Problem 770 (difficulty: 9/10)

Let \(\displaystyle a_1<a_2<\ldots<a_n\) and \(\displaystyle b_1<b_2<\ldots<b_n\) be real numbers Show that

\(\displaystyle \det\begin{pmatrix} e^{a_1b_1} & e^{a_1b_2} & \dots & e^{a_1b_n} \\ e^{a_2b_1} & e^{a_2b_2} & \dots & e^{a_2b_n} \\ \vdots & \vdots & \ddots & \vdots \\ e^{a_nb_1} & e^{a_nb_2} & \dots & e^{a_nb_n} \\ \end{pmatrix} >0. \)

(KöMaL A. 463., October 2008)

Solution:

Apply induction on \(\displaystyle n\). For \(\displaystyle n=1\) the statement is \(\displaystyle e^{a_1b_1}>0\) which is obvious. Now suppose \(\displaystyle n>1\) and assume that the statement is true for all smaller values.

Let \(\displaystyle c_i=a_i-a_1>0\). Then

\(\displaystyle \det\begin{pmatrix} e^{a_1b_1} & e^{a_1b_2} & \dots & e^{a_1b_n} \\ e^{a_2b_1} & e^{a_2b_2} & \dots & e^{a_2b_n} \\ \vdots & \vdots & \ddots & \vdots \\ e^{a_nb_1} & e^{a_nb_2} & \dots & e^{a_nb_n} \\ \end{pmatrix}= \det\begin{pmatrix} e^{a_1b_1} & e^{a_1b_2} & \dots & e^{a_1b_n} \\ e^{a_1b_1}e^{c_2b_1} & e^{a_1b_2}e^{c_2b_2} & \dots & e^{a_1b_n}e^{c_2b_n} \\ \vdots & \vdots & \ddots & \vdots \\ e^{a_1b_1}e^{c_nb_1} & e^{a_1b_2}e^{c_nb_2} & \dots & e^{a_1b_n}e^{c_nb_n} \\ \end{pmatrix} =\)

\(\displaystyle = e^{a_1(b_1+b_2+\dots+b_n)} \det\begin{pmatrix} 1 & 1 & \dots & 1 \\ e^{c_2b_1} & e^{c_2b_2} & \dots & e^{c_2b_n} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nb_1} & e^{c_nb_2} & \dots & e^{c_nb_n} \\ \end{pmatrix},\)

so it is sufficient to prove that the last determinant is positive.

To eliminate the first row, subtract the \(\displaystyle (n-1)\)th column from the \(\displaystyle n\)th column. Then subtract the \(\displaystyle (n-2)\)th column from the \(\displaystyle (n-1)\)th column, and so on, finally subtract the first column from the second column. Then

\(\displaystyle \det\begin{pmatrix} 1 & 1 & \dots & 1 \\ e^{c_2b_1} & e^{c_2b_2} & \dots & e^{c_2b_n} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nb_1} & e^{c_nb_2} & \dots & e^{c_nb_n} \\ \end{pmatrix}=\)

\(\displaystyle = \det\begin{pmatrix} 1 & 0 & 0 & \dots & 0 \\ e^{c_2b_1} & e^{c_2b_2}-e^{c_2b_1} & e^{c_2b_3}-e^{c_2b_2} & \dots & e^{c_2b_n}-e^{c_2b_{n-1}} \\ e^{c_3b_1} & e^{c_3b_2}-e^{c_3b_1} & e^{c_3b_3}-e^{c_3b_2} & \dots & e^{c_3b_n}-e^{c_3b_{n-1}} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ e^{c_nb_1} & e^{c_nb_2}-e^{c_nb_1} & e^{c_nb_3}-e^{c_nb_2} & \dots & e^{c_nb_n}-e^{c_nb_{n-1}} \\ \end{pmatrix} =\)

\(\displaystyle = \det\begin{pmatrix} e^{c_2b_2}-e^{c_2b_1} & e^{c_2b_3}-e^{c_2b_2} & \dots & e^{c_2b_n}-e^{c_2b_{n-1}} \\ e^{c_3b_2}-e^{c_3b_1} & e^{c_3b_3}-e^{c_3b_2} & \dots & e^{c_3b_n}-e^{c_3b_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nb_2}-e^{c_nb_1} & e^{c_nb_3}-e^{c_nb_2} & \dots & e^{c_nb_n}-e^{c_nb_{n-1}} \\ \end{pmatrix}.\)

