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**Deveno** presumably, you mean $\displaystyle B = \{1,x,x^2\}$, that is, $\displaystyle p_0(x) = 1, p_1(x) = x, p_2(x) = x^2$. suppose we have a vector in $\displaystyle P_2(\mathbb{R})$ ginve in C-coordinates, that is:

$\displaystyle v = c_1h_1 + c_2h_2 + c_3h_3$ or $\displaystyle v = [c_1,c_2,c_3]_C$. in other words,

$\displaystyle h_1 = [1,0,0]_C,h_2 = [0,1,0]_C,h_3 = [0,0,1]_C$.

now if we had such a matrix A. then:

$\displaystyle A([h_i]_C)$ would be its i-th column. but that would just be the i-th basis vector of C in B-coordinates. so, for example,

$\displaystyle A([h_1]_C) = [h_1]_B = [3,2,7]_B$, since:

$\displaystyle h_1(x) = 3 + 2x + 7x^2 = 3p_0(x) + 2p_1(x) + 7p_2(x)$

convince yourself that a linear combination of the basis vectors in C is taken to the same linear combination of the images of the basis vectors in B-coordinates, because a matrix is a linear mapping.