$ \newcommand{\matP}{\mathbf{P}} \newcommand{\matD}{\mathbf{D}} \newcommand{\summe}[2]{\sum\limits_{#1}^{#2}} \newcommand{\matQ}{\mathbf{Q}} $
I have a question (also somewhat related to my previous question properties of orthogonal similarity transformation). The following is known:
Let $\matP$ be a permutation matrix and $\matD$ a diagonal matrix. Then $\matP^T \matD \matP$ is a diagonal matrix with permuted diagonal elements.
I present my own proof below.
My question is: Does the reverse statement also hold: Given two diagonal matrices $\matD$ and $\matD'$ which have permuted diagonal entries and are interrelated by an orthogonal similarity transformation $\matQ^T \matD \matQ = \matD'$ (where $\matQ$ is orthogonal), is $\matQ$ always a permutation matrix or are there other solutions?
(Plus: Does the reverse statement depend on whether we have pairwise different diagonal elements or not?)
Thanks!
Proof (forward direction): Let $\pi(i)$ be the permuted index to index $i$. Then \begin{eqnarray} (\matP)_{ij} &=& (\delta_{i,\pi(j)})_{ij}\\ (\matP^T)_{ij} &=& (\delta_{j,\pi(i)})_{ij}\\ (\matD)_{ij} &=& (d_i \delta_{ij})_{ij}. \end{eqnarray} We form the product $\matP^T\matD$ \begin{eqnarray} (\matP^T\matD)_{ij} &=& \summe{k}{} (\matP^T)_{ik} (\matD)_{kj}\\ &=& \summe{k}{} \delta_{k,\pi(i)} d_k \delta_{kj}\\ &=& \delta_{j,\pi(i)} d_j \end{eqnarray} and then multiply by $\matP$: \begin{eqnarray} ((\matP^T\matD)\matP)_{ij} &=& \summe{k}{} (\matP^T\matD)_{ik} (\matP)_{kj}\\ &=& \summe{k}{} \delta_{k,\pi(i)} d_k \delta_{k,\pi(j)}\\ &=& d_{\pi(i)} \delta_{\pi(i),\pi(j)}. \end{eqnarray} We see from $\delta_{\pi(i),\pi(j)}$ that all off-diagonal elements are zero (since $\pi(i) \neq \pi(j)$ for $i \neq j$). The diagonal elements are $d_{\pi(i)}$ and thus the permuted diagonal entries of $\matD$.