Question

1. For each of the following inner product spaces V and linear transformations g, find a value of y € V for which g(x) = (x, y), for all x € V. (i) V = P2(R) with f(t)g(t) and g: V+ R defined by g(s) = f'(0) +2f(1). (ii) V = M2x2(C) with the Frobenius inner product, and g: V+C defined by i g(A) =tr :((1141 - :)4).

Answer 1

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Let f(t) = at & bttc let 9: V-R be defined by 964)3 + 0 + 2001) then gend = (x, fd (0)+2801) SX(t) f(t)dt for alt=1 from I, we have 1 2= scat?+be+c) dt 2=29 +20 C+3 C3 (t)- t from ③ we have Again for 1+2= ś tx (at" tbt tc) at 3= 2b 2h b = 9 Again for r(t) = t from 11 we have Stucat tbt+c) dt 0+2.2 - Stre - 2- 2a + 2/3 3a +5 C = 15 gives 4C - - 6 NOW (2 X 3 C: -3/2

a = 3+92 a = 15/2 : FCE)şt + + - Ž i for the above the have g(x) = {4,6) for all ye P, CIR) Here V= M2 x 2 (C) We know frobenius innen product defined by (A,B) = tv (ATB). A,BEV a, a 2 Let EV az au EV E 9(A) = {4,B> 13 NA EV ie 9(1): tv ATB) ty a az ay VC 212 73 ár ani tazaz tāądz tautu Given that iti i 9(A)=.tr a, 92 43 2 = Citila, tiazt az + (1-ilau Let Ok= bk tick Ee 14 = YktiZk k=1,2,3,4 » ak= bk- ick

Then from O ५ 9(A): E (bknick) (Yk tizx) É (Yetize) bet & (-ivktza) ( KET from @ 9(A) = (1+i) (biticilti (bzti (3) + (britics)+(1-1) (buticu) = (1+i)b,+ 1 (1+i), #betic+ ibg +(-1)C3 +(1-1) by +(41) cu from 3 6 W comparing the coefficient be E Ck we get [-- O&O holdy * Clee Civic Hby, CKER) Yi+iz, =(1+1) -4,-2 civ, tz, = (1+i) it has no solution :: klith this inner product, we cannot get any BEV such that geA) = (A,B) AEV. If we define, inner product (A,B) = tv (AT,B) then g(A) = (bktick) (Yxizk) k FE (Uk Hizk) but (ink-Zx) (k ka companing y r we get,

" tiz, =lti in- z, ciliti) =) Y,=.2,=1 11.= (1+i). Yztizz=1 i4₂-22=; Y = 1. 2= then N=1 Yz tizz= 143- 2z = -1 as Y3=0 , 23=1 then ilg=i Yutizu iti iyu - 2y = iti » Yu= 1 Zu=-1 then nu=lni Then B = (6) such that glal <A,B) #ÄEV