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1.

I'm hesitant to bring this up, since someone might point out why it's trivial, but the question http://mathoverflow.net/questions/118225/how-to-show-a-certain-determinant-is-non-zero-closed was closed yesterday, with a comment by a high rep user that this site is not for homework. (Edit: Since the user removed his comment, I have now edited this post to remove his name.) How sure are we that this is homework? It might very well be I'm missing some obvious proof, but for some reason it doesn't seem that easy to me. (Meanwhile the question was asked and answered at mathstackexchange.com, but it doesn't seem to me the answer is completely correct; I am unable to comment there, but I left a comment on this under the MO question.)

• CommentAuthorabatkai
• CommentTimeJan 7th 2013

I do not have the reputation to decide on these matters, but I also do not believe this is homework. Might be more a language problem as we have experienced it earlier. we can find plenty of questions on the level of the question, which remained open.
2.
Since the question appears not that much trivial. I voted to reopen.
By the way is there any closed form for such determinat ?
If x_i = i , then it is Vandermonde of exp(\lambda_i).
More generally for integers x_i it resembles numenator in Schur polynoms.
When everything is small we get rank 1 + small correction, I think it possible to see that eigenvalues in first order are non-zero and see that is why det \ne 0.
But all these does not seem to answer the question.
3.

I cast a last vote to reopen a few minutes ago. Thanks for your attention, guys.

While I was struggling with and googling for this problem last night, I learned about the concept of alternant matrix, of which this is a special case. It feels like the result should be well-known, and that methods other than Noam Elkies' clever solution ought to be available.

Edit: Huh, and now a vote to close again. Voter, please explain.