Find a basis for a subspace of $\mathbb{R}_3[X]$ Let $D = \{P \in \mathbb{R}_3[X] | P'(1) = 0\}$
Find a basis for $D$.
I already figured that $\{1, X^2 -2X , X^3 - 3X\}$ might be a basis, but I'm struggling to prove it. I already showed that it is linear independent, but I don't know how to show that it spans $D$. I also figured that we can rewrite $D$ as:
$D = \{aX^3 + bX^2 + cX + d | 3a + 2b + c = 0\}$
$D = \{aX^3 + bX^2 - (3a + 2b)X + d \quad|\quad a,b,d \in \mathbb{R}\}$
How do I show this set I found spans $D$?
 A: You pretty much have it right there.
$$aX^3 + bX^2 - (3a + 2b)X + d=a(X^3-3X)+b(X^2-2X)+d(1).$$
Right side is the linear combination of your suspected basis set. This shows your set spans $D$.
A: Consider the “standard” basis $\{1,X,X^2,X^3\}$; then $P(X)\mapsto P'(1)$ is a linear map $\mathbb{R}_3[X]\to\mathbb{R}$ and its matrix relative to the standard basis and the basis $\{1\}$ on $\mathbb{R}$ is
$$
[0\;1\;2\;3]
$$
A basis of the null space can be obtained in the usual way:
$$
\begin{bmatrix}1 \\ 0 \\ 0 \\ 0 \end{bmatrix}
\quad
\begin{bmatrix}0 \\ -2 \\ 1 \\ 0 \end{bmatrix}
\quad
\begin{bmatrix}0 \\ -3 \\ 0 \\ 1 \end{bmatrix}
$$
which gives
$$
\{1, -2X+X^2,-3X+X^3\}
$$
as a basis for the kernel of the linear map.
A: Your system of polynomials has rank $3=\dim D$. This ensures the system spans $D$, without having to compute the coefficients.
It has rank $3$ because in the matrix of column vectors:
$\;\begin{bmatrix}1&0&0\\0&-2&-3\\0&1&0\\0&0&1\end{bmatrix}\;$ there is the unit submatrix of dimension $3$ (remove the 2nd row).
Another basis:
Apply *Taylor's formula for polynomials: if $p(x)\in \mathbf R_3[x]$, then
$$p(x)=p(1)+p'(1)(x-1)+\frac{p''(1)}2 (x-1)^2+\frac{p'''(x)}6 (x-1)^3.$$
There results that a basis for $D$ is the set 
$$\bigl\{1, (x-1)^2, (x-1)^3\bigr\}.$$
