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

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.

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.

• It's obviously a Gram matrix. Commented Jun 28, 2020 at 18:01

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 $$a$$.