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 
 A: It seems to me that this problem comes down to getting the definitions straight.  So here they are:
Symmetric matrix: a matrix $A$ is called symmetric when $A=A^T$.  If $A$ is a matrix of real entries, then we can say that $A$ is an instance of a "Hermitian" matrix
Hermitian matirix a matrix $A$ is Hermitian when $A=A^*$, that is, $A$ is equal to its own conjugate transpose.
Depending on the context, real symmetric matrices and complex Hermitian matrices might be referred to as self-adjoint.
Orthogonal matrix: a matrix $U$ is called orthogonal when $UU^T=1$.  If $U$ is a matrix of real entries, then we can say that $U$ is an instance of a "unitary" matrix
Unitary matrix a matrix $U$ is called unitary when $UU^*=1$.  That is, the inverse of $U$ is its conjugate transpose.
The spectral theorem could then be understood as follows:
Suppose that $A$ is any normal matrix (that is, $AA^*=A^*A$).  We can then find a matrix $P$ such that $PAP^*$ is diagonal.
If $A$ happens to be a real and symmetric matrix, then we can find a $P$ that is not only unitary, but also itself real.  That is, there is a real matrix $P$ such that $PAP^*$ is diagonal.  Since $P$ is real, this is the same as saying $PAP^T$ is diagonal.
A: This is a straightforward consequence of the spectral theorem. Let $u_1=p_1+iq_1,\ \ldots,\ u_n=p_n+iq_n$ be an orthonormal eigenbasis of $A$, where $u_1,\ldots,u_{k_1}$ correspond to the eigenvalues $\lambda_1$. Since $\lambda_1$ is real, we have $Ap_\ell=\lambda_1p_\ell$ and $Aq_\ell=\lambda_1q_\ell$ for $\ell=1,2,\ldots,k_1$. Hence every vector in $\mathcal{B}=\{p_1,\ldots,p_{k_1},q_1,\ldots,q_{k_1}\}$ is either zero or an eigenvector of $A$ corresponding to the eigenvalue $\lambda_1$. Hence the span of $\mathcal{B}$ is precisely the eigenspace of $A$ corresponding to the eigenvalue $\lambda_1$. However, as all vectors in the set $\mathcal{B}$ are real, we can choose from its real span $k_1$ orthonormal eigenvectors. The similar holds for other eigenvalues. Put these orthonormal eigenvectors together, we obtain a real orthogonal matrix $S$ such that $AS=SD$, i.e. $S^TAS=D$.
A: 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).
