Change of basis for polynomials I don't understand how the evaluation for $T(1), T(t),$ and $T(t^2)$ work. Does each $a_n$ correspond to a basis? If it does how does $T(t) = a_1t = a_1$ ?

 A: I think your confusion comes from the difficulty of thinking of a polynomial as both a vector in $\mathbb{P}_2$ and as a function.
Every polynomial of degree at most $2$ looks like
$$
p(t) = a_0 + a_1 t + a_2 t^2
$$
for some real numbers $a_0$, $a_1$ and $a_2$. The "variable" $t$ is really irrelevant and confusing, but we're stuck with it. The polynomial $t^2 = 0 + 0t + 1t^2$ corresponds to $a_0 = 0$, $a_1 = 0$ and $a_2 = 1$.
The function $T$ (what a terrible choice of name in this context) maps polynomials to polynomials using the formula specified in the problem. To see what it does to the polynomial $t^2 = 0 + 0t + 1t^2$ you apply that formula. What comes out is just $0 + 2t + 0t^2$, which of course you write as just $2t$. Similarly
$$
\begin{align}
T(t) &= T(0 + 1t + 0t^2) \\
     &= 1 + (2 \times 0)t + 0t^2 \\
     &= 1 .
\end{align}
$$
The fact that $T$ is a linear transformation (which is easy to check since you're given the formula) reflects the facts (true for more than just polynomials) that the derivative of a sum is the sum of the derivatives and that the derivative of a constant times a function is the constant times the derivative.
