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Let $z_1, \ldots, z_n \in \mathbb{C}\setminus\left\lbrace 0 \right\rbrace$ be distinct complex numbers, $\lambda_1 < \lambda_2 < \ldots < \lambda_n$ positive integers and define $$A = \left( z_i^{\lambda_k}\right)_{i,k = 1,\ldots, n} = \begin{pmatrix} z_1^{\lambda_1} & z_1^{\lambda_2} & \cdots & z_1^{\lambda_n} \\ z_2^{\lambda_1} & z_2^{\lambda_2} & \cdots & z_2^{\lambda_n} \\ \vdots & \vdots & \ddots & \vdots \\ z_n^{\lambda_1} & z_n^{\lambda_2} & \cdots & z_n^{\lambda_n} \end{pmatrix}.$$ Is it true that $A$ is invertible?

I found this related question but it deals with finite fields where in my case the underlying field is the complex plane.

If $\lambda_k = k -1 $ for $k = 1, \ldots, n$, then $A$ is the well-known Vandermonde-Matrix, so in this case the answer is positive. I have tried to calculate the determinant of $A$ analogously to how it's done if $A$ is the Vandermonde-Matrix but without success.

Does anyone have a reference or a proof which answers this question?

Thanks in advance...

Bruno Krams
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  • Could you find a reference that computes the determinant of this matrix? – QGravity Feb 11 '21 at 18:33
  • No. But actually with paulinho's answer I haven't looked for any further reference – Bruno Krams Feb 12 '21 at 11:20
  • Thanks. The determinant of this matrix divided by the usual Vandemonde determinant is actually the definition of Schur function. And that function can be zeros. But I am interested in taking the large-$N$ limit of these polynomials in some specific way so I was looking for a generic expression for the Schur polynomial of $N$ variables. – QGravity Feb 13 '21 at 03:04

1 Answers1

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It is not necessarily invertible. Consider $z_1 = 1, z_2 = -1$, with $\lambda_1 = 2, \lambda_2 = 4$. Then the resulting matrix is $2 \times 2$ matrix of ones, which is clearly not invertible.

To see why this in general is not necessarily invertible, think of the Vandermonde matrix as polynomial interpolation. The normal Vandermonde matrix, for example, gives the $(n - 1)$th degree polynomial going through $n$ distinct points. What you are asking here is to find a polynomial whose coefficients are all zero except the coefficients to the terms $x^{\lambda_1}, x^{\lambda_2}, \cdots, x^{\lambda_n}$. To construct a counterexample, we just need to note that such a polynomial has $\lambda_n - \lambda_1$ nonzero roots, and if $n$ of these $\lambda_n - \lambda_1$ roots are distinct, we could find $n$ distinct values to give to $z_1, \cdots, z_n$ and have the matrix be singular (why?).

paulinho
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