How to find a basis of a vector space? Let $P_4(R)$ denote the set of all polynomials with degree at most 4 and coefficients in $R$. I was attempting to find a basis of $U=\{p\in P_4(R): p''(6)=0\}$. I can find one by taking the most basic approach. Basically start with $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$. Then differentiate this polynomial twice and factor the differentiated version so that one of its root is 6. Then integrate the factored version twice and get the general description of an element of $U$. Then write that in terms of the standard basis of $P_4(R)$ and choose the free variables and construct a basis of $U$.
My question, is there a clever way to find a basis of a given vector space? How could one simplify this process of finding a basis?
 A: $1, x-6, (x-6)^2, (x-6)^3, (x-6)^4$ forms a basis of $P_4(\mathbb R)$, as we can find the Taylor expansion of any polynomial around $6$.
For $p(x)=\sum_{n=0}^4 a_n(x-6)^n$, $p''(6)=0$ is equivalent to $a_2=0$. Now it should be clear that a basis is given by $1, x-6, (x-6)^3, (x-6)^4$
A: Just because you called what you did “the most basic approach”, let me add what I think is the most basic approach. (As opposed to for example the more elegant approach in the first answer.)
$U$ is the kernel of the linear map $p\mapsto p’’(6)$. If you write $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ you express $p$ in terms of the basis of $P_4(R)$ that consists of the monomials. Now $p’’(6)=0$ becomes
$$2a_2+ 36a_3 + 432a_4 = 0.
$$
The solution space of this is a four dimensional sub space of $R^5$ (we do not forget $a_0$ and $a_1$). You write down a basis of this in the way that you have learnt and then use the corresponding elements of $P_4(R)$ as a basis of $U$.
This is the solution that you should be able to come up with without thinking. It is always good to find a clever solution (and looking for a convenient basis is definitely something that one should do), but I also find it important to know when it is not necessary to be clever.
Btw, I suppose that $R=\mathbb R$, otherwise the solution space is only four dimensional if $2\ne 0$.
