ST UDEN T EXAMINATION SERIES XAM NOTE © SOLVED QUESTION PAPER SOLVED QUESTION PAPER-July, 2022 CH. DEVI LAL UNIVERSITY, SIRSA B.A,/B.Sc. 3rd Year (Semester-6) « MATHEMATICS LINEAR ALGEBRA IPAPERAI] (Max. Marks: B.Sc. 40; B.A. 26 2: Attempt five questions in all, selecting one question from each section. Question No. 1 is compulsory. i ~*~ +, [COMPULSORY QUESTION fw that the function T : R* — R® given by T(x, y) = (x - y, 2y - x, -x) is a linear transformation. ol. Here, wehave Pry) = (x-y, 2y-x,-x) Let foe wa by yy) Vey yeR . T(u+0) = Tey +X, + ¥p) (4 + y= Ya — Yor 24 + 2Yg— Hy — Hy —¥D) (4 Yi +p — Yor 2Y4 = %y + 2Yp Fy —%1 — 2) = (> Yor 2 ~ By -¥) + Cy Yay 2a — Xa» —¥) Pet yn) + PCY) = T(u) + T(r). 1.(@) Also, for ae Rand u ¢.R? Tau) = Tax, yy) (ax, a¥4) (ax, — ayy, 2ay, ~ ax, ~ax;) (= Ye 2M, =H) = aT (x,y) =aT(u) Tisa linear transformation. res linear transformation T: R° > R*, whose null space is generated by (0, 1, -3), 0, 4). 1. Let N( be the null space of T generated by (0, 1, ~3) and (0,-3, 4) Also, these vectors are not scalar multiple of each other. :. These vectors are linearly independent. We extend this set to $ = {(0, 1,-3), (0, -3, 4), (1,0, O)} It can be verified that $ is linearly independent set, so Now, _ T(0,1,-3) = (0, 0,0,0) and T(0,-3, 4) = (0,0, 0,0) [ Jury 202: 155 i ) it forms a basis of R°. (0, 1,-3) and (0, -3, 4) « N(T)] ‘Marnemarics (Sem. 67H) Sotvep Quesrion PaPen:a But (1, 0, 0) € N(T), so it is not a linear combination of (0, 1, -3) and (0, 3,4) Let . T(1, 0,0) = (1,0,0,0) If (x, y, 2) © R*is arbitrary, th (x, y,2) = a(0, 1, -3) 1b, -3, 4) 4 (1, 0,0) > xac | y =a~3b {a | 3a + 4b % Solving (2) and (3), we have 243; 4y + 32) be EA) and ge “SE aT (0, 1,-3) + 6T (0, -3,4) et 0,0) (0, 0,0, 32600 0,0,0) +c (1,0, 0, 0) = (¢,0,0,0) = (x, 0, 0,0) [From (1) Hence, T(x, y, 2) = (x, 0, 0, 0) is the required linear transformation. 10 :U— Vand T, : V-> W be two linear transformations, then if, and 7, are one to one onto, so is 7, T, and (T, T,)* = 7," T,'. SISince T, and’F, are one-one onto, therefore by definition Tz' and T;" exist. T(x y,2) Assume that ra i ay T_T, @) = 7,7, @) ye)? iL ‘ = 7, (T, (9) = T3(T, W) % aati) = 7.) =T1@) FAW gl, Bhenetome a xe 2 WL; 7, is one-to-one] > ThT;isone-to-one. , 7 we : d Also, T)T;:U>W If we W, then since T, is onto, there exist v ¢ V such that T, (0) = w. Again as T, : U-> V is onto, there exists u ¢ U such that 7, (u) =». T,(u) =0 = 1, (T, (uw) = T,(0) = = ee yar = TT,isonto. U7 TW) =G Ww) Also, (TT) (Ty1T,') = TAT, (Ty T)) =u =T((T,T,) 7; = 7, (IT;!) =1,T;'=1 Similarly, (Ty'T;') (T,T,) = Ty" (T;'T,) T, TUT) =7,"T, Hence, 7 (1) 7)" = Te Ty 1. (d) Let Sbe a subset of an inner