Can we prove that the eigenvalues of an $n\times n$ orthogonal matrix are $\pm 1$ from the definition of orthogonal matrix alone?
An $n\times n$ matrix $A$ is orthogonal iff $AA^T=A^TA=I$.
Is it possible to use solely this definition, without using the result proven from this definition which says orthogonality preserves length, to show that the eigenvalues of $A$ is $\pm 1$?
Suppose $x$ is an eigenvector with corresponding eigenvalue $\lambda$. Then $x^T A^T Ax = (Ax) \cdot (Ax) = |Ax|^2 = \lambda^2 |x|^2$. But you also know that $x^T A^T A x = x \cdot x = |x|^2$. So $\lambda^2 = 1$ i.e. $\lambda = 1,-1$. Note this says that if the matrix is diagonalizable then the eigenvalues must be $1$ and/or $-1$. This doesn’t in and of itself show that an orthogonal matrix is always diagonalizable (indeed it doesn’t have to be over the reals).