Change of basis of $P_2$ Hi this is an assignment question which I have no clue how so solve to I was wondering if someone could solve one part of it so I  could solve the rest thanks
Let $S = \{1, x, x^2\}$ denote the standard basis for $P_2$, let $B = \{1, 2 + x ,3 + 2x + x^2\}$ and let $C = \{1, 1 + x, 1 + x + x^2 \}$ You may assume that B and C are also bases for $P_2$.
(a) Compute the change-of-basis matrix $[I]_{S \leftarrow B}$ from B to S.
(b) Compute the change-of-basis matrix $[I]_{C \leftarrow S}$from S to C.
(c) Compute the change-of-basis matrix $[I]_{B \leftarrow C}$ from B to C
 A: Let's find a matrix that sends $S \to B$ : You want
$$
A(1) = 1, A(x) = 2+x, A(x^2) = 3 + 2 x + 1 x^2
$$
So treat the original basis as the standard basis in $\mathbb{R}^3$, if you will
$$
1 = (1,0,0), x = (0,1,0), x^2 = (0,0,1)
$$
Then
$$
A = \begin{pmatrix}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{pmatrix}
$$
Now $A^{-1}$ will solve (a). Can you try the rest?
A: I will show (a), you can follow to do the rest:
Pick a point in $P_2$ represented in terms of $B$. If the representation in terms of $B$ is $(\alpha_0,\alpha_1, \alpha_2)^T$ then the actual point is  $x \mapsto \alpha_0 + \alpha_1(2+x)+\alpha_2(3+2x+x^2)$. Now we want to write this in terms of the basis $S$. Rewriting the above gives 
$\alpha_0 + \alpha_1(2+x)+\alpha_2(3+2x+x^2) = (\alpha_0+2\alpha_1+3\alpha_2) + x (\alpha_1+2 \alpha_2)+ x^2 (\alpha_2)$.
Hence to represent the point $x \mapsto (\alpha_0+2\alpha_1+3\alpha_2) + x (\alpha_1+2 \alpha_2)+ x^2 (\alpha_2)$ in terms of $S$, we see we need the coefficients $(\alpha_0+2\alpha_1+3\alpha_2, \alpha_1+2 \alpha_2, \alpha_2)^T$.
To get the transformation matrix, we see what happens to the standard basis $(1,0,0)^T,(0,1,0)^T, (0,0,1)^T$. Using the formula above (for example, the first has $\alpha_0 = 1, \alpha_1 = 0, \alpha_2 = 0$) gives
$(1,0,0)^T, (2,1,0)^T,(3,2,1)^T$, hence the transformation matrix is
$\begin{bmatrix}
1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1
\end{bmatrix}$.
