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I'm lost on determining the matrix for this linear transformation. Any hints would be greatly appreciated.

$$\mathbb{R}[X]_{[\leq3]} \to \mathbb{R}$$ $$f = a_3X^3 + a_2X^2 + a_1X + a_0 \mapsto f(x)=a_3c^3 + a_2c^2 + a_1c + a_0$$

Evaluating polynomials at $0,1,2,3$, we get a linear map $\Phi : \mathbb{R}[X]_{[\leq3]}\to\mathbb{R}^4$ defined by

$$\Phi : \mathbb{R}[X]_{[\leq3]}\ni f \mapsto \begin{pmatrix}   f(0) \\   f(1) \\   f(2) \\   f(3) \\ \end{pmatrix} \in \mathbb{R}^4$$

Question:

Give the matrix expression $A_\Phi$ of $\Phi$ with respect to the basis $B=\{1,X,X^2,X^3\}$ of $\mathbb{R}[X]_{[\leq3]}$ and the standard basis $\{e_1,e_2,e_3,e_4\}$ of $\mathbb{R}^4$.

I am having trouble understanding how to approach this problem and I am looking for some hints in where to begin.

EDIT: Would the matrix look something like this? $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1\\ 1 & 2 & 4 & 8\\ 1 & 3 & 9 & 27\\ \end{pmatrix}$$

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  • $\begingroup$ What is $c$ in the definition of $f$? $\endgroup$
    – amd
    Apr 20 '17 at 1:18
  • $\begingroup$ @amd Just a scalar constant. The question was a little long, so I tried to cut out some stuff. $\endgroup$
    – cdignam
    Apr 20 '17 at 1:57
  • $\begingroup$ So the map $f$ is just the linear functional “evaluate at $c$?” $\endgroup$
    – amd
    Apr 20 '17 at 2:40
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Your linear transformation $\Phi$ is represented by a $4\times4$-matrix. Its columns are the images of the basis vectors $1$, $X$, $X^2$, $X^3$ under $\Phi$. So; what column vectors in $\Bbb{R}^4$ are these basis vectors mapped to?

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  • $\begingroup$ Would the matrix look like the one I edited in above? $\endgroup$
    – cdignam
    Apr 20 '17 at 0:58
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    $\begingroup$ That's exactly right. $\endgroup$ Apr 20 '17 at 1:18

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