product space V. Then show that S* = [L(S)|" Sol. Since S is a subset of inner product space V, so $ < L($) = [L(S)P c st a [Iv W,cW, = Wi WH, where W), wcll Let SH tly Haye My ‘Then any v € L(S) can be expressed as B= a), Hay ty tom +4, Uy = Eau, where a, € F 56 | STUDENT Examiarion Sewies agWe know that S*= (we V: (u,v) =0 forall ue s) Let 1 ¢ S* be arbitrary. => wis perpendicular to S. => uis perpendicular to 1, | = du, u) =0 Now, (u, 0) = (u, Zain) Ea; (u,u)=Ea,0=0 > (u,v) = 0 = wis orthogonal to v. Thus 1 is orthogonal to L(S) [- v € L(S) is arbitrary] > ue [L(S)' . st c(i)" (2) From (1) and (2), it follows that $+ = [L(S)]* . 1. ()/Prove that a linear operator T on a unitary space V is Hermitian iff (T (a), «) is real for every a. Sol” Let T be a linear operator on a unitary space VC) Let Tbe Hermitian, ie, T = T* (Ta), &) = (a, T* (a)) =a, T(a)) = (T(a), a) ie, (T(a),a) = (Tla),a) => (T(a), a) is real for all a € V. Conversely, let (T(a), «) be real for all « ¢ V (T(a), a) € Rforallac V CT) je - re = (Ta), a) = (Tle).a) ext CON oo = (@T*@)) = (Ta), 0) rte) - = (Tla),a)-(Ta),a) =0 Oot? é > (T(a) - T(a); a) = 0 : te > ((T-T*) (a), & 0 > T-T=0 — [« (TQ),x)=0 = T=0, forall x € unitary space] => T=T Hence, T is Hermitian. SECTION -I 2. (a) Prove that the necessary and sufficient conditions for a vector space V(F) tobea direct sum of its subspaces W, and W, are that (a) V=W,+W, (b) W, OW, = {0}. Sol. The conditions are necessary: Let v=W,eW, (1) : W,+W, ..2) [By definition of direct sum] If possible, let oeve WW, > ve W, and veW, Also, » € V and (2) gives =0+v, where 0¢W, and veW, +0, where veW, and 0¢ W, and Mamnewanics (Sem. 61H) SoLven Quesriow Paver-Juty 2022 (C.D.L.U.) | 87which shows that v € V can be expressed in at lea el 1 ear € Vcan be expressed in at least two different ways. This is contrary Hence the only vector which is common to both W, and W, is 0, ie, W, OW; = {0} The conditions are sufficient Let Vv =W,+W, a and W, OW, = {0 o Now (3) implies that each element v € V can be expressed as sum of an element of W, and an element of W3. : We shall now show that this expression is unique. If possible, let paw tw, we Wy, ve W; and v=, +wy, w/e Wy wy € Wy => w, + W, = W,' + wy! > W,-w,' = W,'-W, Since W, is a subspace, wy, wy EW, > w-w'eW, Also, Wy wy EW, > wi-w,e Wy > w,-w, But W,nW, Thus, w,-0 - and wy -w, w, whith sitow's that expression for each v V is unique and hence V is a direct sum of W, ney ie, ae Vv =W,eW, . (b) If Viga vector space of all 2 x 2 matrices over Rand Wis the set of all 2 x 2 diagonal matrices dver R, then show that Wis a subspace of V. Also find the basis of V/W. ig we fe sve Then, W is a subspace of V 