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A Vandermonde-matrix is a matrix of this form:

$$\begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix} \in \mathbb{R}^{(n+1) \times (n+1)}$$.

condition ☀ : $\forall i, j\in \{0, \dots, n\}: i\neq j \Rightarrow x_i \neq x_j$

Why are Vandermonde-matrices with ☀ always invertible?

I have tried to find a short argument for that. I know some ways to show that in principle:

  • rank is equal to dimension
  • all lines / rows are linear independence
  • determinant is not zero
  • find inverse

According to proofwiki, the determinant is

$$\displaystyle V_n = \prod_{1 \le i < j \le n} \left({x_j - x_i}\right)$$

There are two proofs for this determinant, but I've wondered if there is a simpler way to show that such matrices are invertible.

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1  
Simpler than knowing its determinant and checking very easily it can't be zero by the given data? I doubt it... –  DonAntonio Jun 22 '13 at 14:48
2  
You have to prove the determinant first. I'm with you that checking it with the determinant is easy, but I guess there are easier ways than the two proves from proofwiki to show that it is invertible. –  moose Jun 22 '13 at 14:59
    
The Vandermonde determinant is non-zero $\implies$ the vandermonde matrix is invertible –  Ale Jan 23 at 8:25

3 Answers 3

up vote 8 down vote accepted

This is not entirely dissimilar to the answer already posted by Chris Godsil, but I'll post this anyway, maybe it can provide slightly different angle for someone trying to understand this.

We want to show that the matrix
$$\begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix}$$
is invertible.

It suffices to show that the rows of this matrix are linearly independent.

So let us assume that $c_0v_0+c_1v_1+\dots+c_nv_n=\vec 0=(0,0,\dots,0)$, where $v_k=(1,x_k^1,\dots,x_k^n)$ is the $k$-the row written as a vector and $c_0,\dots,c_n\in\mathbb R$.

Then we get on the $k$-th coordinate
$$c_0+c_1x_k+c_2x_k^2+\dots+c_kx_k^n=0,$$ which means that $x_k$ is a root of the polynomial $p(x)=c_0+c_1x+c_2x^2+\dots+c_kx^n$.

Now if the polynomial $p(x)$ of degree at most $n$ has $(n+1)$ different roots $x_0,x_1,\dots,x_n$, it must be the zero polynomial and we get that $c_0=c_1=\dots=c_k=0$.

This proves that the vectors $v_0,v_1,\dots,v_n$ are linearly independent. (And, in turn, we get that the given matrix is invertible.)

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Let $V$ be your Vandermonde matrix. If $p(t)=a_0+a_1t+\cdots+a_nt^n$ and $\alpha$ is the vector of coefficients of $p$, then the entries of $V\alpha$ are the values of $p$ on the points $x_0,\ldots,x_n$.

Now for $r=0,\ldots,n$ choose polynomials $p_r$ of degree $n$ so that $p(x_r)=1$ and $p(x_s)=0$ if $s\ne r$. Then the matrix with the coefficients of the polynomials $p_r$ as its columns is the inverse of $V$.

This is, of course, just a way of viewing Lagrange interpolation.

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Why is the matrix with the coefficients of the polynomials $p_r$ as its columns the inverse of $V$? –  moose Jun 22 '13 at 17:58
    
Because when you multiply it by $V$ you get the identity. The point is that $V$ applied to the coefficients of $p_r$ returns the $r$-th standard basis vector. –  Chris Godsil Jun 22 '13 at 21:15
    
Ok, but when you write "now [...] choose polynomials [...] so that $p(x_r)=1$ and $p(x_s) = 0$ if $s \neq r$" don't you already use that $V$ is invertible? Why can you choose polynomials like this? –  moose Jun 23 '13 at 6:30
    
Well, to get $p_r$ first take the product $q_r(x)=\prod_{i\ne r}(x-x_r)$ which is zero except at $x_r$ and not zero at $x_r$. Now set $p_r(x)=q_r(x)/q_r(x_r)$. –  Chris Godsil Jun 23 '13 at 13:24

For any $n+1$ distinct numbers $x_0, \ldots, x_n \in \mathbb{R}$, let $V(x_0,\ldots,x_n)$ and $D(x_0,\ldots,x_n)$ be a Vandermonde matrix and its determinant:

$$V(x_0,\ldots,x_n) = \begin{pmatrix} x_0^0 & \cdots & x_0^n \\ \vdots & \ddots & \vdots \\ x_n^0 & \cdots & x_n^n \end{pmatrix}\quad\text{ and }\quad D(x_0,\ldots,x_n) = \det V(x_0,\ldots,x_n) $$

It is clear $D(x_0) = 1 \ne 0$. Let us assume $D(x_0,\ldots,x_{m}) \ne 0$ for some $m < n$ and consider what happens to $D(x_0,\ldots,x_{m+1})$.

Viewed as a function of its last argument $x_{m+1}$, $D(x_0,\ldots,x_m,x)$ is a polynomial in $x$ with degree at most $m+1$. Since this polynomial vanishes at $m+1$ different values $x_0, \ldots, x_{m}$ already, it cannot vanish on any other $x$ (in particular at $x_{m+1}$). Otherwise, $D(x_0,\ldots,x_m,x)$ will be identically zero.

We know that $D(x_0,\ldots,x_m,x)$ isn't the zero polynomial. The leading coefficient of $x^{m+1}$ in $D(x_0,\ldots,x_m,x)$ is proportional to $D(x_0,\ldots,x_m)$ which is non-zero by induction assumption.

By induction, we can conclude $D(x_0,\ldots,x_n) \ne 0$ and hence $V(x_0,\ldots,x_n)$ is invertible.

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