Let $W= \text{span}(p_{1},p_{2},p_{3}), W \subseteq R_{3}[x]$
$p_{1}(x)= 4x^2+3x^3$
$p_{2}(x)= 1+2x^2+3x^3$
$p_{3}(x) = 3-2x^2+3x^3$
From $p_{1},p_{2},p_{3}$ choose a basis $B$ for $W$ and calculate the coordinates of $p_1,p_2,p_3$ of the basis $B$.
$p_1$ alone is linearly independent. $(p_1,p_2)$ are linearly independent too because they aren't scalar multiple of each other.
$(p_1,p_2,p_3)$ is linearly dependent because $p_3= 3p_1-2p_2$
So basis $B= \left\{ p_1,p_2 \right\}$
(I got this info from my previous question and it should be correct till here.)
But how to calculate the coordinates? I don't even understand what we are looking for, why in combination with the basis?
Anyway, I would only look for coordinates of $p_1,p_2$ because these belong to the basis.
So $p_1(x)=0$
$0=4x^2+3x^3 \Leftrightarrow 0=x^2(4+3x)$
$x_1=0, x_2=-\frac{4}{3}$
And do the same for $p_2(x)$..?
What would the coordinates be then? I think I did it wrong :(
You can look for coordinates of any vector in the span of the basis (i.e. in the subspace). If $B = \{p_1,p_2\}$ is a basis, then finding coordinates of a polynomial $p$ means finding $a_1, a_2$ such that $$a_1p_1 + a_2p_2 = p$$ The coordinate vector is then ${a_1 \choose a_2}$
From inspection, you can see that the coordinate vector of $p_1$ is ${1 \choose 0}$ and the coordinate vector of $p_2$ is ${0 \choose 1}$. To find the coordinate vector of $p_3$, find the weights of the linear combination of $p_1$ and $p_2$ that gives you $p$.