10 00 : 01/4 lo 1 are linearly independent and span W. These two matrices (vectors) form a basis of W. Since W is a subset of V, so this basis can be extended to form a basis of V. Let us consider an extend basis of V. ie. | ° is linearly independent and any matrix Zo OPO va Oleh c dj can be written as a linear combination of these matrices. We have V/W={W+u:ue VI. 01 0 0 We claim that W + [0 A and W+ [" | forms a basis of V/W. 01 00 For Waly gltWrely f=” pee 58 | STUDENT Examvarion Series [Verify it!] | [Verify it!) egwhere W on R.HLS. is the additive identity in V/W. s wef? af? lw wef gewe[? . [Pale [e ohmmee[o ole => a=0, b=0 01 0 0 - Thus W+], y]andW+], 4) are linearly independent. u Let W + v be arbitrary element « V/W. Weo=W+ [: i] sbede v. cd oleh a -(welE oor a) (wos P(E 9) (vee ae (veel dl] “(oe a))e(vels al which shows that v + W ¢ V/Wisa linear combination of on 0 0 i Ww Hence, basi VW {i ome ol Find basis of the sub-space spanned by the vectors (-3, 1, 2); (0, 1, 3); (2, 1, 0); (41,1). Let (-3, 1, 2), (0, 1, 3), (2, 1,0), (1, 1, 1) Let W be a subspace spanned by S. Then the basis of subspace W will be the largest (maximal) linearly independent subset of S. Clearly, ((-3, 1, 2)} being a singleton set of non-zero vectors is linearly independent. Also, {(-3, 1, 2); (0, 1, 3)] isa linearly independent set of vectors because neither vector is a multiple of the other. Let us now consider the set of three vectors {(-3, 1, 2), (0, 1, 3), (2, 1, 0)) Let a(-3, 1, 2) + (0, 1, 3) + c(2, 1,0) = Maruewancs (Sem. 61) Sowveo Quesniow Parer—Jury 2022 (C.D.L.U.) | 59s Sol. -3a + 0b +2c=0 > atb+ 2a+3b+0c=0 In matrix form, the above system of equations is AX = O. -3 0 2][a] fo 11 1|/b] =|0 23 ole} [0 3 0 2/ Also, lAl=| 1 1 1j=ll40 | 230 | > Aisnor-singular and so the only solution of 1) is @=0, b=0) ¢= 0 ‘The vector {(-3, 1,2); (0 1,3); (1, 0)] is a linearly independent set. ‘Also as the elements of $ € R®. % W ¢ R° and dimW (Wea, 0,) + (W400) +t (Wa, 2,) = W > We (0, 0, +0) 0, + one #04,0,) = W > a, Dy ay 0, to +4,0,€ W = 0,0, +a;0, + +4, 0, i a linear combination OF Wy Dy 0 $y Vy toe HA, D, = By Wy + By Wy + + Og Wye Sj, 0,0, to + dy by, Wy, = 0 > and £ Oy Dyy see Dy Wyy Wy toy are => Tis linearly independent. Again, let W + v be any element of V/W. Here v € V can be expressed as @ Jine# combination of Wy, Woy... Wy, Dy, Dy ie, V= A, Wy + dy Wy + mm +a, Wy + by Vy + Dy Vy + + Dy Oy 60 | STUDENT Exammnarion Series|. Let U and V be finite dimensional y; VF MYA, Home Hay, YA) + (Lh, Dy + Dy Dy + + By U,) We (by 0 by 0, +o +b, v,) = CW, 0) + (Wt by 02) + oso + (Wb, 0) © by (W +0) 4 by (W404) 400 +b, (W + 0,) which shows that Tspans V/W. Thus, Tis