TITLE: About positive semidefiniteness of one matrix
QUESTION [10 upvotes]: It is not clear how to prove that the matrix $(\min(i,j))_{i,j=1,\dots,n}$ is (or is not) positive semidefinite. There are some facts from Horn and Johnson's book Matrix Analysis: if $A \in M_n$ is positive semidefinite, then $a_{ii}a_{jj} \ge a_{ij}^2, i,j=1,2,\dots, n$ (source). It seems that this condition is not enough for the matrix to be positive semidefinite. Are there any techniques to check that all eigenvalues of the matrix are positive, or that the minor determinants are non-negative?

REPLY [5 votes]: Another alternative solution: it can be shown that the inverse of the min-matrix is the perturbed Toeplitz tridiagonal matrix
$$\begin{pmatrix}2&-1&&&\\-1&2&-1&&\\&-1&\ddots&\ddots&\\&&\ddots&2&-1\\&&&-1&1\end{pmatrix}$$
whose characteristic polynomial is
$$(-1)^n\left((x-1)U_{n-1}\left(\frac{x-2}{2}\right)-U_{n-2}\left(\frac{x-2}{2}\right)\right)$$
(where $U_n(x)$ is the Chebyshev polynomial of the second kind), and has the eigenvalues
$$\mu_k=2+2\cos\left(\frac{2k\pi}{2n+1}\right),\qquad k=1\dots n$$
Thus, the eigenvalues of the min-matrix are the reciprocals of the $\mu_k$,
$$\lambda_k=\frac14\sec^2\left(\frac{k\pi}{2n+1}\right)$$
and these are easily seen to be positive, thus showing that the min-matrix is positive definite.