If $A$ is a symmetric matrix in $\mathbb{R}$, why is $PAP^t$ diagonal? In Linear Algebra Why is following correct:
Given a symmetric matrix $A$ on the field of the real numbers, why is that true that there exists an unitary matrix $P$ such that $PAP^t$ is a diagonal matrix?
I know that from the Spectral Theorem there exists a unitary matrix $P$ such that $PAP^{-1}$ is diagonal. However how can I conclude that $PAP^t$ is diagonal as well?
Thank you 

Take a look at the real Schur decomposition. This gives you an orthogonal (meaning real unitary) $U$ and a quasitriangular (blockdiagonal with diagonal blocks of order at most $2$) $T$, with each block of order $2$ having conjugate pairs of eigenvalues. Since $A$ is real symmetric, we know that all the eigenvalues of $A$ are real, so $T$ is diagonal.
By the way, in the above paper, what you're asking is a corollary (on the last page).