a basis of V/W and so dim V/W =n Hence dim V/W = dim V—dim w, => Woe We (an, [es dim V=m+n= dim W +n] SECTION ~ 11 (a) Prove that two finite dimensional ve ifand only if they have the same di ctor spaces over the same field are isomorphic mension, ector space over the same field F. Let U= V. Then we have to show that dim U= dim V Since U so there exist a one-to-one, onto linear transformation T from U > V Te A= {tty ty ony ty] 18 a basis of U, then dim U =n Consider, B = (T(u,), T(t), nee Tlt,)} We claim that Bis a basis for V Now, a,T(u,) + a9 T(u,) + cF > Tau, + au + [+ Tis linear] > Oyu; + au, + - > 4, =a AisL1] Hence, Bis linearly independent. Now, we have to show that B spans V: Leto ¢ V be any arbitrary, then since T is onto, so there exists u ¢ U, such that Tu) = 0 Also, A is a basis of U, so u ¢ U can be expressed as = by, + dyn, + + bu, b € F Thus, > Tb, + byw + = by: T(t) + BT (uy) + +b, Tlu,) = 0 = ve Vcanbe expressed as a linear combination of elements of B. 2 Bspans V. Also, B is linearly independent, therefore B is a basis of V. As B contains n elements, therefore dim V = n Hence dim U = dim V Conversely: let dim U = dim V Let if (iy, yy ney Uy} and [Oy Vy, «4, By} are bases of U and V respectively, then there exists a linear transformation, T: U > V such that T(u,) = 0; Let ue s WS atl, tally tome hat, EU > Tu) That, + ally + ayy) Tat) + 7 (ug) + +4, They) [v TisaLT] 1D + yy + 05 + + 4,0, [+ Tt) =2] Thus, T(u) © V Next, we show that T is one-to-one: Let T(u) = T(w’) for u, uw’ € U Matnemarics (Sem. 67H) Sotveo Question Paper—Juty 2022 ('Thay + ate, +o + gly) = PUM + Milly + + My) {u, u’ « Lave linear combination of elements of basis sun > ay T(uy) + a,Puy) +o 44,7) = a), Tuy) $44, Tig) +r a’, Ty) Tina > (a= a)) T (ty) + = a5) T (tg) + rr + (0, ~ 4) T (,) = 0 > a-a,= 0 for £2 1,2,3, 00) > 4,24, i=1,2,3, ie, weil Then T is one-to-one, Finally, we show that T is onto: Now, veV = 2 = bv, + bpp, + + Dy Tuy) + byt) +--+ 6, TCU, = Tb yu + Bll +e + Byt,) ie, 1 = byt, + Dally + + Byly Tis onto Heny uzv. ‘ind a linear transformation T : R’ —» R* whose range is generated by (1, 2, 0,- 4) and (2,0,-1,~3). Sol. We know that {¢,,¢,¢) is a standard basis of R°, where e,=(1,0,0), €=(0,1,0) and ¢,=(0,0,1) ». The range of T, i.e., R (T) is generated by {T(e,), T(e), T(es)} But it is given that R (T) is generated by {(1, 2, 0,~4), (2,0,-1,~-3)} Let Te) = (1,2,0,-4), T(¢,) = (2,0,-1,-3), and T (e,) = (0,0,0,0) Now, for each (x, y, z) € R®, we get (x,y, 2) = x(1,0,0) +y 0, 1,0) +2 (0,0,1) = xe, tye +2e T(x, y,2) = T (xe, + ye + 265) = xT (e,) + ¥T (ce) + 27 (es) = x(1,2,0,-4) +y (2,0,-1,-3) +2 (0,0,0,0) = (x+2y, 2x,-y,-4t-3Y), which is the required linear transformation. 