# How to see that polynomial is fixed by finitely many values.

Let's say we have a polynomial $$p(z)=\sum_{i=0}^n c_iz^i$$. If we have $$n+1$$ known values $$(z_i,w_i)$$. We find \begin{align*} \sum_{i=0}^n c_iz_0^i&=w_0\\ &\vdots\\ \sum_{i=0}^n c_iz_n^i&=w_n \end{align*}

Which comes down to row-reducing $$$$\left( \begin{array}{cccc|c} 1c_1&z_0c_2&\ldots&z_0^n c_{n+1}&w_0\\ 1c_1&z_1c_2&\ldots&z_1^nc_{n+1}&w_1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 1c_1&z_nc_2&\ldots&z^n_nc_{n+1}&w_n \end{array} \right)\label{matrix}$$$$ For $$i\neq j$$ it holds that $$z_i\neq z_j$$. why do the rows have to be linearly independent, aka there has to be a unique solution for all the $$c_i$$. Therefore $$p$$ being determined by $$n+1$$ values?

Without using the fundamental theorem of algebra!

• The deerminant of this linear system is a Vandermonde determinant, which happens to be equal to the product of the $c_i-c_j$ for all $1\le i<j\le n$n hence is non-zero. – Bernard Oct 10 '18 at 0:31
• The main ingredient for doing this as a linear algebra problem is the Vandermonde matrix en.wikipedia.org/wiki/Vandermonde_matrix – Will Jagy Oct 10 '18 at 0:32
• @Bernard you mean the product of the $z_i-z_j$, I think... – Chris Custer Oct 10 '18 at 0:52
• but ci and cj might be the same, so I think you mean zi-zj, since theyre different – AkatsukiMaliki Oct 10 '18 at 0:53
• @ChrisCuster: Yes, I meant the product of the $z_i-z_j$. Just a slip because of the unusual name for coefficients (for my excuse, it was rather late here when I wrote this comment…) – Bernard Oct 10 '18 at 8:12

The $$c_i$$'s are uniquely determined because if you have an other polynomial $$q$$ of degree at most $$n$$ such that $$\forall i, q(z_i)=w_i$$, then $$p-q$$ is a polynomial of degree at most $$n$$ which has $$n+1$$ distinct roots, so it has to be $$0$$, that is $$p=q$$.