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The generalized Vandermonde matrix that I am considering is one where the rows of a matrix correspond to the powers of different elements of the field, but the powers need not be consecutive integers as for the case of a traditional Vandermonde matrix. Is there a result that characterizes when such a matrix is invertible?

Thanks!

signum
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  • Try calculating the det. – Vim Feb 10 '15 at 08:42
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    This is tougher. The ratio of such a determinant and the Vandermonde determinant is known as a Schur polynomial. A lot is known about them, but the question of deciding whether such a determinant can vanish is tricky. At least for me. The question comes up naturally when studying (cyclic) codes defined by certain check equations. In that context there are some related general results with a narrow scope (Hartman-Tzeng), and a number of ad hoc results. – Jyrki Lahtonen Feb 11 '15 at 06:30
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    Thanks @JyrkiLahtonen for your input! The question did come up when studying cyclic codes, but I couldn't find any relevant reference. Seems like it is indeed a hard problem. – signum Feb 11 '15 at 19:37

1 Answers1

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According to Wolfram MathWorld,

![enter image description here

(https://mathworld.wolfram.com/GeneralizedVandermondeMatrix.html)

Now keep in mind I think this corresponds to $\mathbb{C}$, so it does not necessarily apply to finite fields.

However, if it is somehow possible to factor the determinant into a product as in the regular Vandermonde matrix, if all terms in the product are nonzero, then the product is nonzero by the closure property of the multiplicative group (all nonzero elements) of a finite field.

Edit: I rewrote this answer twice. The first time because I didn't realize the Wolfram article may or may not correspond to $\mathbb{C}$. The second time because I found a formula for the determinant of the generalized Vandermonde matrix from another stack exchange answer. However, that formula apparently corresponded to some different form of the matrix, that they claimed was also called a generalized Vandermonde matrix, so I don't think my second answer applied to the question here.

Kevin
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    Interesting, but not clear what this means over a finite field. – Igor Rivin Jul 24 '22 at 16:14
  • Good point. I rewrote the answer to try and better answer the question. – Kevin Jul 24 '22 at 18:11
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    How can it be always positive when it, like the usual Vandermonde, changes its sign whenever you swap two of the variables? Also, positive applies only to reals, I think? – Jyrki Lahtonen Jul 26 '22 at 04:04
  • @JyrkiLahtonen: The $a_i$ is required to increase with the row number $i$. – Hans Jul 08 '25 at 22:26
  • @Hans I was talking about the variables, not the exponents. – Jyrki Lahtonen Jul 09 '25 at 11:18
  • @JyrkiLahtonen: In the notation of the linked webpage of MathWorld with matrix entry $a_i^{b_j}$, $a_i$ is the base variable and $b_j$ is the exponent. Both of them should be strictly increasing. c.f. https://math.stackexchange.com/a/4549001/64809 – Hans Jul 09 '25 at 19:11
  • @Hans, I see your point, now. I was really thinking about the case of a finite field, where there is no ordering on the variables. $\Bbb{C}$ and $\Bbb{F}_q$ have in common the property that there is no $<$-relation. – Jyrki Lahtonen Jul 09 '25 at 20:10
  • I see that we are thinking about two different domains of the numbers. The culprit is this answer which mistook the domain of the MathWorld answer for both the complex number and a finite field. If you are interested in a somewhat loosely related problem, check out https://math.stackexchange.com/a/2547040/64809 . I am interested in your thoughts. – Hans Jul 09 '25 at 20:32