \lambda_n$ (not necessarily all distinct). Suppose that $$ Av_1=\lambda v_1$$ and $$Av_2=-\lambda v_2,$$ where $\pm \lambda \in \{\lambda_1, \ldots, \lambda_n\}$.

Since $A$ is symmetric, $\langle Ax,y \rangle=\langle x,Ay \rangle$ for any $x, y \in \mathbb{R}^n$. I have been trying to compute things like $\langle A(v_1-v_2),v_1+v_2 \rangle=\langle v_1-v_2,A(v_1+v_2) \rangle$ to get a contradiction, but this isn't working. Some help? Maybe there is a counterexample to my question?

Thank you.

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    $\begingroup$ How about $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$? $\endgroup$ – carmichael561 Apr 12 '17 at 20:08
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    $\begingroup$ How about $A=\begin{pmatrix}\lambda&0\\0&\mu\end{pmatrix}$? $\endgroup$ – Hagen von Eitzen Apr 12 '17 at 20:14

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