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Problem A. 463. (October 2008)

A. 463. Let a1<a2<...<an and b1<b2<...<bn be real numbers. Show that

\det\left(\matrix{
e^{a_1b_1} & e^{a_1b_2} & \dots & e^{a_1b_n} \cr
e^{a_2b_1} & e^{a_2b_2} & \dots & e^{a_2b_n} \cr
\vdots & \vdots & \ddots & \vdots \cr
e^{a_nb_1} & e^{a_nb_2} & \dots & e^{a_nb_n} \cr}
\right) >0.

(5 pont)

Deadline expired on 17 November 2008.


Solution. Apply induction on n. For n=1 the statement is ea1b1>0 which is obvious. Now suppose n>1 and assume that the statement is true for all smaller values.

Let ci=ai-a1>0. Then

\det\left(\matrix{
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} \\
}\right)=
\det\left(\matrix{
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} \\
}\right)
=

=
e^{a_1(b_1+b_2+\dots+b_n)}
\det\left(\matrix{
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} \\
}\right),

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

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


\det\left(\matrix{
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} \\
}\right)=

 = \det\left(\matrix{ 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}} \cr
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}} \cr
\vdots & \vdots & \vdots & \ddots & \vdots \cr
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}} \cr
}\right) =

 = \det\left(\matrix{
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}} \cr
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}} \cr
\vdots & \vdots & \ddots & \vdots \cr
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}} \cr
}\right).

Consider the function

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

Then

 \det\left(\matrix{
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}} \cr
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}} \cr
\vdots & \vdots & \ddots & \vdots \cr
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}} \cr
}\right)=f(b_2)-f(b_1).

By Lagrange's mean value theorem, there exists a b1<x1<b2 such that f(b2)-f(b1)=(b2-b1)f'(x1), i.e.

\det\left(\matrix{
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}} \cr
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}} \cr
\vdots & \vdots & \ddots & \vdots \cr
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}} \cr
}\right) =


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

Repeating the same argument for each column, it can be btained that there exist real numbers xi\in(bi,bi+1) (1\lei\len-1) such that

\det\left(\matrix{
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}} \cr
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}} \cr
\vdots & \vdots & \ddots & \vdots \cr
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}} \cr
}\right) =

 = \prod_{i=1}^{n-1}(b_{i+1}-b_i)\cdot
\det\left(\matrix{
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}} \\
}\right)=

 = \prod_{i=1}^{n-1}(b_{i+1}-b_i)\cdot \prod_{i=2}^nc_i\cdot
\det\left(\matrix{
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}} \\
}\right).

By the induction hypothesis, this is positive.


Statistics:

4 students sent a solution.
5 points:Nagy 235 János, Nagy 314 Dániel, Tomon István.
1 point:1 student.

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