Linear Algebra: Determine if the set of polynomials span P2 The polynomial given is { ${1 - 2x + x^2, 2 - 3x, x^2 + 1, 2x^2 + x}$}
What I do not understand is, how am I going to solve the augmented matrix if it only spans to P2 (meaning the highest power will be 2?) but I am given 4 equations.
 A: It's clear that the system $\{x^2,x,1\}$ generates $P2$, since
$$P2 = \{ax^2+bx+c\ |\ a,b,c\in\Bbb{R}\} = span(x^2,x,1).$$
So if we can show that the system
$$\begin{align}&\text{I. } &1 - &2x &+ x^2 \\ &\text{II. } &2 - &3x & \\ &\text{III. } &1\ \ \  &\ \  + &x^2 \\ &\text{IV. } & &x+ &2x^2\end{align}$$
generates $\{x^2,x,1\}$, then we are done.
First of all, it's easy to see that $\text{II.} = \text{I.} + \text{III.} - \text{IV.}$, so we only need $\text{I., III. }$ and $\text{IV.}$ to generate $\{x^2,x,1\}$, and we can use Gauss-elimination to show this:
$$\begin{matrix}\text{I.} \\ \text{III.} \\ \text{IV.}\end{matrix} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &1 &0 &1 \\ &0 &1 &2 \end{bmatrix}} \stackrel{r_2-r_1}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &2 &0 \\ &0 &1 &2 \end{bmatrix}} \stackrel{r_2/2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &1 &0 \\ &0 &1 &2 \end{bmatrix}}  \longrightarrow \\ \stackrel{r_3-r_2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &1 &0 \\ &0 &0 &2 \end{bmatrix}} \stackrel{r_3/2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &-2 &1 \\ &0 &1 &0 \\ &0 &0 &1 \end{bmatrix}} \stackrel{r_1+2r_2}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &0 &1 \\ &0 &1 &0 \\ &0 &0 &1 \end{bmatrix}} \stackrel{r_1-r3}{\longrightarrow} \stackrel{\begin{matrix}\ \ &1\ \ &x\ \ &x^2\ \ \end{matrix}}{\begin{bmatrix}&1 &0 &0 \\ &0 &1 &0 \\ &0 &0 &1 \end{bmatrix}}$$
Following the steps that we took (backwards):
$$\begin{align}x^2 &= \frac{2x^2}{2} = \frac{(2x^2+x)-(x)}{2} = \frac{(2x^2+x)-(2x/2)}{2} = \\ &= \frac{(2x^2+x)-((1-2x+x^2)-(1+x^2))/2}{2} = \\ &= \frac{\text{IV.}-(\text{I.}-\text{III.})/2}{2} = \frac{\text{IV.}}{2} - \frac{\text{I.}}{4} + \frac{\text{III.}}{4}\end{align}$$

$$\begin{align}x &= \frac{2x}{2} = \frac{(1-2x+x^2)-(1+x^2)}{2} = \frac{\text{I.}}{2} - \frac{\text{III.}}{2}\end{align}$$

$$\begin{align}1 &= (1-2x+x^2)+2(x)-(x^2) = \\ &= \text{I.}+2\left(\frac{\text{I.}}{2} - \frac{\text{III.}}{2}\right)-\left(\frac{\text{IV.}}{2} - \frac{\text{I.}}{4} + \frac{\text{III.}}{4}\right) = \\ &= \frac{9}{4}\text{I.}-\frac{5}{4}\text{III.}-\frac{1}{2}\text{IV.}\end{align}$$
So we can express $\{x^2,x,1\}$ out of these, meaning that we could exress any $p \in P2$ with linear combinations of the given polynomials.
