# How to prove that generalized Vandermonde matrix is invertible?

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

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.

• 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 (en.wikipedia.org/wiki/… ). Commented Oct 9, 2022 at 17:55
• I have followed your notation from the picture. Shall I change $\alpha$ to $\lambda$ ? Commented Oct 9, 2022 at 18:41

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 $$n=2.$$ 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 $$c_1x^{\alpha_1}+c_1x^{\alpha_2}+\ldots +c_nx^{\alpha_n}$$ vanishes at the points $$x_1 Hence the function $$f(x)=c_1+c_2x^{\alpha_2-\alpha_1}+\ldots +c_nx^{\alpha_n-\alpha_1}\quad (*)$$ vanishes at $$x_1 By the Rolle theorem there exist $$u_1 such that the function $$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 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 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. Let $$k$$ be a positive integer such $$\alpha_1<\ldots <\alpha_n Then $$\alpha_j Consider the function $$g(t):=F((1-t)\alpha_1+t(k+1) ,(1-t)\alpha_2+t(k+2),\ldots,(1-t)\alpha_{n}+t(k+n))$$ Then $$g(0)=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$$ and $$g(1)=F(k+1,k+2,\ldots, k+n)>0.$$ As $$g(t)$$ does not vanish we conclude $$g(0)>0.$$
• 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. Commented Oct 9, 2022 at 18:05
• 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. Commented Oct 9, 2022 at 18:23