5. (a) Let 5 = {v, vy, v,) Is a basis of a vector space V, (0). Find the dual basis of S where vy = (14,1), ¥, = (1,4, -1), vy = (1,1, -1). Also Find f, (8), fo 0-08) wh [> TisLT) x= (0, 1,0). Sol. If$={0,,0,,0,) isa basis of a vector space V;(C), then the corresponding dual basis is [fy-/» such that : _ a fl if i=j o GO) = By V9 ig ii Let SY, 2) = aX + ayy + O32 62 | STUDENT Exammarion SeriesAO M2) = Bx by + bye fy Wr 2) = Ox + ey + ez By (1), we have f,()) = Lf, @) = 0, fo) = 0 Now, AO) =f, 1,1) =a, +a, +0,=1 A) = 1, -1) =a, +4, ~0,=0 As) = f\(1, 1,1 By] Solving these equations, we get a 1 a=0 2 So, FiQ yz) = fos) Again by (1), we have Al) = 0, £@)=1, f(@,) =0 ve Sol) = ful, 1,1) =b, +b 4b = Ar) = LAA, -1) = by +b, — fxs) = fy, -1,-1) = Solving these equations, we get” [By (1)] i by = O,by= 5, by $0, Alou = Fy-2 Again by (1), we have f(01) = 0, fy(,)=0, f,(0,) = 1 Now, fy) = f(L, 1,1) =e, +) +0, =0 f,@) = fx, 1,-1) = ¢, + e-c5=0 Sys) = fy(L 1, 1 j-¢3=1 [By ()] Solving these equations, we get 1 1 f= 51 = — 5, 20 1 So, filtime2) = 5(e-y) Hence, the dual basis is {fy fy, fl = Fi (+2), 2 y-2), Ze -»} Finally, if x = (0, 1, 0), then f,(0,1,0) = 3100) =0 1 .! f,(0, 1,0) = pee =3 1 1 £,0,1,0) = pea y zo , f,) be the dual basis of {v}}. Then ~ V,} be a basis of V (F). Let {f,, fy © Any vector uc Vcan be expressed as u= >> f,(u)v, mi ee Manumarics (Sem. 61m) Souveo QuesTion Parer—Juty 2022 (C.D.L.U.) | 63- > (ii) Any linear functional « V* can be expressed as f=) /(¥,) f rot Sol. Since {0 yon, By) is a basis of V(F) and [fy for mf is the dual basis of {2, we have 0 if i#j fife) = By = {; if i=j ( We have to show that u= "fi (u) 2 a 2,) isa basis of V, thus any u ¢ Ucan be expressed as ajeF Since {24,2 MH Ay +My + oon FOP ye =) jen t., Then from (1), we get u = )” f(u) 2; ia (ii) Since {fy fye-rf,] is a basis for V*, thus any f € V* can be expressed as f= Df t bafy tne tf or f= Ds ~Q) a fo) = Yb f=Db oa & i > fy) = Dy + ByByy + nn +B By = by1 + by 0b nn +B, O= Dy In general, fo,) = b Putting these values in (2), we get f= ”f (2) f,- Et SECTION - III 6. {ef Let T:R° — R*bealinear transformation defined by T (x,y,2)= (2x+y-2, 3x-2y +4 Find the matrix of T with respect to ordered basis B, = {(1,1,1),(1, 1,0), (1, 0,0)} B, = {(1,3), (1, 4) of R? and R® respectively. Also, verify tht (7B, B,llu, By} = [T(u),B,). and 64 | STUDENT Exammariow Sens_ given linear transfortnation is gol. The 8 T(v.y, 2) = (e+ y—2, 3x 2y +42) T(Q,1,1) = (2,8) m TA10 © BD T(,0,0) = (2,3) Assume tat (2.5) = a(1,3) +b (1,4) = (a+b, 30+ 4b) (1) = 0(,3) 4d (14) = (4d, 3c+ Ad) (2,3) = (1,3) +f (1,4) = (e+ fer 4) = atb=2 ga+4b = 5 c+d =3 3c+4d = 