I'm trying to solve the following question from Hefferon's Linear Algebra, but I cannot match the solution manual:
The question is: Find the span of each set and then find a basis
We're given this set: $\{2 - 2x, 3 + 4x^2\}$ in $\mathcal{P}_2$
The solution ultimately expresses the span and basis of the set in terms of $\mathcal{P}_2$.
It starts by setting up a linear equation:
$\begin{pmatrix} 2 & 3 | a_1 \\ -2 & 0 | a_2 \\ 0 & 4 | a_3 \\ \end{pmatrix}$
And then does row reduction such that:
$c_1 = -a_1/2$
$c_2 = a_0/3 + a_1/3$
$0 = a_2 - 4(a_0/3 + a_1/3)$
This all makes sense.
What I then cannot figure out is how to get the span that it finds.
It says the span should be:
$\{(−a_1 + (3/4)a_2) + a_1x + a_2x^2| a_1, a_2 \in \mathbb R\}$
I'm confused by this solution. $a_0$ is expressed in terms of $a_1$ and $a_2$. I'm wondering why $a_0$ is chosen to be replaced by $a_1$ and $a_2$.
So I have two questions:
- Why not one of the other parameters?
- What is it called when a basis or span is found in this way in terms of the parameters of the original vector space?
Let $S:=\{2 - 2x, 3 + 4x^2\}$.
Note that $S$ is a linearly independent set. To see this, note that for any $a,b\in \mathbb R$, if $a(2-3x)+b(3+4x^2)=0$
Then, $(2a+3b)-3a x+4bx^2=0$, which holds iff $a=0=b. \quad\square$
To find the span of $S$, note that
$\text{span }S=\{r(2-2x)+s(3+4x^2): r,s\in \mathbb R\}=\{(2r+3s)-2rx+4sx^2: r,s\in \mathbb R\}\tag 1$
Note that the other parameters can be used too. In $(1)$, set for example $2r+3s=u, s=v$ then $2r=u-3v$ and $(1)$ becomes $$\text{span }S=\{u-(u-3v)x+4vx^2: u,v\in \mathbb R\}$$
Since $S$ is a linearly independent set, $S$ is a basis for span $S$.