Show that $A=(a_{ij})_{1\le i,j\le5}$ is positive definite, where $a_{ij}=\frac1{n_i+n_j+1}$

Question

Problem

Let $A=(a_{ij})_{1\le i,j\le5}$ be a matrix such that
$$a_{ij}=\frac1{n_i+n_j+1}\,,$$ where $n_i,n_j\in\Bbb{N}$. Show that $A$ is positive definite if $n_1>n_2>n_3>n_4>n_5$ or, $n_1<n_2<n_3<n_4<n_5$.

As far as my knowledge is concerned, I have following two equivalent criteria for a matrix $A$ to be positive definite.

• all eigenvalues are positive.

• there is a matrix $B$ such that $A=B^tB$.

But nothing is helping me here. Any hint how to solve this!! Thank you.

Answer 1

Consider the space of polynomials with real coefficients defined on the unit interval $[0,1]$, with the inner product $\langle f,g\rangle=\int_0^1 f(x)g(x)\,dx$. We have $$\langle x^{n_i},x^{n_j}\rangle=\int_0^1 x^{n_i+n_j}\,dx=\displaystyle\frac1{n_i+n_j+1}=a_{ij}$$

And in general, any matrix $M$ of the form $M_{ij}=\langle v_i,v_j\rangle$, where $v_i,v_j$ are vectors in a fixed inner product space is positive definite. This is because $$a^tMa=\sum_{ij}a_ia_j\langle v_i,v_j\rangle =\left\langle \sum_i v_ia_i,\sum_i v_ia_i\right\rangle\ge 0$$ for any vector $a$.