1 e+f=2 Be+4f=3 Solve these equations, we have a=3, b=-1, c=1 d=-8, e=5, f=-3 then T(1, 1, 1) = (2,5) =3(1,3) + (-1) (1,4) T(1,1,0) = (3, 1)= 11 (1,3)-8 (1,4) T(1, 0,0) = (2,3)=5(1,3)-3 (14) nis -8 °,] () 3 Thus, IT: By By) = i ‘ Also, we shall find [u, B,]: Assume that u = (a, b, c) € R° be arbitrary then u can be expressed as linear combination of the elements of basis set B,, then u = (a,b,c) =a, (1,1, 1) + by (1, 1,0) + & (1, 0,0) a =a, tbc b =a, +h, c=a Solving the above equations, wehave a,=c b, =b-c ¢ =a-b ‘ then u = ¢(1, 1, 1) + (b-c) (1, 1,0) + (ab) (1,0, 0) \ 2) then from (2), we have > c [u,B,] = |b-c 3) a-b Finally, we shall find [T(w), By] From (2), we get T(u) = cT (1, 1,1) + (b-c) T (1, 1,0) + @-6) T (1,0, 0) ‘Matnemarics (Sem. 61H) Souven Question Paren-Jury 2022 (C.D.L.U.) | 65_ 130,39) + 1) 4D] + (b= 0) 1G, 3) 8 (1, 49) ¥ @-1)15.01,3)-3(, gy (5a + 6b = Be) (1, 3) + (= 3a 5h +70) (1, 4) 5a+6b *| = 3a-5b +7e (4) Therefore, [7 (), Bl ¢ 15 3 61 Hence, (7: By Bll B= |") g _ 5a + 6b ~ 8 * [-3a-5b+7e = [T(u), By] [By (4)) (b) Show that linear operator T on R* defined by T(x, y, 2) = (x - 3y ~ 22, y - 4z, 2) is invertible and find T~*. To show that Tis invertible, we have to show that T is one-one and onto. bee a-b ad Sol. Tis one-one: Let uy, u, € B®. . ty = Cy Yar 2s) aNd t= py Yor 2) For T(u)) = TQ) = Tey Yur 2) = Ty Yor 22) (3p — Dey My ~ 42,2) = y= 3a ~ 22) Ya — Ayr Zr) = 17 3y, ~ 22 = ¥—3Yy—2zy y- 42 =o Ar = Hy Wad AAS > (Ya 2) = Cay Yor 22) = My = T is one-one Tis onto: For (x, y, 2) € R°, T is onto if there exists (r, 5, t) ¢ R®, such that T(r,s,t) = (x,y, 2) > (r-3s~2t,5-4t, t) = (x, y,2) > r-35-2t =x, s-4t=y, t=z Now, xyz2eR > 1,5,teR (5) eR “a ©. For (x, y, 2) © R°, there exists (r, s, 1) R®, such that T(r,s,t) = T(x + 3y + 142, y + 42, z) = (x, y, 2) Thus T is onto. T being one-one and onto is invariable. Thus, T(r,8, 0) = (y,2) => T(x, y,2) = (1,8, t) = (x + 3y + 14z, y+ 4 . Find the eigen values and the basis for eigen space for linear transformation 7: —+ R® defined by T(x, y, z) = (x-y, 2x + 3y + 2z, x+y +22) Gol. ‘The given operator is T: RR? such that T(x, y,2) = (xy, 2x + 3y + 22, x4 +22) B = {(1,0,0), (0, 1,0), (0,0, 1)} be standard basis for R* Let 66 | STUDENT Exammarion Sewies ae1,0) + 10,0, 1) & 1,0) + 1(0,0, 1) (0, 0, 1) T(1,0,0) = (1,2, I) = 1(1, 0, 0) + 20, TO, 1,0) = 13,11) (10,0) +3(0, (0,0, 1) = (0, 2,2) =0(1, 0, 0) + 200, L, 0) + 2 1-10 [TB] =A=|2 3 2 P 11 2 oa ‘The corresponding characteristic equation is ( EO N-%--1 0 lA-Ml =| 2 3-2 2]=0 - 1 1 2-a] > (1-2) Q2-5A +6) =0 = (1-A)Q-3)-2) =0 4 =1,2,3 — = Hence, 1, 2, 3 are the eigen values of A. If X is the eigen vector corresponding to eigen value 