# Using Vandermonde Determinant to Prove an Existence of a Polynomial

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.

• Why do you think the proof might be easy using that determinant? Did someone give you a hint? – darij grinberg Nov 14 '18 at 3:18
• I just think that multiplying that matrix to a vector of variables would give the linear system that shows $f(a_1)=b_1, ..., f(a_n)=b_n$ – dmsj djsl Nov 14 '18 at 3:19
• Good, but what vector? And what should the $x_i$ be in the matrix? – darij grinberg Nov 14 '18 at 3:22
• 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? – dmsj djsl Nov 14 '18 at 3:48
• 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? – darij grinberg Nov 14 '18 at 3:49

Consider the map $$\phi:\def\R{\Bbb R}\R[x]_{ defined by $$\phi(P)=(P[a_1],P[a_2],\ldots,P[a_n])$$, where $$\R[x]_{ 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 $$\R[x]_{ is essentially just matrix $$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) 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.