for any symmetric real matrix $S$, the following eigendecomposition exists:
$$ S = Q \Lambda Q^{\top} $$
where $Q$ is a unitary matrix, consisting of the eigenvectors of $S$ wikipedia .
By definition of unitary, we have $Q^{\top}Q=QQ^{\top}=I$.
Given an orthonormal set of eigenvectors, $Q^{\top}Q=I$, is trivial. How can one show $QQ^{\top}=I$ ?
Best Answer
Hmm, I think, the more interesting part in the original question is in the direction: "why is it with a symmetric matrix S, that the diagonalization $ \small S = Q \Lambda Q^{-1} $ provides a unitary matrix Q , such that $ \small Q^T = Q^{-1} $ ?" Which, of course, can be answered by considering the equality of S with its transpose: $\small S = S^T=(Q^T)^{-1} \Lambda Q^T = Q \Lambda Q^{-1} \to Q=(Q^T)^{-1} $ and $ \small Q^{-1}=Q^T $