Consider the matrix $ A = \left( \begin{matrix} 1/2 & 1/2 & 0 & 0 & 0 & 0 \\ 1/3 & 1/3 & 1/3 & 0 & 0 & 0 \\ 0 & 1/2 & 0 & 1/2 & 0 & 0 \\ \vdots & \ & \ddots & & \ddots & \\ 0 & \cdots & 0 & 1/3 & 1/3 & 1/3 \\ 0 & \cdots & 0 & 0 & 1/2 & 1/2 \end{matrix} \right) $
This question is about the singular vectors of $A$, i.e., the eigenvectors of $A^T A$. Ill denote them by $v_1, \ldots, v_n$. Suppose $n$ is even, and let $P$ be the permutation matrix which flips the vector around the midpoint, i.e., $P$ flips entries $1$ and $n$, entries $2$ and $n-1$, and finally entries $n/2$ and $n/2+1$.
Observation: A few numerical experiments in MATLAB suggest that exactly half of the $v_i$ satisfy $Pv=-v$ and the remainder satisfy $Pv=v$.
My question: Can someone provide an explanation for why this is the case?
Update: I suppose I should note that it isn't surprising that we can find eigenvectors which satisfy either $Pv=v$ or $Pv=-v$. Indeed, we can go through the list of $v_1, \ldots, v_n$ and if $Pv_i \neq v_i$, then because $Pv_i$ is also an eigenvector of $A^T A$ with the same eigenvalue, we can replace $v_i$ with v_i'=v_i+Pv_i, v_i''=v_i-Pv_i; then Pv_i' = v_i', Pv_i'' = -v_i''; finally we can throw out the redundant vectors. What is surprising to me is that the orthogonal basis returned by the MATLAB eig command has exactly half of the vectors which satisfy $Pv=v$ and exactly half which satisfy $Pv=-v$.