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

1 Answer 1


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 $S$.


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