How to express the span and basis in terms of a vector space

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?
• By $\mathcal{P}_2$, do you mean vector space of all polynomials over the field of real numbers $\mathbb R$?
– Koro
Mar 12, 2022 at 4:20
• @Koro Yes, I wasn't sure how to write it in latex. Mar 12, 2022 at 11:46

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 $$S$$.