Question

Q3. Consider the vector space B, consisting of all polynotninls of degree at most two together with the zero polynomial. Let S = {p(t).p2(t)} be a set of polynomials in P, where P.(t) = -2+3 pa(t) --21-24 + 3 (a) Determine whether the set S = {p(), pea(t)} is linearly independent in Py? Provide a clear justification for your solution. (8 pts) (b) Determine whether the set S = {p(t), pa(t)} spans the vector space B? Provide a clear justification for your solution. (7 pts) (c) Is it possible to find a polynomial ps(t) in P, so that {p.(t), p2(0).pz(t)} is a basis for the vector space P2 (Justify your answer).(5 pts)

Answer 1

Sol - + Given that P, is vector space of all polynomials of degree at most two and the set s={ Pilt), balt)} , where pilt) ++3, Plt) = -27-2+3 Let x and ß be two scalers such that & pilt) + ß (t)=0 → (-++3) + B(-2+2-2t+3) = 0 → 6- X-2B) +-2ßt + 32+38 = 0 1-6-2$)+2-28+ +(3 « +38) = 0f +0.0 +0.1 Equating the coefficients, we have - 2-28=0 ? - 28=0 3X+3B=0 solving above system, we get d = B = 0 Thus, & pilt) + ß belt)=0 x=8=0 Hence the set s = {bilt), g(t)} is linearly independent in Pz

(6) Any spanning Set of P must consists of at least three linearly independent polynomials. Thus the set S={bi(t), p(t)} can not span the vector space P, as it contains only two polynomials. (c) suppose Palt) = 1 and consider the linear combination & pilt) + B Bit) + Xp314) = 0 → X (-++3)+B(-2+-2t+3) + 7.1 = 0 66-2B) ť–2B+ +32 +3B+V = 0.t + o.t +0.1 setting the coefficients equal to o, we have 2-2ß=0 -2.p = 0 34+3 8+8=0 solving above system, we have a=8=8=0