# 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

• +1: nice question badge on the way, but this should be awarded to your friend... Nov 22, 2013 at 16:37
• You could try and show your matrix is positive definite, hence nonsingular, and hence it has a nonzero determinant. The inequality you'd get would be interesting enough. But of course this is a stronger thing to show. Nov 22, 2013 at 16:50
• @nayrb The matrix is clearly symmetric by the commutativity of sums of real numbers. So choose an arbitrary column vector of real numbers of the same dimension, {ai} where i is between 1 and n. Take the transpose and carry out the required multiplication. Nov 23, 2013 at 6:14
• Just curious,what stimulates your friend to raise such a question?
– zy_
Nov 23, 2013 at 10:59

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 $u_i = 0$.
• It's very nice!+1,Thank you ,this problem Finally solved! Nov 23, 2013 at 10:50
• That's quite interesting-I wouldn't have done the gamma function substitutions into the sum yielding the integral,I'd have just done the direct computation. Since n is finite,is it necessary? Nov 23, 2013 at 18:52
• @Mathemagician1234 It is not necessary but doing this way, we are dealing with functions that is more familiar and make it easier to avoid mistakes (at least for me). Nov 23, 2013 at 19:07
• @achille See,that's the advantage of SOME memorization in mathematics-it gives you a large toolbox you can use upon instant recognition of an identity that simplifies computations. Understanding is always better,but since computation is still just as critical a part of mathematics,that "mindless application" toolbox still comes in handy as a time saver. Nov 23, 2013 at 22:53
• @math110 : How does this solve the question? What happened to the log? Aug 30, 2015 at 12:13