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Given $$A = \left( z_i^{\lambda_k}\right)_{i,j = 1,\ldots, n} = \begin{pmatrix} z_1^{\lambda_1} & z_1^{\lambda_2} & \cdots & z_1^{\lambda_n} \\ z_2^{\lambda_1} & z_2^{\lambda_2} & \cdots & z_2^{\lambda_n} \\ \vdots & \vdots & \ddots & \vdots \\ z_n^{\lambda_1} & z_n^{\lambda_2} & \cdots & z_n^{\lambda_n} \end{pmatrix}.$$ with $z_1 < z_2 < ... < z_n \in \mathcal{R}_+$ and $\lambda_1 < \lambda_2 < ... \lambda_p \in \mathcal{R}$, how to show that A is invertible?

I think I probably need to show that the determinant is not zero but I am not sure how.

I do some derivation and get

enter image description here

Not sure how to continue. Also not sure if this is the right way.

There is a similar question here Generalized Vandermonde-Matrix but the conditions are different.

fixingmath
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    The determinant is nonzero but it's quite complicated: if the $\lambda_i$ are non-negative integers it's given by an ordinary Vandermonde determinant times a Schur polynomial (https://en.wikipedia.org/wiki/Schur_polynomial#Definition_(Jacobi's_bialternant_formula) ). – Qiaochu Yuan Oct 09 '22 at 17:55
  • I have followed your notation from the picture. Shall I change $\alpha$ to $\lambda$ ? – Ryszard Szwarc Oct 09 '22 at 18:41

1 Answers1

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We assume $\alpha_j\in \mathbb{R}$ and $x_j>0$ for $j=1,2,\ldots, n.$ The proof will go by induction on $n.$ The conclusion is obviously true for $n=1.$ Assume the conclusion is true for $n-1.$ Assume $\det A=0.$ Equvalently the columns of $A$ are linearly dependent. Therefore there exist constants $c_1,c_2,\ldots , c_n$ not all equal $0,$ such that the function $$c_1x^{\alpha_1}+c_2x^{\alpha_2}+\ldots +c_nx^{\alpha_n}$$ vanishes at the points $0<x_1<x_2<\ldots < x_n.$ Hence the function $$f(x)=c_1+c_2x^{\alpha_2-\alpha_1}+\ldots +c_nx^{\alpha_n-\alpha_1} \tag{*}$$ vanishes at $0<x_1<x_2<\ldots < x_n$ with $\alpha_i-\alpha_1>0, \forall i>1$ and strictly increasing, satisfying the induction hypothesis. By the Rolle theorem function $$f'(x)=c_2(\alpha_2-\alpha_1)x^{\alpha_2-\alpha_1-1}+\ldots +c_n(\alpha_n-\alpha_1)x^{\alpha_n-\alpha_1-1}$$ vanishes at $\{u_1,u_2,\ldots,u_{n-1}\}$ where $(x_i)_{i=1}^n$ and $(u_i)_{i=1}^{n-1}$ interlace, i.e. $0<x_1<u_1<x_2<u_2<\ldots<x_{n-1}<u_{n-1}<x_n$. By induction hypothesis we conclude that $c_j(\alpha_j-\alpha_1)=0$, $2\le j\le n.$ Thus $c_2=\ldots =c_n=0.$ By $(*)$ we get $f(x)=c_1,$ i.e. $c_1=0,$ a contradiction. The induction step is thus completed. $\blacksquare$

Remark The conclusion does not hold when the nodes are not necessarily positive. For example $ x_1=0$ and $\alpha_k>0.$ If we require all nodes nonzero, take $x_1=-x_2$ and $\alpha_k=2k.$ Then the first two rows of the matrix are equal.

By the intermediate value property it can be proved that the determinant is positive. Indeed, the function $$F(\alpha_1,\alpha_2,\ldots,\alpha_n)=\det A(\alpha_1,\alpha_2,\ldots,\alpha_n)$$ is continuous. Consider the function $$g(t):=F\big((1-t)\alpha_1,(1-t)\alpha_2+t,\ldots,(1-t)\alpha_{n}+t(n-1)\big).$$ $(1-t)\alpha_i+t(i-1), \forall i\in\{1,2,\ldots,n\}$ increases with $i$ thus satisfies the condition for the exponents same as $\alpha_i$ and $i-1$. Then $g(0)=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$ and $$g(1)=F(0,1,\ldots, n-1)=\prod_{1\le i<j\le n}(x_j-x_i)>0.$$ As $g(t)$ is continuous and does not vanish we conclude $g(0)>0.$

Hans
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Ryszard Szwarc
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  • I think it should be mentioned that the domain of $f$ is $\mathbb{R}^{+}$ to ensure continuity and differentiability of $f$ and allow the use of the Rolle theorem. This is why $z_1,\ldots,z_n\in\mathbb{R}^{+}$ is important. – Reinhard Meier Oct 09 '22 at 18:05
  • @ReinhardMeier I do not give up positivity of the nodes. Just exponents do not need to be positive. – Ryszard Szwarc Oct 09 '22 at 18:08
  • I know. But the fact that the points $z_1,\ldots,z_n$ are positive allows you to define a differentiable function $f$ on a domain that contains all those points in the first place. I just thought that this is important enough to be mentioned. – Reinhard Meier Oct 09 '22 at 18:23
  • @ReinhardMeier It is mentioned in OP. It is important, as otherwise fractional powers are not well defined. I will mention that assumption in the answer. – Ryszard Szwarc Oct 09 '22 at 18:30