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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!

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    $\begingroup$ 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 $p'(\xi)=0$. $\endgroup$ – Jens Schwaiger Mar 2 at 4:59
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    $\begingroup$ @JensSchwaiger Pleaes your comment an answer. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 2 at 5:02
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Following a request I transform my comment above into an answer:

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 $ξ∈(x_i,x_{i+1}$) such that $p′(ξ)=0$.

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  • $\begingroup$ 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)? $\endgroup$ – dxdydz Mar 2 at 17:00
  • $\begingroup$ 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 $x_i$. $\endgroup$ – Jens Schwaiger Mar 2 at 22:19
  • $\begingroup$ 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 $\endgroup$ – dxdydz Mar 3 at 4:48
  • $\begingroup$ @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. $\endgroup$ – Jens Schwaiger Mar 3 at 6:36

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