# Proof determinant of transpose Vandermonde matrix is $\prod_{1\le i\lt j\le n}(\alpha_i-\alpha_j)$

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?

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

• If you understand the other proofs, just use the fact that $\det (A^T)=\det (A)$. – Git Gud Dec 24 '13 at 13:49
• @Git Gud e.g. this proof is one of the transpose matrix and also comes to the solution with the interchanged indices – sj134 Dec 24 '13 at 13:52
• The formula with with interchanged indices is incorrect, as can be seen in the $2\times 2$ case. – Olivier Bégassat Dec 24 '13 at 13:59
• @Olivier Bégassat thanks, I thought so because my initial step failed as well... I just wanted to make sure thats why I asked here. – sj134 Dec 24 '13 at 14:05
• 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). – 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.