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Let $T : \mathbb{R}[x]_{\leq 1} \rightarrow \mathbb{R}[x]_{\leq 2}$ be the linear map defined by the differential operator $T = (x-2) \frac{d}{d x} - \text{id}$.

Now my exercise asks me to find the matrix representing $T$ with respect to the standard coordinate systems

$(1 \; x): \mathbb{R}^2 \rightarrow \mathbb{R}[x]_{\leq 1} \quad \text{ and } \quad (1 \; x \; x^2) : \mathbb{R}^3 \rightarrow \mathbb{R}[x]_{\leq 2}$

I guess what I am mostly confused about, is the order in which I should compose the maps. Am I looking for a map from $\mathbb{R}^2$ to $\mathbb{R}^3$, or am I looking for a map from $\mathbb{R}[x]_{\leq 1}$ to $\mathbb{R}[x]_{\leq 2}$?

Currently, the more I think about it, the more confused I get, so I am looking for a little clarity most of all.

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For $i \ge 0$, let $p_i$ denote the polynomial $x^i$, with the convention $x^0=1$. Then for $0 \le i \le 1$ we have $ T(p_0)=-p_0=-1\cdot p_0+0\cdot p_1+0\cdot p_2,\quad T(p_1)=x-2-x=-2\cdot p_0+0\cdot p_1+0\cdot p_2. $ So the matrix of $T$ wrt the bases $\{p_0,p_1\}$ and $\{p_0,p_1,p_2\}$ is given by $ \left[ \begin{array}{cc} -1&-2\cr 0&0\cr 0&0 \end{array} \right] $

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    I had a thought about your argument, and if other people come here, I just want to extend your answer. By combining the coordinate-system matrices with the matrix for $T$, we obtain a map from $\mathbb{R}[x]_{\leq 1}$ to $\mathbb{R}[x]_{\leq 2}$, which goes $\mathbb{R}[x]_{\leq 1} \rightarrow \mathbb{R}^2 \rightarrow \mathbb{R}^3 \rightarrow \mathbb{R}[x]_{\leq 2}$. We start by mapping the polynomial $a_0+a_1x$ to its coordinate vector $(a_0,a_1)$, which we multiply by $T$ (giving a vector in $\mathbb{R}^3$), and then we map it back to a polynomial $b_0+b_1x+b_2x^2$.2012-08-20