When determinant of matrix of exponents is non-negative

Question

This is problem I heard briefly, not even sure if it's well-stated, but here we are:

Prove that if $x_i>0, y_j>0$ then $\det e^{x_iy_j}\ge 0$. I think there was also a hint to use "generalized Vandermonde determinant", but I can't figure out even what this generalized determinant is.

May be you have some ideas? thanks in advance!

score 2 · Answer 1 · answered Jul 23 '23 at 20:14

2

This property is erroneous : take the case of a $2 \times 2$ matrix with :

$$x_1=3, x_2=2 , y_1=1, y_2=2$$

The determinant of the matrix is

$$e^{x_1y_1}e^{x_2y_2}-e^{x_1y_2}e^{x_2y_1}$$

$$=e^{3}e^{4}-e^{6}e^{2}=e^{7}-e^{8} \color{red}{< 0}$$

answered Jul 23 '23 at 20:14

Jean Marie

81,803

Any comment ?... – Jean Marie Jul 24 '23 at 08:07
Well, I see now, but... what was the task then?.... – Big Coconut Jul 24 '23 at 08:16
It cannot be the matrix with $M_{ij}=e^{x_i\color{red}{+}y_j}$ because this matrix has rank $1$ therefore with determinant always $0$. – Jean Marie Jul 24 '23 at 17:24

When determinant of matrix of exponents is non-negative

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