Finding a Basis in $\mathbb{P}_2$ doesn't make sense I am having trouble understanding creating a basis in $\mathbb{P}_2$. Maybe I don't completely understand what a basis is, but I'm trying to review for a test and I've come across a problem that I do not understand at all. It says

Give an example of a basis $\mathcal{B}= \left\{\vec{p_1},\vec{p_2},\vec{p_3}\right\}$ of $\mathbb{P}_2$ such that $\left[t^2\right]_\mathcal{B} = (1,1,0)^T$ 

So of course this "function" is in the form $ax^2+bx+c$. To me it seems like $t^2$ needs to be in the first 2 terms of the basis, so I would come up with something like $t^2+t,t^2-t,1$. But the answer they've provided is $\{1+t, -1-t+t^2,1\}$ and I don't understand why there isn't a $t^2$ in the first term and also why everything has to cancel out except the 1 in the last term? It doesn't seem like $t^2$ would be $(1,1,0)$ if there is not a $t^2$ term in the first term of the basis if that makes sense.
 A: There are infinitely many correct answers here.
What is necessary is that, first, $\mathcal{B}=\{b_1,b_2,b_3\}$ is a basis, meaning that the three basis vectors in it are linearly independent and together they span the entirety of $\mathbb{P}_2$.
Second, we wanted $[t^2]_{\mathcal{B}} = 1\cdot b_1+ 1\cdot b_2 + 0\cdot b_3$.  That is to say, we want $b_1+b_2 = t^2$.
We could have chosen literally anything for $\mathcal{B}$ so long as it satisfies those requirements.  We can be creative in our approaches as well.
For example, we could have had $\mathcal{B} = \{1+42t^2, -1 - 41t^2, t\}$.  We could have just as well had $\mathcal{B}=\{t^2+t, -t, 1\}$.
Note, your answer is slightly off as the sum of your first two terms does not result in $t^2$.  You had $(t^2+t)+(t^2-t) = 2t^2$.  Had you answered instead $\{\frac{1}{2}t^2+t, \frac{1}{2}t^2-t,1\}$ that would have been just as correct as the provided solution.
A: Suppose, for instance, that $B=\{t^2-1,1,t\}$. Then $B$ is a basis of $P_2$ and$$t^2=1\times(t^2-1)+1\times1+0\times t.$$So, $[t^2]_B=(1,1,0)$.
This is not the only answer, of course. Another possibility is, say, $B=\left\{\frac12t^2+t,\frac12t^2-t,1\right\}$.
