Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question
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
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
up vote 3 down vote accepted

$T$ is onto: Langrange Polynomial.

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

share|cite|improve this answer

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.

share|cite|improve this answer
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

Hint: What is in the kernel of your map?

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.