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In one of my questions, we have $L: P_3(\mathbb{R}) \to \mathbb{R}^3$, and $L$ is defined as $L(p) = \left( \begin{array}{c} p(-2) \\ p(1) \\ p(3) \end{array} \right)$. What does $L(p)$ mean? Since the basis for $P_3(\mathbb{R})$ is $\{1, x, x^2, x^3\}$, then does it mean that for $L(1), L(x), L(x^2),$ and $L(x^3)$, everything will equal to $-2, 1, 3$ regardless of the variable given?

edit: the question is to find a matrix representing the linear transformation $L$, with bases $\{1, x, x^2, x^3\}$ and the standard basis for $\mathbb{R}^3$, $\{e_1, e_2, e_3\}$. I know how to do these kinds of questions, I'm just not sure what $L(p)$ is in this case.

Actually I'm unsure about this. Is the following solution correct?

$ L(1) = (-2, 1, 3) \\ L(x) = (-2, 1, 3) \\ L(x^2) = (4, 1, 9) \\ L(x^3) = (-8, 1, 27) $

$A = \left( \begin{array}{cccc} -2 & -2 & 4 & -8 \\ 1 & 1 & 1 & 1 \\ 3 & 3 & 9 & 27 \end{array} \right) $

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To which question are you referring? –  JavaMan Dec 11 '12 at 9:05
    
The function $L$ takes a polynomial and evaluates it at the points $-2$, $1$ and $3$. –  wj32 Dec 11 '12 at 10:29
    
L is a function defined on $P_3(\mathbb{R})$ taking polynomials $p \in P_3(\mathbb{R})$ into $\mathbb{R}^3$. For any specific $p in P_3(\mathbb{R})$, $L(p)$ is the value of the function $L$ evaluated at the polynomial $p$. The equality you have there tells you now to get the value of $L$ at $p$ for any $p$. –  kigen Dec 11 '12 at 15:50
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The solution is incorrect - read the comment for details about what $L(p)$ is.

If you'll excuse me writing row vectors for column vectors, then:

$L(1)$ is the image of the polynomial $p(x)=1$ under $L$. $p(-2)=p(1)=p(3)=1$ in this case, so $L(1)=(1,1,1)$.

$L(x)$ is the image of the polynomial $p(x)=x$. In this case $L(x)=(-2,1,3)$.

And so on. If you understand what $L(p)$ means now then you should be able to finish.

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So then $L(x^2) = (4, 1, 9)$ and $L(x^3) = (-8, 1, 27)$? And the matrix of the linear transformation would be $A = \left( \begin{array}{cccc} 1 & -2 & 4 & -8 \\ 1 & 1 & 1 & 1 \\ 1 & 3 & 9 & 27 \end{array} \right)$. –  user52610 Dec 11 '12 at 18:16
    
Yup. That's exactly it. Good! –  kigen Dec 12 '12 at 3:42
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