How prove this matrix $\det (A)=\left(\frac{1}{\ln{(a_{i}+a_{j})}}\right)_{n\times n}\neq 0$

Question:

let $a_{i}>1,i=1,2,3,\cdots,n$,and such $a_{i}\neq a_{j}$,for any $i\neq j$

define the matrix

$$A=\left(\dfrac{1}{\ln{(a_{i}+a_{j})}}\right)_{n\times n}$$

show that: $$\det(A)\neq 0$$

My try: I know this matrix $A$ is similar this Cauchy determinants： http://en.wikipedia.org/wiki/Cauchy_matrix

and

$$\det(A)=\begin{vmatrix} \dfrac{1}{\ln{(a_{1}+a_{1})}}&\dfrac{1}{\ln{(a_{1}+a_{2})}}&\cdots&\dfrac{1}{\ln{(a_{1}+a_{n})}}\\ \dfrac{1}{\ln{(a_{2}+a_{1})}}&\dfrac{1}{\ln{(a_{2}+a_{2})}}&\cdots&\dfrac{1}{\ln{(a_{2}+a_{n})}}\\ \cdots&\cdots&\cdots&\cdots\\ \dfrac{1}{\ln{(a_{n}+a_{1})}}&\dfrac{1}{\ln{(a_{n}+a_{2})}}&\cdots&\dfrac{1}{\ln{(a_{n}+a_{n})}} \end{vmatrix}$$

but I can't,Thank you.and this problem is my frend ask me.

this is he ask me is second problem .and I think this problem is interesting.

Now this problem is up $21$. that's mean this problem is hard.I hope someone can solve it.Good luck!Thank you

Solution 1:

For any $s > 0$, let $A(s) \in M_{n\times n}(\mathbb{R})$ be the matrix with entries

$$A(s)_{ij} = \frac{1}{(a_i + a_j)^s}$$

For any non-zero $u \in \mathbb{R}^n$ with components $u_i, i = 1\ldots n$, we have

$$u^T\!A(s)\,u = \sum_{1\le i,j \le n} \frac{u_i u_j}{(a_i+a_j)^s} = \sum_{i\le i,j \le n} \frac{u_i u_j}{\Gamma(s)}\int_0^\infty t^{s-1} e^{-(a_i+a_j)t} dt\\ = \frac{1}{\Gamma(s)}\int_0^\infty t^{s-1} \left(\sum_{i=1}^n u_i e^{-a_i t} \right)^2 dt > 0 $$ because $u_i$ not all zero implies as a function of $t$, $\displaystyle \sum_{i=1}^n u_i e^{-a_i t}$ not identically zero$\color{blue}{^{[1]}}$.

Notice for $x > 1$, $\frac{1}{\log x}$ can be expressed as an absolute convergent integral:

$$\frac{1}{\log x} = \int_0^{\infty} \frac{1}{x^s} ds$$

As a result, we have

$$u^T A u = \sum_{1 \le i, j \le n} \frac{u_i u_j}{\log (a_i+a_j)} = \sum_{1 \le i, j \le n } \int_0^\infty \frac{u_i u_j}{(a_i+a_j)^s} ds = \int_0^\infty u^T\!A(s)\,u\;ds > 0$$

This implies $A$ is positive definite and hence invertible.

Notes

• $\color{blue}{[1]}$ To justify $\displaystyle f(t) = \sum_{i=1}^n u_i e^{-a_i t}$ not identically zero, we need to use the fact $a_i$ are all distinct. If $f(t)$ vanishes on $t = 1, \ldots, n$, then we can construct a Vandermonde matrix with entries $e^{-a_i j}, 1 \le i, j \le n$ and use it to conclude all $u_i = 0$.