# Linear transformation with change of ordered basis

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?

• 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 $T$? – Theo Bendit Mar 21 at 5:01

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 $$(T_{C\to P_2})^{-1}D_B^{P_2}=\begin{pmatrix}0&1&2&-6\\0&1&2&-9\\0&1&0&-6\end{pmatrix}$$