Vandermonde Determinant for NxN Matrices

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The problem I have is this: Show that

\begin{bmatrix} 1 & 1 & 1 \\ λ_{1} & λ_{2} & λ_{3} \\ λ_{1}^{2} & λ_{2}^{2} & λ_{3}^{2} \end{bmatrix}
Has determinant

$$ (λ_{3} - λ_{2}) (λ_{3} - λ_{1}) (λ_{2} - λ_{1}) $$

And generalize to the NxN case (proof not needed)Obviously solving the 3x3 was not hard, I simply expanded the expression for the determinate given and showed it to be the same as the one i calculated using the rule of Sarrus.

However for the nxn case I'm not sure how to proceed. I tried entering bigger and bigger matrices into wolfram but there was no clear pattern. Is there a non computational way to solve this?
 
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Have you tried an induction over ##n##?
 
grassstrip1 said:
The problem I have is this: Show that

\begin{bmatrix} 1 & 1 & 1 \\ λ_{1} & λ_{2} & λ_{3} \\ λ_{1}^{2} & λ_{2}^{2} & λ_{3}^{2} \end{bmatrix}
Has determinant

$$ (λ_{3} - λ_{2}) (λ_{3} - λ_{1}) (λ_{2} - λ_{1}) $$

And generalize to the NxN case (proof not needed)Obviously solving the 3x3 was not hard, I simply expanded the expression for the determinate given and showed it to be the same as the one i calculated using the rule of Sarrus.

However for the nxn case I'm not sure how to proceed. I tried entering bigger and bigger matrices into wolfram but there was no clear pattern. Is there a non computational way to solve this?
I don't have a complete answer, but a 4x4 with the 4th row being cubic will result in a 6th power polynomial form. The number of combinations of 4 things taken 2 at a time is 6, making for the 6 factors. (I don't know how to choose the minus or plus sign for the terms, but if it does generalize, this might be helpful.)...editing... One additional observation is if two columns of a matrix are identical, I think the determinant is zero. This is consistent with setting the product of the factors equal to zero.
 
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Thank you! The formula for the vandermonde determinant was exactly what I needed!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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