Positive determinant

Let $a_1<a_2<\cdots<a_n$ and $b_1<b_2<\cdots<b_n$ be positive real numbers. Prove that:

$$ \det \begin{pmatrix} e^{a_1 b_1} &e^{a_1 b_2} &\cdots &e^{a_1b_n} \\
e^{a_2 b_1}& e^{a_2 b_2} &\cdots &e^{a_2 b_n} \\
\vdots & \vdots & \ddots &\vdots \\
e^{a_n b_1}&e^{a_n b_2} &\cdots &e^{a_n b_n}
\end{pmatrix}>0$$

Solution



We are applying induction to $n$.

$\color{gray} \bullet$ For $n=1$ the determinant is equal to $e^{a_1 b_1}$ which is clearly positive.
$\color{gray} \bullet$ For $n>1$ we assume that the hypothesis is true. Let us denote $c_i=a_i-a_1, \, i>1$. Then:

$$\begin{aligned} \det \begin{pmatrix}
e^{a_1 b_1} &e^{a_1 b_2} &\cdots &e^{a_1b_n} \\
e^{a_2 b_1}& e^{a_2 b_2} &\cdots &e^{a_2 b_n} \\
\vdots & \vdots & \ddots &\vdots \\
e^{a_n b_1}&e^{a_n b_2} &\cdots &e^{a_n b_n}\end{pmatrix} &=\det \begin{pmatrix}
e^{a_1 b_1} &e^{a_1 b_2} &\cdots &e^{a_1 b_n} \\
e^{a_1 b_1}e^{c_2 b_1}&e^{a_1 b_2}e^{c_2 b_2} &\cdots &e^{a_1 b_n}e^{c_2 b_n} \\
\vdots & \vdots & \ddots & \vdots \\
e^{a_1 b_1}e^{c_n b_1}&e^{a_1 b_2}e^{c_n b_2} & \cdots & e^{a_1 b_n}e^{c_n b_n}
\end{pmatrix} \\
&= e^{a_1 \left ( b_1+b_2+\cdots+b_n \right )}\det \begin{pmatrix}
1 & 1 &\cdots &1 \\
e^{c_2 b_1}&e^{c_2 b_2} &\cdots &e^{c_2 b_n} \\
\vdots & \vdots & \ddots & \vdots \\
e^{c_n b_1}& e^{c_n b_2} &\cdots &e^{c_n b_n}
\end{pmatrix}
\end{aligned}$$

So, it is sufficient to prove that the last det is positive.

To eliminate the 1st row we substract the $n-1$th column from the $n$th , then $n-2$th column from the $n-1$th and so on. Hence:

$$\begin{aligned} \det \begin{pmatrix}
1 & 1 &\cdots &1 \\
e^{c_2 b_1}&e^{c_2 b_2} & \cdots & e^{c_2 b_n}\\
\vdots& \vdots &\ddots &\vdots \\
e^{c_n b_1}& e^{c_n b_2} &\cdots &e^{c_n b_n}
\end{pmatrix} &=\det \begin{pmatrix}
1 &0 &\cdots &0 \\
e^{c_2 b_1}& e^{c_2 b_2}-e^{c_2 b_1} &\cdots &e^{c_2 b_n}-e^{c_2b_{n-1}} \\
\vdots & \vdots & \ddots & \vdots \\
e^{c_n b_1}&e^{c_n b_2}-e^{c_n b_1} &\cdots & e^{c_n b_n}- e^{c_n b_{n-1}}
\end{pmatrix} \\
&= \det \begin{pmatrix}
e^{c_2 b_2}-e^{c_2 b_1} &e^{c_2 b_3} -e^{c_2b_2} &\cdots & e^{c_2 b_n}-e^{c_2 b_{n-1}}\\
e^{c_3 b_2}- e^{c_3 b_1}&e^{c_3 b_3}- e^{c_3 b_2} &\cdots &e^{c_3 b_n} -e^{c_3 b_{n-1}}\\
\vdots & \vdots & \ddots &\vdots \\
e^{c_n b_2}-e^{c_n b_1}& e^{c_n b_3}-e^{c_n b_2} &\cdots & e^{c_n b_n}-e^{c_n b_{n-1}}
\end{pmatrix}
\end{aligned}$$

Now consider the function:


$$f(t)=\det \begin{pmatrix} e^{c_2 t} &e^{c_2b_3}-e^{c_2 b_2} &\cdots &e^{c_2 b_n}- e^{c_2 b_{n-1}} \\
e^{c_3 t}&e^{c_3 b_3}-e^{c_3 b_2} & \cdots & e^{c_3 b_n}- e^{c_3 b_{n-1}}\\
\vdots & \vdots & \ddots &\vdots \\
e^{c_n t}&e^{c_n b_3} -e^{c_n b_2}&\cdots &e^{c_n b_n}- e^{c_n b_{n-1}}
\end{pmatrix}$$

