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Let the linear transformation $ D: P_3 \rightarrow P_2 $ be defined as $ D(f(x)) = f'(x) $.

$ B = \lbrace 1, x, x^{2}, x^{3} \rbrace, \\ C = \lbrace 2+x+x^{2}, -2-x^{2}, 1-x \rbrace $

Find the matrix $ [D]_B^C $ for $ D $ relative to the basis $ B $ in the domain and $ C $ in the codomain.


So $ [D]_B^{P_2} = \begin{bmatrix} 0&1&0&0\\ 0&0&2&0\\ 0&0&0&3 \end{bmatrix} $

and

$ T_{P_2 \rightarrow C} = T\left( \begin{bmatrix}1\\x\\x^2\\\end{bmatrix} \right) = \begin{bmatrix}2+x+x^2\\-2-x^2\\1-x\end{bmatrix}$, $ [T]_{P_2}^C= \begin{bmatrix} 2&1&1\\ -2&0&-1\\ 1&-1&0 \end{bmatrix} $


$ T(D(x)) = [T]_{P_2}^C \cdot [D]_B^{P_2} = \begin{bmatrix} 0&2&2&3\\ 0&-2&0&-3\\ 0&1&-2&0 \end{bmatrix} $

Why doesn't this work?

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  • $\begingroup$ I don't understand what you're doing here. What does $[D]_B^{P_2}$ represent? I thought $P_2$ was the space of polynomials of degree at most $2$, not a basis. Also, what is $T$? $\endgroup$ – Theo Bendit Mar 21 at 5:01
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The matrix from basis $C$ to $B$ is given by $$T_{C\to P_2}=\begin{pmatrix}2&-2&1\\1&0&-1\\1&-1&0\end{pmatrix}$$ because if you take, say, the $(1,0,0)$ vector, it is mapped to $(2,-2,1)$ in terms of the basis $B$. So the matrix you require is $$(T_{C\to P_2})^{-1}D_B^{P_2}=\begin{pmatrix}0&1&2&-6\\0&1&2&-9\\0&1&0&-6\end{pmatrix}$$

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