# Show that Vandermonde matrix $𝑉_𝑚$ is invertible

Let $$x_1 < x_2 < . . . < x_m$$, and let $$y_1 , y_2 , . . . , y_m$$ be real numbers. There exists exactly one polynomial $$p$$ of degree $$≤ m − 1,$$ such that $$p(x_i) = y_i, 1 ≤ i ≤ m.$$ This is equivalent to showing that the Vandermonde matrix $$V_m$$ is invertible. Prove it using the following steps.

1. Explain: It suffices to prove that $$p(x) = 0$$ is the only solution when all the $$y_i$$ are zero.
2. Assume that $$p$$ were a polynomial of degree $$≤ m − 1$$ with $$p(x_i ) = 0$$ for all $$i$$. Show that there are $$m − 1$$ different points in which the derivative $$p'$$ is zero.

My attempt:
1. $$p(x)=\sum_{j=1}^mc_jx^{j-1}=V_m\vec{c}=\vec{y}.$$ This is a system of linear questions. So, $$V_m$$ is invertible if columns are linearly independent. In other words the only solution to $$V_m\vec{c}=\vec{0}$$ is $$\vec{c}=\vec{0}.$$
I'm stuck on the second point, any help would be appreciated!

• Rolle's theorem.: If $p$ is differentiable and If $p(x_i)=p(x_{i+1})=0$ (and $x_i<x_{i+1}$) there is some $\xi\in(x_i,x_{i+1})$ such that $p'(\xi)=0$. – Jens Schwaiger Mar 2 at 4:59
• @JensSchwaiger Pleaes your comment an answer. – GNUSupporter 8964民主女神 地下教會 Mar 2 at 5:02

Rolle's theorem: If $$p$$ is differentiable and if $$p(x_i)=p(x_{i+1})=0$$ (and $$x_i) there is some $$ξ∈(x_i,x_{i+1}$$) such that $$p′(ξ)=0$$.
• So, the polynomial of the degree $m-1$ has at most $m-1$ roots, therefore, shouldn't it have $m-2$ different points where the derivative is 0 (since between any two zeros of the function there is one point where the derivative is 0)? – dxdydz Mar 2 at 17:00
• We want to show with induction that $p$ as above is$\equiv 0$. The induction hypothesis and Rolle show that $p'\equiv 0$. Thus $p$ is constant and therefore $\equiv 0$ since $p$ by assumption has the zeroes $x_i$. – Jens Schwaiger Mar 2 at 22:19
• Can you clarify it a little bit more? So from what I get, we want to prove with induction that polynomial $p(x)=0$. So base case is trivial $(n=0):$ 0=0. Now we make the induction hypothesis that polynomial of degree $m-1 \ \ p(x_i)=0$ for all $i$.We want to show that polynomial of degree $m \ \ p(x_i)=0$ for all $i$. By Rolle's theorem there exist $m-1$ different points where $p'=0$. How do we know that $p'$ is 0 everywhere (so that p is constant)? Correct me if I'm wrong – dxdydz Mar 3 at 4:48
• @dxdydz: $p'$ has degree $m-2$ and vanishes at $m-2$ points $\xi_i\in(x_i,x_{i+1})$. Thus by induction hypothesis $p'$ has to vanish everywhere. – Jens Schwaiger Mar 3 at 6:36