Linear Algebra: Polynomials Basis Consider the polynomials $$p_1(x) = 1 - x^2,\;p_2(x) = x(1-x),\;p_3(x) = x(1+x)$$
Show that $\{p_1(x),\,p_2(x),\,p_3(x)\}$ is a basis for $\Bbb P^2$.
My question is how do you even go about proving that these polynomials are even independent? Are there certain rules I should know?

You need to show two things:
(i) $S = \{p_1, p_2, p_3\}$ form a linearly independent set, and
(ii) $\operatorname{span} S = \mathbb{P}^2$
The first part is relatively easy, let $\vec p_1, \vec p_2, \vec p_3$ be the coefficient vectors of $p_1, p_2$, and $p_3$ respectively. 
Then, $S$ is linearly independent if the system
$$\begin{bmatrix}
\vec p_1 &
\vec p_2 &
\vec p_3 &
| &
\vec 0 \\
\end{bmatrix}$$
has only the trivial solution. Since we are working with a $3$ by $3$ homogeneous system, a clever approach would be to show that the determinant is zero. Otherwise, you may row reduce and proceed as usual.
To show that $\operatorname{span} S = \mathbb{P}^2$, take an arbitrary polynomial $p \in \mathbb{P}^2$ and show that $p \in \operatorname{span} S$. Once again, we let $\vec p$ be the coefficient vector of $p$. It is clear that $p$ is of the form $at^2 + bt + c$. Hence,
$$ \vec p = \begin{bmatrix}
a \\
b \\
c \\
\end{bmatrix}$$
It remains to show that the system
$$ \begin{bmatrix}
\vec p_1 & \vec p_2 & \vec p_3 & \mid & \vec p\\
\end{bmatrix} $$
is consistent (i.e. has at least one solution). I will leave that to you.
Edit: You may also note that $\dim \mathbb{P}^2 = |S|$. Then, you only need to prove one of (i) or (ii) to show that $S$ is a basis for $\mathbb{P}^2$.
