Find the matrix of $T(p(t)) = tp'(t)$ with respect to the basis $B = \{1, 1 + t, t^2\}$ of $P_2$.

Question

A function $T : P_2 \to P_2$ is defined by $T(p(t)) = tp'(t)$.

Find the matrix of $T$ with respect to the basis $B = \{1, 1+t, t^2\}$ of $P_2$.

The expression below is the only thing I could come up with. If someone could explain to me how to go about solving this, that'd be great. Thank you.

$T(a + bt + ct^2) = t(b + 2ct) = bt + 2ct^2$

Answer 1

The columns of the matrix of a linear operator with respect to a basis are given by the operator applied to the basis elements. We compute

\begin{align} T(1) &= 0\\ T(1+t) &= t\\ T(1+t+t^2) &= t+2t^2, \end{align} and hence the matrix is $$ \pmatrix{0&1&0\\0&1&1\\0&0&2}. $$

Answer 2

Hint

Let $(T)_{\mathcal S\mathcal S}$ the matrix of $\mathcal T$ in the standard basis (i.e. $\mathcal S=\{1,t,t^2\}$), and $P$ is the change of basis from $\mathcal S\to \mathcal B$, then $$(T)_{\mathcal B\mathcal B}=P(T)_{\mathcal S\mathcal S}P^{-1}.$$