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I have $a_1, ..., a_n, b_1, ..., b_n \in \mathbb{R}$. I must show that there is a unique polynomial $f(x) = c_0 + c_1x +...+ c_{n-1}x^{n-1} \in \mathbb{R}[x]$ of degree less than $n$ such that $f(a_1) = b_1, ..., f(a_n) = b_n$

$a_i$ is distinct and $b_i$ are just any real numbers.

Now, I am not sure how to quite do this, but I think that the proof might be easy if you use the Vandermonde determinant \begin{align*} A= \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{pmatrix} \end{align*} \begin{align*} \det(A) = \prod_{j > i} (x_j - x_i). \end{align*}

However, I am not sure how to carry out this proof. Is my thought process correct? Could somebody tell me how should I go on from here? Thanks.

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  • $\begingroup$ Why do you think the proof might be easy using that determinant? Did someone give you a hint? $\endgroup$ – darij grinberg Nov 14 '18 at 3:18
  • $\begingroup$ I just think that multiplying that matrix to a vector of variables would give the linear system that shows $f(a_1)=b_1, ..., f(a_n)=b_n$ $\endgroup$ – dmsj djsl Nov 14 '18 at 3:19
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    $\begingroup$ Good, but what vector? And what should the $x_i$ be in the matrix? $\endgroup$ – darij grinberg Nov 14 '18 at 3:22
  • $\begingroup$ I was thinking about keeping the $x$s in the matrix and multiplying by the vector of $c_i$. Is this not okay? If this is correct, I am still stuck on how to prove that this polynomial is unique. Am I missing something obvious? $\endgroup$ – dmsj djsl Nov 14 '18 at 3:48
  • $\begingroup$ No. What does keeping the $x$s mean? You don't have any $x$'s given. You want $b_i = f\left(a_i\right) = c_0 + c_1 a_i^1 + \cdots + c_{n-1} a_i^{n-1}$ for all $i$; does this suggest anything about what matrix and what vectors you should take? $\endgroup$ – darij grinberg Nov 14 '18 at 3:49
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Consider the map $\phi:\def\R{\Bbb R}\R[x]_{<n}\to\R^n$ defined by $\phi(P)=(P[a_1],P[a_2],\ldots,P[a_n])$, where $\R[x]_{<n}$ is the subspace of $\R[x]$ of polynomials of degree less than$~n$, and $P[a]$ denotes the evaluation of the polynomial $P$ at $x=a$. Your are asked to show that $\phi$ is a bijection: every point $(b_1,\ldots,b_n)$ equals $\phi(P)$ for a unique $P$.

Now $\phi$ is a linear map, since every evaluation map $P\mapsto P[a_i]$ is linear. Also, clearly $\phi(x^k)=(a_1^k,a_2^k,\ldots,a_n^k)$ for $k=0,1,\ldots,n-1$, so the matrix $M$ of $\phi$ with respect to the basis $[1,x,x^2,\ldots,x^{n-1}]$ of $\R[x]_{<n}$ is essentially just matrix $A$ of the question, where one just needs to set $x_i:=a_i$ for $i=1,2,\ldots,n$. The evaluation of $\det(A)$ that you cited, and the fact that the $a_i$ are all distinct (i.e., $a_j-a_i\neq0$ whenever $i<j$) show that $\det(M)\neq0$. Then $M$ is an invertible matrix, which corresponds to the fact that $\phi$ is a bijective linear map, and you are done.

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