Basis and change of co-ordinate matrix for $\mathbb P_n$ 
a) Find a basis for $\mathbb P_2$ such that $$\begin{bmatrix}p(0) \\ p(1) \\ p(2) \end{bmatrix}$$ are the co-ordinates of any polynomial $p$ with respect to that basis.
b) Find a basis for $\mathbb P_2$ such that $$\begin{bmatrix}p(0)\\p'(0)\\p''(0)\end{bmatrix}$$ are the co-ordinates of any polynomial $p$ with respect to that basis.
c) Find the co-ordinates of $p(x) = x^2$ in both these co-ordinate systems.
d)Find the change of co-ordinate matrix between these co-ordinate systems. That is, find a matrix $A$ such that $$A\begin{bmatrix}p(0) \\ p(1) \\ p(2) \end{bmatrix} = \begin{bmatrix}p(0)\\p'(0)\\p''(0)\end{bmatrix}$$ For any $p$ in $\mathbb P_2$.

For a) I thought of the bases $a_0,a_0 +a_1 +a_2,a_0 +2a_1 +4a_2$
for b) I though of the bases $a_0,a_1+2a_2 x,2a_2$
Not sure how to find the co-ordinates of $p(x) = x^2$ in these co-ordinate systems though (I suspect my bases are wrong). Also not sure how to do part d).
 A: Hint:
This kind of problems are easily solved remembering that:

the components of a vector with respect to  a basis are the inner products of  the vector with the corresponding element of the basis.

This implies that, if $p(x)=p_0+p_1x+p_2x^2$ is the vector $\vec p$ that in the standard basis has  components $\vec p=(p_0,p_1,p_2)^T$, and we want a basis where the components are $(p(0),p(1),p(2))^T=(p_0,p_0+p_1+p_2,p_0+2p_1+4p_2)^T$ we have to find the vectors (of the basis) $\{\vec u,\vec v, \vec w\}$ such that:
$$
\langle \vec u,\vec p\rangle=p_0 \qquad \langle \vec v,\vec p\rangle=p_0+p_1+p_2 \quad \langle \vec w,\vec p\rangle=p_0+2p_1+4p_2
$$
can you see what the vectors (and polynomials) $\{\vec u,\vec v, \vec w\}$ are in the standard basis? 
And in a similari way you can solve the other problem.
A: Each of the elements in a basis for $P_2$ must be a member of $P_2$ i.e. a quadratic polynomial.
For part (a), the $p(0)$ element of the basis should be a quadratic polynomial such that $p(0) = 1$ and $p(1)=p(2)=0$. $(x-1)(x-2)$ satisfies the second condition, and to satisfy the first condition we can divide by two, so the first element of the basis is $\frac{(x-1)(x-2)}{2}$. You can use a similar method to find the other two basis elements.
The co-ordinates of $x^2$ in the part (a) basis are just the values of $p(x)=x^2$ at 0, 1 and 2 i.e. $(0,1,4)$.
For part (b) the basis is $(1, x, \frac{1}{2}x^2)$. Note that the factor of $\frac{1}{2}$ is required so that $p''(0)=1$. The co-ordinates of $x^2$ in this basis are then $(0,0,2)$.
For part (d) you can form each column of the 3x3 matrix A by finding the co-ordinates of each of the basis elements from part (a) relative to the part (b) basis. For example, the co-ordinates of $p(x)=\frac{(x-1)(x-2)}{2}$ relative to the part (b) basis are $(p(0),p'(0),p''(0)) = (1,\frac{-3}{2},1)$, so the first column of A is $(1,\frac{-3}{2},1)$.
You can validate your result for part (d) by checking that your A matrix correctly converts the co-ordinates of $x^2$ (from part (c)) from one basis to the other.
