Question

Q3 14 Points Consider the vector space P2(R). Let T1, T2, T3 be 3 distinct real numbers and 21, 22, az be three strictly positive real numbers. Define (p(x), q(x)) = Li_1 Qip(ri)q(ri) Q3.1 5 Points Show that this P2 (R) together with (-:-) is an inner product space. Please select file(s) Select file(s) Save Answer Q3.2 2 Points Give a counter example that (-, - ) is not an inner product when T1, 12, 13 are still distinct real numbers, both a2, az > 0 but aj < 0. r1 = Enter your answer here 12 = Enter your answer here 7'3 = Enter your answer here ai = Enter your answer here a2 = Enter your answer here a3 = Enter your answer here Show/Explain why it is a counter example. (1pt) Enter your answer here Save Answer

Answer 1

If you have any doubts in the solution please ask me in comments here i use basic definition of innner product space

* Solution- consider vector Space V- P (IR); P2 (R) till degree 2, set whose weff is polynomial of are from IR. PO ":81,82,83 EIR € 8,82,83 are distinct; a,, 92, 93 > 0 EIR, 3 < P(x), qcx)) 5 di Pri)q171). 0 i=) 3.1 TO Show P₂(IR) together with Saoud is inner product space. V is vector space then <-,-) is inner product space at <-,-) satisfies following propesties, iff c) < u.tv, W) = <u, W> + <V, w) # U, V, W EU. N QEF <2u, W) = 2<u,w> <u, W) HU, WE U. < w.us WVEVO iff V=0, V.V>=0 u. <viv> >O (N) * properties : 1, Let P(x), 4(x), M(x) E P2 (IR) then < RAV P(x) + mix), 4(x)) = < P+m)2, 21x)), OTT num

T < P(1) m (x) ) %(x)> ะ IMO Q, (p4 m) (71) 9 (ช) Q: ( p(r) + m(vi)) % (2) ป. Q: P ri) () + 2 9, m(r) 9 () เE | (=) - < P(x), (x)) + ((x), 9(x)). < P(X) + m(x) , 4 (x)) - < P(x), 9(8)) + (m(x) , 9(x)>. 3 Properties : 2; ": ๔e IR 2 x P(x), 9 (x) > 2 at (AP) (21) 9 (0) . I aix Plri)q(71) - 2 S Q: (P(70) (9() x < P(x) , 9(x)) < < P(x)), 9 (x)) << P(x) , (x)). properties :3; 2, e P, (IR) 3 3 < P(), 9 (x)) Qu P(7) 9() - > Q:9 (r) P(1) พ. (Iris an abelian group), < 9 (x), p(x)) ai 267) Pori) [= < P(x), 9 (8) > > < ดู tz), p(4) > properties :4, y p e 9, (IR) 3 3 <P (x), P(x)) - > Q; P (2) P(2j) = E ai ( Proi) 2. (-) (=) Vi= , 2, 3 , ai to เ = 1, 2, 3 ": (P(r) 30 ร่ ๔ (Prr.) > 0 As >0 [ 1 - 1, 2, 3. S <P(x), P(x) > > 0 ,

Now <P(x), P(x)) = 0 E ai (Ploid) = 0 3 جے ¿ Qi ( PIV:))2 = 0 (P(703= 0 Hi=1,2,3 i=1 *; ai >o. So 81,82,83 P(x). sroots are of Pa) EP2 (IR) : 713233 distinct are a SO that only possiblity two degree. polynomial P(X) 3 distinct has rest iff P(x) = 0 P(x) = 0 VXER. iff P(x) -- <pra), pax)) By ③ , ③, 4 is - ai Ploi) qiri) < Plze), q12) > inner product space for P2 (IR)- i=1 Hence powove: 3.2 solution: To show <-,-) is Not inner product space: when 2,082,83 EIR { distinct , az, az > 0 a,<o > Let P(x) = x2, = 1. 7, inz, da distinct ) 72 = 73 o s 7 = -1, P(72) = 0 ; P103) = 1 P(0,) = 1, 93 -2, 12, Qz = d. a, Now assume 3 ai (P(+1) Now < P(x), P(x) >= i- < P(X), P(x)) = < x2, x2) = 9, (P(7))) + q (P(2))2 + 2y (P(73))2

K x², «²> -12 (0)2 + 1 (0)2 + 2(1)3 - - 10 <o > <P(x), PW > < 0 V pia) e V = P2 (IR) which that the property contradicts < P(V), P(x) > >o. <-,-) is not inner product Space. o Hence proove > 31 = -1 Answer а, - 12 i Answer Answer 1 72 Answer Az O = 2 - Answer 1 j Az Auswer 73 P(x) Here P(x) = x? E Pz(IR) Because x² - is counter example 2 coeff. 1 EIR. where x² has degree 22 € P (IR). for x² <-,-) is not innes certain choice of ai, si product Space is x² P(X) = counter example. Ansurer