Solution
For an arbitrary polynomial of degree 2,
p(x)=ax2+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={1−x,3−x2,x}.
Then, for the three constants, r, s and t , we can write
p(x)=r(1−x)+s(3−x2)+t(x)ax2+bx+c=r−rx+3s−sx2+txax2+bx+c=−sx2+(−r+t)x+(r+3s)
This gives,
a=−s⇒a=0⋅r−1⋅s+0⋅tb=−r+t⇒b=−1⋅r+0⋅s+1⋅tc=r+3s⇒c=1⋅r+3⋅s+0⋅t
To check that a solution exists, we set up the augmented matrix and row reduce it,
⎣⎡0−11−103010rst⎦⎤
which is row reduced to
⎣⎡1000100013r+t−r3r+s+t⎦⎤
Hence we have
a=3r+tb=−rc=3r+s+t
Therefore, we conclude that Therefore, S is a spanning set for P2
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