Consider the function

\(\displaystyle f(t)=\det\begin{pmatrix} e^{c_2t} & e^{c_2b_3}-e^{c_2b_2} & \dots & e^{c_2b_n}-e^{c_2b_{n-1}} \\ e^{c_3t} & e^{c_3b_3}-e^{c_3b_2} & \dots & e^{c_3b_n}-e^{c_3b_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nt} & e^{c_nb_3}-e^{c_nb_2} & \dots & e^{c_nb_n}-e^{c_nb_{n-1}} \\ \end{pmatrix}.\)

Then

\(\displaystyle \det\begin{pmatrix} e^{c_2b_2}-e^{c_2b_1} & e^{c_2b_3}-e^{c_2b_2} & \dots & e^{c_2b_n}-e^{c_2b_{n-1}} \\ e^{c_3b_2}-e^{c_3b_1} & e^{c_3b_3}-e^{c_3b_2} & \dots & e^{c_3b_n}-e^{c_3b_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nb_2}-e^{c_nb_1} & e^{c_nb_3}-e^{c_nb_2} & \dots & e^{c_nb_n}-e^{c_nb_{n-1}} \\ \end{pmatrix}=f(b_2)-f(b_1).\)

By Lagrange's mean value theorem, there exists a \(\displaystyle b_1<x_1<b_2\) such that \(\displaystyle f(b_2)-f(b_1)=(b_2-b_1)f'(x_1)\), i.e.,

\(\displaystyle \det\begin{pmatrix} e^{c_2b_2}-e^{c_2b_1} & e^{c_2b_3}-e^{c_2b_2} & \dots & e^{c_2b_n}-e^{c_2b_{n-1}} \\ e^{c_3b_2}-e^{c_3b_1} & e^{c_3b_3}-e^{c_3b_2} & \dots & e^{c_3b_n}-e^{c_3b_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nb_2}-e^{c_nb_1} & e^{c_nb_3}-e^{c_nb_2} & \dots & e^{c_nb_n}-e^{c_nb_{n-1}} \\ \end{pmatrix} = \)

\(\displaystyle = (b_2-b_1)\det\begin{pmatrix} c_2e^{c_2x_1} & e^{c_2b_3}-e^{c_2b_2} & \dots & e^{c_2b_n}-e^{c_2b_{n-1}} \\ c_3e^{c_3x_1} & e^{c_3b_3}-e^{c_3b_2} & \dots & e^{c_3b_n}-e^{c_3b_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ c_ne^{c_nx_1} & e^{c_nb_3}-e^{c_nb_2} & \dots & e^{c_nb_n}-e^{c_nb_{n-1}} \\ \end{pmatrix}.\)

Repeating the same argument for each column, it can be obtained that there exist real numbers \(\displaystyle x_i\in(b_i,b_{i+1})\) (\(\displaystyle 1\le i\le n-1\)) such that

\(\displaystyle \det\begin{pmatrix} e^{c_2b_2}-e^{c_2b_1} & e^{c_2b_3}-e^{c_2b_2} & \dots & e^{c_2b_n}-e^{c_2b_{n-1}} \\ e^{c_3b_2}-e^{c_3b_1} & e^{c_3b_3}-e^{c_3b_2} & \dots & e^{c_3b_n}-e^{c_3b_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nb_2}-e^{c_nb_1} & e^{c_nb_3}-e^{c_nb_2} & \dots & e^{c_nb_n}-e^{c_nb_{n-1}} \\ \end{pmatrix} = \)

\(\displaystyle = \prod_{i=1}^{n-1}(b_{i+1}-b_i)\cdot \det\begin{pmatrix} c_2e^{c_2x_1} & c_2e^{c_2x_2} & \dots & c_2e^{c_2x_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ c_ne^{c_nx_1} & c_ne^{c_nx_2} & \dots & c_ne^{c_nx_{n-1}} \\ \end{pmatrix}= \)

\(\displaystyle = \prod_{i=1}^{n-1}(b_{i+1}-b_i)\cdot \prod_{i=2}^nc_i\cdot \det\begin{pmatrix} e^{c_2x_1} & e^{c_2x_2} & \dots & e^{c_2x_{n-1}} \\ \vdots & \vdots & \ddots & \vdots \\ e^{c_nx_1} & e^{c_nx_2} & \dots & e^{c_nx_{n-1}} \\ \end{pmatrix}. \)

By the induction hypothesis, this is positive.

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