Answer to Question #305188 in Linear Algebra for misfit

Question

Show that the set {1 − 𝑥, 3 − 𝑥 2 , 𝑥} spans 𝑃2

Answer 1

Solution

For an arbitrary polynomial of degree 2,

$p(x)=ax^2+bx+c$

To prove that S is a spanning set, we need to show that $p(x)$ can be written as linear combination of the elements of $S=\left \{ 1-x, 3-x^2,x \right \}$.

Then, for the three constants, $r$, $s$ and $t$ , we can write

$p\left( x \right) = r\left( {1 - x} \right) + s\left( {3 - {x^2}} \right) + t\left( x \right)\\ a{x^2} + bx + c = r - rx + 3s - s{x^2} + tx\\ a{x^2} + bx + c = - s{x^2} + \left( { - r + t} \right)x + \left( {r + 3s} \right)$

This gives,

$a = - s \Rightarrow a= 0 \cdot r - 1 \cdot s + 0 \cdot t\\ b = - r + t \Rightarrow b= - 1 \cdot r + 0 \cdot s + 1 \cdot t\\ c = r + 3s \Rightarrow c = 1 \cdot r + 3 \cdot s + 0 \cdot t\\$

To check that a solution exists, we set up the augmented matrix and row reduce it, 

$\begin{bmatrix} 0 & -1 & 0 & r\\ -1 & 0 & 1 & s\\ 1 & 3 & 0 & t \end{bmatrix}$

which is row reduced to

$\begin{bmatrix} 1 & 0 & 0 & 3r+t\\ 0 & 1 & 0 & -r\\ 0 & 0 & 1 & 3r+s+t \end{bmatrix}$

Hence we have

$a=3r+t\\ b=-r\\ c=3r+s+t$

Therefore, we conclude that Therefore, S is a spanning set for P2