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So we're studying Linear Algebra and our teachers are doing a terrific job in making us memorize solutions. Got hard stuck on this question:

Let $t_1, t_2,...,t_n$ be complex numbers. Show that:

\begin{gather} \det \begin{bmatrix} 1 & t_1 & t_1^2 & \cdots & t_1^{n-1}\\ 1 & t_2 & t_2^2 & \cdots & t_2^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & t_n & t_n^2 & \cdots & t_n^{n-1} \end{bmatrix} = \begin{vmatrix} 1 & t_1 & t_1^2 & \cdots & t_1^{n-1}\\ 1 & t_2 & t_2^2 & \cdots & t_2^{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & t_n & t_n^2 & \cdots & t_n^{n-1} \end{vmatrix} =\prod_{1 \leq i < j \leq n} (t_j - t_i) \end{gather} First thing is, I'm not even sure what this product notation exactly means. I also should mention that we are yet to study eigenvalues and eigenvectors, so the solution shouldn't use them. I genuinely am not getting anywhere with this question, so any hints would be appreciated.

emacs drives me nuts
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Flan Ben Flen
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  • This is known in the literature as a Vandermonde Matrix. If you only want a hint, then recall that many statements which involve $n$ can be approached with induction. – JMoravitz Jun 01 '22 at 14:52
  • Presumably. You want $1\leq i<j\leq n,$ not $1<i\leq j<n,$ because the value is trivially zero if you allow $i=j.$ – Thomas Andrews Jun 01 '22 at 14:56
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    "First thing is, I'm not even sure what this product notation exactly means" Expanded out you have $\prod\limits_{1\leq i<j\leq n} (t_j-t_i) = (t_2-t_1)(t_3-t_1)(t_4-t_1)\cdots(t_n-t_1)(t_3-t_2)(t_4-t_2)\cdots (t_n-t_2)(t_4-t_3)(t_5-t_3)\cdots (t_n-t_{n-1})$. That is to say, it is the product of all differences of terms $(t_j-t_i)$ satisfying $1\leq i<j\leq n$ – JMoravitz Jun 01 '22 at 15:09
  • Thank you so much Jmoravitz and Thomas Andrews! Yes this does answer my question. – Flan Ben Flen Jun 02 '22 at 06:39

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