Consider a matrix $A=(a_{ij})_{5\times 5}$ , $1\le{i,j}\le5$ such that $a_{ij} = \dfrac{1}{n_{i}+n_{j}+1}$ where $n_{i}$ and $n_{j}$ are natural numbers. Then which of the following cases A is positive definite matrix?
- $n_{i} = i$ for all i=1,2,3,4,5
- $n_{1}<n_{2}<.....<n_{5}$
- $n_{1}=n_{2}=.....=n_{5}$
- $n_{1}>n_{2}>.....>n_{5}$
My Attempt : for option third, observe that, all entries of matrix will be equal, this gives A has zero as eigen value, so not positive definite. For other options, I have only way that putting Particularl values of $n_{i}$ and $n_{j}$and find out eigen values of A or using sylvester's criterion . But this approach of problem solving is much long as well as much calculation. Any better approach , please help me. Thanks
Assuming $n_k>-1$, $$ A = \int B(x)\,dx \qquad\text{where}\qquad B(x)_{ij}=x^{n_i}\cdot x^{n_j}\qquad $$ and $B(x)=C(x)\cdot C(x)^T$ and $C(x)_{ij} = x^{n_i}$. Like any real matrix of the $CC^T$ form, $B$ is positive semi-definite, and so it is $A$, which is a principal minor of a Hilbert matrix. The order of $n_1,\ldots,n_5$ cannot affect the positive-definiteness of $A$ or the lack of it, since you may always assume $n_1\leq n_2\leq n_3\leq n_4\leq n_5$, up to mapping $A$ into $\Pi^{-1} A \Pi=\Pi^{T} A \Pi$ with $\Pi$ being a permutation matrix.