It is easy to note that:

$$\det \begin{pmatrix} e^{c_2 b_2}-e^{c_2 b_1} &e^{c_2 b_3} -e^{c_2b_2} &\cdots & e^{c_2 b_n}-e^{c_2 b_{n-1}}\\
e^{c_3 b_2}- e^{c_3 b_1}&e^{c_3 b_3}- e^{c_3 b_2} &\cdots &e^{c_3 b_n} -e^{c_3 b_{n-1}}\\
\vdots & \vdots & \ddots &\vdots \\
e^{c_n b_2}-e^{c_n b_1}& e^{c_n b_3}-e^{c_n b_2} &\cdots & e^{c_n b_n}-e^{c_n b_{n-1}}
\end{pmatrix} =f(b_2)-f(b_1)$$

From the Lagrange Mean Value Theorem there exists an $x_1: b_1<x_1<b_2$ such that $\displaystyle f'(x_1)= \frac{f(b_2)-f(b_1)}{b_2-b_1}$, that is:

$$\det \begin{pmatrix} e^{c_2 b_2}-e^{c_2 b_1} &e^{c_2 b_3} -e^{c_2b_2} &\cdots & e^{c_2 b_n}-e^{c_2 b_{n-1}}\\
e^{c_3 b_2}- e^{c_3 b_1}&e^{c_3 b_3}- e^{c_3 b_2} &\cdots &e^{c_3 b_n} -e^{c_3 b_{n-1}}\\
\vdots & \vdots & \ddots &\vdots \\
e^{c_n b_2}-e^{c_n b_1}& e^{c_n b_3}-e^{c_n b_2} &\cdots & e^{c_n b_n}-e^{c_n b_{n-1}}
\end{pmatrix}= $$
$$=\left ( b_2-b_1 \right )\det \begin{pmatrix}
c_2 e^{c_2 x_1} & e^{c_2 b_3} -e^{c_2b_2} & \cdots & e^{c_2 b_n}-e^{c_2 b_{n-1}}\\
c_3 e^{c_3 x_1}& e^{c_3 b_3}- e^{c_3 b_2} & \cdots &e^{c_3 b_n} -e^{c_3 b_{n-1}} \\
\vdots & \vdots & \ddots &\vdots \\
c_n e^{c_n x_1}&e^{c_n b_3}-e^{c_n b_2} & \cdots& e^{c_n b_n}-e^{c_n b_{n-1}}
\end{pmatrix}$$

Repeating the same arguement for each column we have that:

$$\det \begin{pmatrix} e^{c_2 b_2}-e^{c_2 b_1} &e^{c_2 b_3} -e^{c_2b_2} &\cdots & e^{c_2 b_n}-e^{c_2 b_{n-1}}\\
e^{c_3 b_2}- e^{c_3 b_1}&e^{c_3 b_3}- e^{c_3 b_2} &\cdots &e^{c_3 b_n} -e^{c_3 b_{n-1}}\\
\vdots & \vdots & \ddots &\vdots \\
e^{c_n b_2}-e^{c_n b_1}& e^{c_n b_3}-e^{c_n b_2} &\cdots & e^{c_n b_n}-e^{c_n b_{n-1}}
\end{pmatrix} =$$
$$= \prod_{i=1}^{n-1}\left ( b_{i+1}-b_i \right )\det \begin{pmatrix}
c_2 e^{c_2 x_1} & e^{c_2 b_3} -e^{c_2b_2} & \cdots & e^{c_2 b_n}-e^{c_2 b_{n-1}}\\
c_3 e^{c_3 x_1}& e^{c_3 b_3}- e^{c_3 b_2} & \cdots &e^{c_3 b_n} -e^{c_3 b_{n-1}} \\
\vdots & \vdots & \ddots &\vdots \\
c_n e^{c_n x_1}&e^{c_n b_3}-e^{c_n b_2} & \cdots& e^{c_n b_n}-e^{c_n b_{n-1}}
\end{pmatrix}=$$
$$= \prod_{i=1}^{n-1}\left ( b_{i+1}-b_i \right )\prod_{i=2}^{n}c_i \det \begin{pmatrix}
e^{c_2 x_1} &e^{c_2 x_2} &\cdots & e^{c_2 x_{n-1}}\\
\vdots & \vdots & \ddots &\vdots \\
e^{c_n x_n}&e^{c_n x_2} &\cdots &e^{c_n x_{n-1}}
\end{pmatrix}$$

which from the induction hypothesis is positive and the exercise comes to an end!

This problem can be found in  Kömal A. , October 2008!