Let V=P2(R) be the vector space of all polynomials of degree at most 2. Consider
ID: 2962133 • Letter: L
Question
Let V=P2(R) be the vector space of all polynomials of degree at most 2. Consider the elements:p1(x)=x-1, p2(x)=(x^2)-1, p3(x)=(x^2)-x, p4(x)=1
Which of the following lists are spanning sets for v? Explain why or why not.
a.) (p1(x))
b.) (p1(x), p2(x))
c.) (p1(x), p2(x), p3(x))
d.) (p1(x), p2(x), p3(x), p4(x))
I was told that the answer was that a, b, and c are not spanning sets of V, but d is a spanning set of V. Why would this be and can you show me why?
And I thought that since a and b are linearly independent that would mean that they are a spanning set of V?
Thank you for any feedback! Let V=P2(R) be the vector space of all polynomials of degree at most 2. Consider the elements:
p1(x)=x-1, p2(x)=(x^2)-1, p3(x)=(x^2)-x, p4(x)=1
Which of the following lists are spanning sets for v? Explain why or why not.
a.) (p1(x))
b.) (p1(x), p2(x))
c.) (p1(x), p2(x), p3(x))
d.) (p1(x), p2(x), p3(x), p4(x))
I was told that the answer was that a, b, and c are not spanning sets of V, but d is a spanning set of V. Why would this be and can you show me why?
And I thought that since a and b are linearly independent that would mean that they are a spanning set of V?
Thank you for any feedback!
Explanation / Answer
take the polynomial P(x) = 1 which can not be written in terms of P1(x) eliminating option a
take the polynomial P(x) = x^2+x = a(x^2-1)+b(x-1)
=> equating coeeficents gives a =1, b = 1, 0 = -2 => x^2+x is not in the span of {p1(x),p2(x)} eliminating option A
take the polynomial P(x) = 1 = p(x-1)+q(x^2-1)+r(x^2-x)
=>
equating coeeficents gives
q+r =0, p-r =0,1= -p-q, adding the three equation gives 0 = 1 => no solution =>
p(x)= 1 is not in the span of {p1(x),p2(x),p3(x)} eliminating option C
let p(x) = ax^2+bx+c = p(x-1)+q((x^2)-1)+r((x^2)-x)+s(1)
=>
a = q+r, b = p-r,c = -p-q+s
3 equations in 4 unknowns => more than 1 solution => P(x) is in the span of {p1(x),p2(x),p3(x),p4(x)}
=>
answer is option D
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