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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 \begin{equation} \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} \end{equation} 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!

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    $\begingroup$ 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. $\endgroup$ – Bernard Oct 10 '18 at 0:31
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    $\begingroup$ The main ingredient for doing this as a linear algebra problem is the Vandermonde matrix en.wikipedia.org/wiki/Vandermonde_matrix $\endgroup$ – Will Jagy Oct 10 '18 at 0:32
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    $\begingroup$ @Bernard you mean the product of the $z_i-z_j$, I think... $\endgroup$ – Chris Custer Oct 10 '18 at 0:52
  • $\begingroup$ but ci and cj might be the same, so I think you mean zi-zj, since theyre different $\endgroup$ – AkatsukiMaliki Oct 10 '18 at 0:53
  • $\begingroup$ @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…) $\endgroup$ – Bernard Oct 10 '18 at 8:12
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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 $p=q$.

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  • $\begingroup$ Im using this to prove the fundamental theorem of algebra, so I can't say this. $\endgroup$ – AkatsukiMaliki Oct 10 '18 at 0:45

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