My teacher gave us a homework problem that is considered an "advanced topic" that we have not really delved deeply into yet. I'm struggling to connect the dots in the question to formulate my answer.
I assume that I will be constructing a system of linear equations that gives solutions d, -b, -c, and a. My question is how the basis U = (1, t, t^2, t^3) relates to T(at^3 + bt^2 + ct + d) so that I can move forward and solve this problem.
On
The matrix is $$M= \begin {bmatrix} 0&0&0&1\\0&-1&0&0\\0&0&-1&0\\1&0&0&0\end{bmatrix} $$
Note that $$M\begin {pmatrix} a\\b\\c\\d\end{pmatrix} =\begin {pmatrix} d\\-b\\-c\\a\end{pmatrix}$$
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To find the matrix of your transformation, see what the transformation does to each basis element.
For instance: to find the third column of the matrix, see what $T$ does to the third element of $U$, $t^2$. If $M$ denotes the matrix of the transformation, then we find that (in the notation described here) $$ M \pmatrix{0\\0\\1\\0} = M [t^2]_U = [T(t^2)]_B = \left[\pmatrix{0&-1\\0&0}\right]_B = \pmatrix{0\\-1\\0\\0} $$ So, the matrix we're looking for has the form $$ M = \pmatrix{?&?&0&?\\?&?&-1&?\\?&?&0&?\\?&?&0&?} $$
We have the basis $U=\{ u_1, u_2, u_3, u_4 \}$ and $B=\{ b_1, b_2, b_3, b_4 \}$. Then $T$ acts like this: \begin{align} T(u_1) &= T(1) = T(\underbrace{0}_a\cdot t^3 + \underbrace{0}_b\cdot t^2 + \underbrace{0}_c\cdot t + \underbrace{1}_d\cdot 1) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = b_1 \\ T(u_2) &= T(t) = T(\underbrace{0}_a\cdot t^3 + \underbrace{0}_b\cdot t^2 + \underbrace{1}_c\cdot t + \underbrace{0}_d\cdot 1) = \begin{bmatrix} 0 & 0 \\ -1 & 0 \end{bmatrix} = -b_3 \\ T(u_3) &= T(t^2) = T(\underbrace{0}_a\cdot t^3 + \underbrace{1}_b\cdot t^2 + \underbrace{0}_c\cdot t + \underbrace{0}_d\cdot 1) = \begin{bmatrix} 0 & -1\\ 0 & 0 \end{bmatrix} = -b_2 \\ T(u_4) &= T(t^3) = T(\underbrace{1}_a\cdot t^3 + \underbrace{0}_b\cdot t^2 + \underbrace{0}_c\cdot t + \underbrace{0}_d\cdot 1) = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = b_4 \end{align}