2, then AX = AX (A-ADX =O 1-2 -1 0 ][x] fo 2 3-2 2 |ly/| =|0 1 1 2-aflz} [0 : The corresponding eigen vector is a solution of 0 -1 Offx] fo 2 2 aly] =]o 11 iz} [0 y=0, 2x+2y+2z2=0, x+y4z=0 Thus, {(1, 0, 1)} can be taken as basis for eigen space, when 2 When 2. = 2: The corresponding eigen vector is a solution of -1 -1 0ffx) fo 2 1 2iyl =|0 1 1 Of[z} [0 Applying R, ; (-1) and R, , (1), we get “1-1 Ofx] [0 1 0 2ify| =]o 0 0 olz} lo « Marneuanics (Sem. 61m) Sowven Question Paren-Juty 2022 (C.D.fy Thus, {(1, - 2, - 1)} is a basis for eigen space, when A = 3. 7. (Bb) IEB = {Vy Vz) - V,} is an ordered basis of V such that [T, B] = A is a diagonal matri, Sol. It is given that [T, B] = A is a diagonal matrix. ank T. then Rank A 4 0 0 0 0.2.0 0 Let As 7 0.0.0 0-4) such that Rank A =r By definition of rank of a matrix, the first r rows of A are non-zero and all other rows are of zeros. Now B=[0,. Also Let > > > “Yuy: yy dy», dy, Will be all non-zero and d,, ; = Ug sy Dy] 8 an ordered basis of V TW) = di, 61, 2, oo T(v,)} c Range T 1 Op {T (2), - MT (0) + 0. +4, T(v,) = 0,4, € F adv, + =0, foralli=1,2 4, = 0 forall, 2, .000f S = (T(0)), ., T (,)} isa linearly independent set. T(v) © Range T. so BD, + Dy Oy gt veV = v= bv, + T (v) = by T (0) + nnn. +b, T (0,) S spans Range'T dim (Range T) = r Rank T = r Ran A = Rank T. eee 68 | STUDENT Examwarion Semres dy =O. Sf + b,v, 4,#0)>» a) came ee 8. (a) HE (iy ty» oy U,) be an orthonormal subset of an inner product space V(F), then CA prove that > |(u,1,)) < jul? forall ue Vv. mt 1. Since {24), May «+ W,} iS. an orthonormal subset of an inner product space V(F). 0 duu) =1 if i=] and (ii) (uu) = 0 if i#j Let v= 1S (uu) 4 (1) Obviously, 7 € V ” Now, ol? = @, 0) = (+S aes) a = (usu)—D'(u ui) (ui) ci : (rt Bd) $e (a Sn) a a fa 2) [Using (1) step by step] We know that for all u,v € Vanda € F, (u, av) = @ (u,v) Thus, (u,u—~ Yu, u))1y) = (us)~ (um) (um) ete. --@) Eo Fast Using (3) in (2), we get Wel? = (su) — YY um) (m5) fA * 8 1" -S len) un) +S (um) $form) wm) Post im ja or lol? = [uP -@ u;) (u, u)— uu) Cu, u;) [By (1)] I . [uF 25 (eayP +S iumyh ‘Maruemarics (Sem, 67H) Souvep Question PAveR-Juty 2022 (C.D.L.U.) | 69¥ a n ‘ = beh -Liuu)e m or Wot? = Jue Diu, u)? vel ; Also, fol? > Oforalloe V [uf Shu u)P 20 = gus Shunyh = Lead sta frat e y a 8. (b) Show that (u,v) = 2xy gy + %1 Uo + *2i + %2 Bar defines an inner product ony, where w= (,.%,),V= WiJ2) € ¥2(O- Sol. Let us verify that all the three axioms for an inner product space. @ (o,u) = Mm B+W Rt yaH +¥2% J ea = (a) = MRNA A) = WnthathHurth = mhrnhtentnh = (u,v) (i) (uu) = 24 +m % tH +H [By taking v = u in the definitr 22 Mx l?+ lal? + x + % = Lal? + Lay l? + Lal? + OX + Fe) = Lxpl?