Find bases for subspaces spanned by vectors. The standard basis for $P_2(\mathbb  R)$, the vector space of quadratic polynomials of the form $ax^2+bx+c$ is the set $S=\{1,x,x^2\}$. Find bases for the subspaces of $P_2(\mathbb R)$ spanned by the following set of vectors: 
a) {$-1+x-2x^2,3+3x+6x^2,9$}
b) {$1+x,x^2,-2+2x^2,-3x$}
I really have a hard time grasping basic linear algebra so I hope someone can guide me through this question?
 A: The knowledge of a basis of a vector space $V$ allows us to reduce computations in $V$ to computations on matrices. If $B=\{v_1,\dots,v_n\}$ is a basis for $V$, we can define a map $C_B\colon V\to\mathbb{R}^n$ by setting
$$
C_B(v)=\begin{bmatrix}\alpha_1 \\ \vdots \\ \alpha_n\end{bmatrix}
\text{ if and only if }
v=\alpha_1v_1+\dots+\alpha_nv_n
$$
The map $C_B$ is linear and bijective, so a set $\{w_1,\dots,w_m\}$ of vectors in $V$ is linearly independent if and only if the set
$$
\{C_B(w_1),\dots,C_B(w_m)\}
$$
is linearly independent in $\mathbb{R}^n$. But for this we have Gaussian elimination and row echelon forms.
So, we take $B=\{1,x,x^2\}$, so that the set of vectors in $\mathbb{R}^3$ corresponding to $\{-1+x-2x^2,3+3x+6x^2,9\}$ is
$$
\left\{
\begin{bmatrix}-1\\1\\-2\end{bmatrix}\,,
\begin{bmatrix}3\\3\\6\end{bmatrix}\,,
\begin{bmatrix}9\\0\\0\end{bmatrix}\,
\right\}
$$
and you just need to find the row echelon form of the matrix
$$
\begin{bmatrix}
-1 & 3 & 9 \\
1 & 3 & 0 \\
-2 & 6 & 0
\end{bmatrix}
$$
Without pivot reduction, it turns out to be
$$
\begin{bmatrix}
-1 & 3 & 9 \\
0 & 6 & 9 \\
0 & 0 & 18
\end{bmatrix}
$$
which allows you to say that the first set is linearly independent.
The matrix to consider for case b is
$$
\begin{bmatrix}
1&0&-2&0\\
1&0&0&-3\\
0&1&2&0
\end{bmatrix}
$$
You'll easily find out that the fourth column is a linear combination of the first three, which are a linearly independent set.
A: I'll do for (a), you do for (b): suppose
$$a\cdot(-1+x-2x^2)+b\cdot (3+3x+6x^2)+c\cdot9=0\;,\;\;a,b,c\in\Bbb R\implies$$
$$=-a+3b+9c+\left(a+3b\right)x+\left(-2a+b\right)x^2$$
Since you know $\;\{1,x,x^2\}\;$ is a basis, the above trivial linear combination is possible iff
$$\begin{cases}-a&+3b+9c&=&0\\{}\\\;\;a&+3b&=&0\\{}\\\!\!\!-2a&+b&=&0\end{cases}$$
Solving the above linear system, one gets:
$$2II+III=0\implies7b=0\implies b=0\implies a=0\implies c=0$$
and thus the given set's lin. independent.
