Is a square commutation matrix positive semidefinite

Question

Let $A \in \mathbb{R}^{n \times n}$ and denote the commutation matrix, made up of 0 and 1 such that each row and each column has exactly one 1, as $K_{n} \in \mathbb{R}^{n^2 \times n^2}$ , which is such that:
\begin{equation}
\operatorname{vec}(A^T) = K_{n} \operatorname{vec}(A)
\end{equation}
It is known (cf. Harville D.A., Matrix Algebra from a Statistician's Perspective) that such a matrix is symmetric,orthogonal, and with determinant $\pm 1$.
Moreover, is it positive semidefinite ?

Answer 1

$K_2=\pmatrix{1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1}$ is indefinite. In fact, every commutation matrix except $K_1$ is indefinite. In particular, for every $\mathcal I=\{(j-1)n+i,\ (i-1)n+j\}$ with $i\ne j$, its principal submatrix $$ A(\mathcal I,\mathcal I)=\pmatrix{0&1\\ 1&0} $$ is indefinite.