the below is a transpose Vandermonde matrix determinant equality. I have seen a lot of proofs of its determinant being $=\prod_{1\le i\lt j\le n}(\alpha_j-\alpha_i)$, but this ones indices are interchanged.

How can this be shown by induction?

$$\det \begin{pmatrix} 1 & 1 & \cdots & 1 \\ \alpha_1 & \alpha_2 & \cdots & \alpha_n \\ \vdots & \vdots & \quad & \vdots \\ \alpha_{1}^{n-1} & \alpha_{2}^{n-1} & \cdots & \alpha_{n}^{n-1} \end{pmatrix} = \prod_{1\le i\lt j\le n}(\alpha_i-\alpha_j)$$

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    $\begingroup$ If you understand the other proofs, just use the fact that $\det (A^T)=\det (A)$. $\endgroup$ – Git Gud Dec 24 '13 at 13:49
  • $\begingroup$ @Git Gud e.g. this proof is one of the transpose matrix and also comes to the solution with the interchanged indices $\endgroup$ – sj134 Dec 24 '13 at 13:52
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    $\begingroup$ The formula with with interchanged indices is incorrect, as can be seen in the $2\times 2$ case. $\endgroup$ – Olivier Bégassat Dec 24 '13 at 13:59
  • $\begingroup$ @Olivier Bégassat thanks, I thought so because my initial step failed as well... I just wanted to make sure thats why I asked here. $\endgroup$ – sj134 Dec 24 '13 at 14:05
  • $\begingroup$ Another way to see it was wrong that the indices were swapped around is that if this were true, $\binom n2$ would be even for all $n \ge 2$ (there are $\binom n2$ factors in the product, each off by a sign). $\endgroup$ – Patrick Da Silva Dec 24 '13 at 15:01

So the product should definitely be $\prod_{1\leq i<j\leq n}(\alpha_j-\alpha_i)$; this makes a difference from what you wrote whenever the number$~\binom n2$ of factors in the product is odd, namely when $n\equiv2$ or $n\equiv3\pmod4$.

For a proof by induction (which starts with the trivial case $n=0$ for which the equation says $1=1$), one can proceed as follows. First subtract the first column from each of the other columns. This makes entry the entry at position $(i+1,j)$ equal to $\alpha_j^i-\alpha_1^i=(\alpha_j-\alpha_1)\sum_{k=0}^{i-1}\alpha_j^{i-1-k}\alpha_1^k$ for $i=0,1,\ldots,n-1$ and $j>1$, which in particular is$~0$ for $i=0$. Developing by the first row, and factoring $\alpha_j-\alpha_1$ out of column $j$ for $j=2,3,\ldots,n$, we see that we will be able to conclude if we can show by induction that $$ \det\left(\biggl(\sum_{k=0}^{i-1}\alpha_j^{i-1-k}\alpha_1^k\biggr)_{i=1,\ldots,n-1\atop j=2,\ldots,n}\right)=\prod_{2\leq i<j\leq n}(\alpha_j-\alpha_i). $$ In the matrix of the left hand side, subtract from each row (except from the first row) $\alpha_1$ times the previous row (proceed from bottom to top, so it is an unchanged row that is subtracted each time). This removes precisely the terms with $k>0$ from the summation, and what is left is the determinant of the matrix $(\alpha_j^{i-1})_{i=1,\ldots,n-1\atop j=2,\ldots,n}$, a Vandermonde matrix whose determinant is indeed given by the right hand side according to the induction hypothesis.

Other approaches using induction (notably ones using a linear algebraic interpretation of the matrix) are certainly possible, this is just the first idea that came up for me.



try to make the first identity of each row vanish using the row right above it,from the last row to the first,and you can easily use the hypothesis to conclude the result


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