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 $-1$
