If eigenvalues are $\pm 1$, then matrix is orthogonal If eigenvalues of a matrix are $1$, or $-1$, or both of them, does that necessarily means that the matrix is orthogonal?
The only thing I'm certain about is that the matrix is non-singular, due to absence of zero among the eigenvalues.
 A: The eigenvectors with eigenvalue $1$ have to be orthogonal to the eigenvectors with eigenvalue $-1$ in order for the matrix to be orthogonal. A $2\times2$ counterexample using this idea is not difficult to construct. For example, let $M$ be a linear operator on $\mathbb{R}^{2}$ for which
$$
                    M\left[\begin{array}{c}1 \\ 1\end{array}\right]=\left[\begin{array}{c}1 \\ 1\end{array}\right] \\
                    M\left[\begin{array}{c}0 \\ 1\end{array}\right]=-\left[\begin{array}{c}0 \\ 1\end{array}\right].
$$
Then $M$ has eigenvalues $1$ and $-1$, but the eigenvectors corresponding to these eigenvalues are not orthgonal. To find $M$,
\begin{align}
        M\left[\begin{array}{c}x \\ y\end{array}\right]
          & = M\left(x\left[\begin{array}{c}1 \\ 1\end{array}\right]+(y-x)\left[\begin{array}{c}0 \\ 1\end{array}\right]\right) \\
          & = x\left[\begin{array}{c}1 \\ 1\end{array}\right]+(x-y)\left[\begin{array}{c}0 \\ 1\end{array}\right] \\
          & = \left[\begin{array}{cc}1 & 0\\2 & -1\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]
\end{align}
So a simple counterexample is
$$
     M = \left[\begin{array}{cc}1 & 0 \\ 2 & -1\end{array}\right].
$$
It is easy to check that $M$ has the correct eigenvectors with eigenvalues $\pm 1$. And $M$ is not orthogonal:
$$
             M^{\perp}M = \left[\begin{array}{cc}1 & 2 \\ 0 & -1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ 2 & -1\end{array}\right]
           = \left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]\ne I.
$$
