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Let $t_1,t_2,t_3$ be distinct real numbers and consider the linear map

$$T : \mathbb{R}[x]_{\leq 2} \rightarrow \mathbb{R}^3, \quad \quad p(x) \mapsto \begin{pmatrix}p(t_1)\\p(t_2)\\p(t_3)\end{pmatrix}$$

I want to show that $T$ is an isomorphism. Since it is given that $T$ is linear, I just need to show that $T$ is bijective. My initial approach was to solve the system

$$ \begin{pmatrix} a + bt_1 + ct_1^2\\ a + bt_2 + ct_2^2\\ a + bt_3 + ct_3^2 \end{pmatrix} = \begin{pmatrix} r_1\\ r_2\\ r_3 \end{pmatrix} $$

for $a,b,c$ in terms of $r_1,r_2,r_3$, thus determining the inverse map $T^{-1}$ mapping a point in $\mathbb{R}^3$ to a polynomial of degree $\leq 2$.

I solved this system using Maple, and got a solution which was defined when $t_1,t_2,t_3$ where not all equal, which is fine by their definition, but I am wondering if there is a nicer argument, especially since the solution is rather ugly.

(this is not homework)

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Personally, I find the answer by Yuki most intuitive, and understandable. By considering the current votes, it seems like the majority disagrees (though only slightly). I can't find a question on meta about this "dilemma". –  utdiscant Aug 14 '12 at 3:02
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You should accept the answer you consider most helpful. If anyone stumbles into this question, he will be able to see the one with the top score (or next-to-top score, if the accepted answer is the top-scoring one) right below the accepted one anyway. There are even badges here for having nonaccepted answers with higher scores than accepted ones. ;) Personally, I find Jacob Schlather's suggestion to be the most elegant. –  tomasz Aug 14 '12 at 4:47
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4 Answers

up vote 3 down vote accepted

Hint:
$T$ is onto: Langrange Polynomial.

Injectivity: if $p$ is an polynomial of degree $\le2$, knowing values in 3 distinct points, you know $p$.

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Just take $1,x,x^2$. They transform into $\begin{pmatrix}1\\ 1 \\ 1\end{pmatrix},\begin{pmatrix}t_1\\ t_2 \\ t_3\end{pmatrix},\begin{pmatrix}t_1^2\\ t_2^2 \\ t_3^2\end{pmatrix}$. To show that they are linearly independent, calculate the determinant of $\begin{pmatrix}1&t_1&t_1^2\\1&t_2& t_2^2 \\1&t_3& t_3^2\end{pmatrix}$, which is, incidentally, Vandermonde matrix.

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The result heavily used here, and a rather important one, is: an operator $\,T:V\to V\,\,\,,\,\,V\,$ a vector space, is an isomorphism iff the image of some basis is again a basis (or more general: the image of any linearly independent is again lin. indep.) +1 –  DonAntonio Aug 14 '12 at 2:30
    
@DonAntonio: Not really: rather than using the result, I've explicitly shown that it is of rank 3. :) –  tomasz Aug 14 '12 at 2:31
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Hint: What is in the kernel of your map?

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I think it may be easier to show that the associated matrix always has non-zero determinant. This is a known result.

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