Question

How to solve all of this linear Algebra

8. (24 points total) LetV be the vector space{P2, +, *}with standard function addition and scalar multiplication Define an Inner product: <p | q>= p(0)q[O) + p(1)q(1)+ p(2)q(2). Let B = {x,x,1} a. Explain why this inner product satisfies the positive property b. Explain how you know that B forms a basis c. State the conclusions of Cauchy-Schwartz and the Triangle inequalities in terms of this inner product d. Use Gram-Schmidt and B (using the vectors in that order) to produce an orthonormal basis for P2, e. Find the angle (leave your answer in terms of the inverse cosine) between f(x) 2x2 -1 and g(x) = x2 + x + 1. Is it acute or obtuse? f. What is the orthogonal complement to Span({x2, x})?

Answer 1

I will be assuming the underlaying field to be the set of real numbers (5th part seems more sensible with this assumption) T dealing with P2, so the all the polynomials are of a. Note that we are SO degree less then 2. Therefore they can not have more than 2 roots, so if p(x)E P2, is not 0. Let p(a) 0 where a is one of the numbers from 0,1,2}. This gives the given inner product satisfies the positive property SO we know that atleast one of the numbers, p(1), p(2), p(0) plp= (p(0))+(p(1))+(p(2)) (p(a) > 0. Therefore us b. If p(r) E P2, then p(r) definition) Hence B spans P2. Now if aa a1ao a2, a1, ao E R then a2 all its coefficents 0(again by definition). Hence B is a definition! a2 a1 ao for some a, a1, ao E R(by 0 for some = 0 since the zero polynomial has basis set(just by

Cauchy Swartz: C. 2 (dlon6) i) i-0 i-0 i-0 lPll for (<p p) Triangle Inequality: We use l|g|| llpll -(r)(r)wr) 2 ((P) (P+g)) i-0 i-0 i-0 d. First we put po(r) element we evaluate q () ( | Po(a)>)Po(x) we normalize 1(x) to get pi (x) and we put p (r) into our set Now evaluate g(r) 1-(1| Po(r) >Po(r) (1 Pi()> V17). To add the second 172Now in the set(2 9 17 5 37 2 and orthonormalize 34 17 17 it to get p2() 2(r). Therefore {po(), pi(r), p2(x)} form the orthonormal basis for P2

I will now solve part f) here itself utilizing the calcuations of this part. Note that the spanfx, x2} spantpo(r), p ()}. But we span{po(), pi(r)}, it is simply sparn{p2(x) as it can be done directly know the orthogonal complement of 51, lgll flg>= 51 and |f|| V59, giving us the angle e, < between them 51 0, the angle between 1 COS Since cos() 59 them is acute