Let $B = ${$1-t^2, t-t^2, 2-2t+t^2$}. Check that $B$ is a basis for $P_2$ and find $[3+t-6t^2]_B$ Let $B = ${$1-t^2, t-t^2, 2-2t+t^2$}. Check that $B$ is a basis for $P_2$ and find $[3+t-6t^2]_B$
In mathematical notation, what exactly does $[3+t-6t^2]_B$ mean? That $[3+t-6t^2]$ is in the subspace of $B$?
 A: To check that a set $B=\{v_1,...,v_n\}\subset V$ is a basis of $V$, you have to prove two things:
(1) The vectors of $B$ are lineraly independent, i.e. if $\sum_{i=1}^na_iv_i=0$ then $a_1=...a_n=0$.
(2) The vectors of $B$ span $V$, i.e. for all $v\in V$ there exist $a_1,...,a_n$ such that $\sum_{i=1}^na_iv_i=v$.
If you know that $\dim V=n$ then (1) and (2) are equivalent, so you can prove either one of them.
Here $V=P_2=\mathbb{R}_{\leq2}[t]$ (this is the more standart notation for the space of polynomial with real coefficients
of degree at most 2). To check that $p_1,p_2,p_3$ are lineraly independent, just expand $\sum_{i=1}^3a_iv_i=0$ and check
that this really implies that $a_1=a_2=a_3=0$.
The notation $[3+t-6t^2]_B$ means the coordinates of $3+t-6t^2$ w.r.t basis $B$. In other words, if $B=(p_1,p_2,p_3)$ then $[3+t-6t^2]_B=\left(\begin{array}{c}a_1\\a_2\\a_3\end{array}\right)$ such that $a_1p_1+a_2p_2+a_3p_3=3+t-6t^2$
A: You know that $\{1,t,t^2\}$ is a basis of $\mathbb P_2$ by construction, since all the vectors in $\mathbb P_2$ are assumed to be linear combinations of those $3$ (i.e. the set $\{1,t,t^2\}$ generates $\mathbb P_2$) and of course if $a_1 + a_2 t + a_3 t^2 = 0$ then all the coefficients vanish (i.e. $\{1,t,t^2\}$ is a linearly independent set). You don't need to do this work, but it is a useful thing to know that $\mathbb P_n$ has dimension $n+1$ in general (in this case, that $\mathbb P_2$ has dimension $3$).
Now since $\mathbb P_2$ has dimension $3$, all you have to prove to show that $\{1 - t^2, t - t^2, 2 - 2t + t^2\}$ is a basis of $\mathbb P_2$ is that it is linearly independent, OR that it is generating. Once you have shown one, the dimension of $\mathbb P_2$ will imply that this set is a basis of $\mathbb P_2$. To show that it is linearly independent, you can show that if 
$$
a_1(1-t^2) + a_2(t-t^2) + a_3(2-2t+t^2) = 0,
$$
then $a_1 = a_2 = a_3 = 0$. If you do the math, 
$$
a_1(1-t^2) + a_2(t-t^2) + a_3(2-2t+t^2) = (a_1 + 2 a_3) 1  + (a_2 - 2a_3) t + (-a_1 - a_2 + a_3) t^2.
$$
Showing that this polynomial is always zero amounts to solving the system of linear equations
$$
-a_1 - a_2 + a_3 = 0, \quad a_2 - 2a_3 = 0, \quad a_1 + 2 a_3 = 0.
$$
and proving that the unique solution is the trivial one. I leave this work to you, just switch to the matrix form of the system and find the row-echelon form.
If you wished to show that the set was generating instead, take an arbitrary polynomial $p(t) = b_0 + b_1 t + b_2 t^2 \in \mathbb P_2$. You need to show that there exists $x_1, x_2, x_3 \in \mathbb R$ such that 
$$
x_1(1-t^2) + x_2(t-t^2) + x_3(2-2t+t^2) = p(t) = b_0 + b_1 t + b_2 t^2, 
$$
i.e. that the system 
$$
x_1 + 2x_3 = b_0, \quad x_2 - 2x_3 = b_1, \quad -x_1 - x_2 +x_3 = b_2
$$
has a solution. Again, to show this, switch to matrix form and find the row-echelon form.
The last question you have been asked, i.e. what is $[3 + t - 6t^2]_B$, actually only demands that you solve the system 
$$
x_1 + 2x_3 = 3, \quad x_2 - 2x_3 = 1, \quad -x_1 - x_2 +x_3 = -6
$$
so that you can write $3+t -6t^2 = x_1(1-t^2) + x_2(t-t^2) + x_3(2-2t+t^2)$, i.e.
$$
[3 + t - 6t^2]_B = 
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}.
$$
All this notation says is "what are the coefficients of $p(t)$ when expressed as a linear combination of the elements of the basis $B$". Since the coefficients are uniquely determined when $B$ is a basis, you can just find the coefficients and put them in column vector form.
Hope that helps,