+ byl? + layl? + 2Re (a4 %) Since x, x, € C Let x, = 0, +iBy = + iB, * Ini? =a2+B2, Ixyl?=0F +B and Re(x; X) = Re [(a + iB,) (.~ iB.)) = Re {(ch oy +B Bp) +1 (2 By ~ BI] = 04 4 + B By ‘Thus, we have(u, u) = (a + Bz) + (af + By) + (az + Ba) + 2 (ch oa + BB) = a? +B? + (a, +a) + (By +B.) 20 and (u,u)=Oiffa,=0, B,=0, 0, +0=0, B,+B,=0 ie, iff a, = 0, B,=0; a,=0, B,=0 ie, iff x, =0, %=0 ice, iff u=0. (iif) Let w = (2,2) € V2 (C) anda, b € C be arbitrary. Now, au + bv = 4 (xy) +b Yy Yo) = (ax, + by, ax, + by) oes 70 | STUDENT Examwarion Sensand (au bo, w) = 2(ary + by) E+ (ax + by) 22 + (ax, + bya) % + (ax, + bys) 2 a (2x, B44) Hy +82 B42) 4 (Quy Ey + My By + a Bi + Wo 2) a(u,w)+b(v, 0) Hence, V3 (C) is an inner product space. 7 a = 1), 0) = 9, (a) Let Tbe a linear operator on a unitary space V(C), prove that T= 0, iff (1(4), 4 for alla « V. Sol. Let T be a linear operator on a unitary space V(C) Let a € V be arbitrary and T= 0 Then (Tia), a) = (O(a), 0) = 0 Conversely, let (T(c), «) = 0 where a € V is arbitrary (1) = (T(a. + B), (a + B)) = 0, BEV (T(a) + T(B), a + B) + TisaLT. = Tia+)=T(a) +76) (T(a), a + B) + (T(B), «+ B) (T(a), a) + (T(a), B) + (T(B), ) + (T(B), B) 0 + (T(a), B) + (T(B), a) +0 (T(a), B) + (T(B), @) = 0 Now (2) is true for all a, Be V Replacing 8 by if in (2), we have euUD (2) (T(a), i) + (T(B), oe) al 7 (T(a), B) +i (T(B), a) [TisaL.T = T(aa)=aT(a)] a ~i(T(a), B) + i (TB), &) = oo —(T(a), B) + (T(B), a) a? (T(a), B) ~ (T(B), a) ‘Adding (2) and (3), we have (T(a), B) + (T(a), B) 3) es (T(a), B) = 0 for all a, Be V wa(4) Taking B = T(a) in (4), we have (T(a), T(a)) = 0 for alla e V T(a) = Oforallae V [v@u=0 > u=0) T=0, 9. (b) Let fbe a linear functional on a finite dimensional inner product space V(F). Then there exists a unique vector v < V such that f(u) =u, v) Vu ¢ V. I. Let f: V—> F bea linear functional on V. Let $ = (iy, lyy Hy, ney ly) be an orthonormal basis of a finite dimensional inner product space V(F) so that f duyu) =1 if ij and (ii) (uy 4) =O if ij. Define v= Suu; ea) m7 Obviously, veV 2) Leta map g: V > F such that g(a) = (a, v) Va e V Leta, b F and (a, B) € V. Then (a0. + bB) = (aa + bB, v) Marnemarics (Se. 61H) SoLveo Question Parer—Juty 2022 (C.D.L.U.) [71a. | ‘a (a, v) +b gisa linear function on V for sj Sm", we have g(t) = (uy 2) [From = (4 $574) trom ® = Df) (uj-m) =f) it [Using (1) and (2) because (ty, tay -.- g(u) = flu), weS. Thus fand g are both equal on $ and so on Vie.,f=g. Uniqueness: If possible let w « V such that fu) = (uw) VueV , u,) is an orthonormal basis of y] [Given] flu) =o vueVv (u, w) = (u, 0) (u, w) - (us, D) (u, w-0) w-0 wee proves the uniqueness. on Euuuyy E