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Given the matrix: $$A_{n}=\begin{bmatrix} e^x&e^{2x}& \cdots &e^{nx}\\ e^x&2e^{2x}& \cdots &ne^{nx}\\ \vdots & \vdots & \ddots & \vdots \\ e^x&2^{n-1}e^{2x}& \cdots &n^{n-1}e^{nx}\end{bmatrix}$$ I need to prove that $\det(A_n) \neq 0$.


I've found det for $A_3$ and $A_4$. But how to find det for common case $A_n$? (or at least show that it is not zero)enter image description here

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  • $\begingroup$ maybe with induction and use of Laplace $\endgroup$ – user409387 Oct 30 '17 at 18:45
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\begin{align}\det(A_n)&=e^{x+2x+\cdots+nx}\begin{vmatrix}1&1&\ldots&1\\1&2&\ldots&n\\1^2&2^2&\ldots&n^2\\\vdots&\vdots&\ddots&\vdots\\1^{n-1}&2^{n-1}&\ldots&n^{n-1}\end{vmatrix}\\&=e^{\frac{n(n+1)}2x}\prod_{1\leqslant i<j\leqslant n}(i-j)\\&\neq0.\end{align}The second equality comes from the fact that the matrix from the first line is a Vandermonde matrix.

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  • $\begingroup$ I have another question to this task. If e^x, e^2x, ... , e^nx - fundamental system of solutions of Linear differential equation how to find form of equation? $\endgroup$ – Николай Журба Oct 30 '17 at 19:13
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The determinant is $$e^{x+2x+\cdots+nx}\left|\matrix{1&1&\cdots&1\\1&2&\cdots&n\\ \vdots&\vdots&\ddots&\vdots\\1&2^{n-1}&\cdots&n^{n-1}}\right|.$$ This is a Vandermonde determinant and is nonzero.

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  • $\begingroup$ I have another question to this task. If e^x, e^2x, ... , e^nx - fundamental system of solutions of Linear differential equation how to find form of equation? $\endgroup$ – Николай Журба Oct 30 '17 at 19:17
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    $\begingroup$ @НиколайЖурба Write $D$ for $d/dx$. Then $f(D)y=0$ has solutions $e^{a_1x},\ldots,e^{a_nx}$ where $f(t)=(t-a_1)\cdots(t-a_n)$. $\endgroup$ – Lord Shark the Unknown Oct 30 '17 at 19:20
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Hint: Your determinant is the Wronskian (see https://en.wikipedia.org/wiki/Wronskian) of the following set $$\{e^{x},e^{2x},\ldots,e^{nx}\}.$$

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