# Eigenvalues of symmetric orthogonal matrix

Can we say that Eigenvalues of symmetric orthogonal matrix must be $$+1$$ and $$-1$$?

Since eigenvalues of symmetric matrices are real and eigenvalues of orthogonal matrix have unit modulus. Combining both result eigenvalues of symmetric orthogonal matrices must be $$+1$$ and $$-1$$.

Please clarify whether I am correct? Is there any other approach to solve this problem?

Thanks

Yes, you're right. Also note that if $A^\top A=I$ and $A=A^\top$, then $A^2=I$, and now it's immediate that $\pm 1$ are the only possible eigenvalues. (Indeed, applying the spectral theorem, you can now conclude that any such $A$ can only be an orthogonal reflection across some subspace.)

Suppose $$A$$ being symmetric and orthogonal, then we have $$A = A^T$$ and $$A^T A = I$$.

Let $$\lambda$$ be an eigenvalue of $$A$$. Then we can derive

\begin{align} Ax &= \lambda x \\ A^T A x &= A^T \lambda x \\ x &= A \lambda x \\ \frac{1}{\lambda} x &= Ax = \lambda x\\ \frac{1}{\lambda} &= \lambda \end{align}

So $$\lambda$$ has to be $$\pm 1$$.

A symmetric orthogonal matrix is involutory.

Involutory matrices have eigenvalues $$\pm 1$$ as proved here: Proof that an involutory matrix has eigenvalues 1,-1 and Proving an invertible matrix which is its own inverse has determinant $$1$$ or $$-1$$