Question #305188

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


1
Expert's answer
2022-03-05T07:27:16-0500

Solution


For an arbitrary polynomial of degree 2,


p(x)=ax2+bx+cp(x)=ax^2+bx+c


To prove that S is a spanning set, we need to show that p(x)p(x) can be written as linear combination of the elements of S={1x,3x2,x}S=\left \{ 1-x, 3-x^2,x \right \}.


Then, for the three constants, rr, ss and tt , we can write


p(x)=r(1x)+s(3x2)+t(x)ax2+bx+c=rrx+3ssx2+txax2+bx+c=sx2+(r+t)x+(r+3s)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=sa=0r1s+0tb=r+tb=1r+0s+1tc=r+3sc=1r+3s+0ta = - 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, 


[010r101s130t]\begin{bmatrix} 0 & -1 & 0 & r\\ -1 & 0 & 1 & s\\ 1 & 3 & 0 & t \end{bmatrix}


which is row reduced to


[1003r+t010r0013r+s+t]\begin{bmatrix} 1 & 0 & 0 & 3r+t\\ 0 & 1 & 0 & -r\\ 0 & 0 & 1 & 3r+s+t \end{bmatrix}


Hence we have


a=3r+tb=rc=3r+s+ta=3r+t\\ b=-r\\ c=3r+s+t


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




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