MATH 218 Problems Solution Manual Chapter 3.4, Fall 2011

3.4 Coordinates and Change of Basis Pe 77 sé€R>, then dim(SNT) =1, cooreo Exercise Set 3.4 If B = {vi,..+,Vn} is an ordered basis for a vector space, then for each vector v there are scalars €1,C2,+++,Cn Such that v = e1vi + cove +++++CnVn. The unique scalars are called the coordinates of the vector relative to the ordered basis B, written as If B is one of the standard bases of R”, then the coordinates of a vector are just the components of the vector. For example, since every vector in R® can be written as Fy 1 0 0: v=|]y}=2}]O0]+y}] 1] +2] 0-], zo 0 0 L x the coordinates relative to the standard basis are [v]z = | y | . To find the coordinates relative to an ordered . z a basis solve the usual vector equation evi + cava te + CnVn = V for the scalars cj, ¢2,...,@,. The order in which the vectors are given makes in. difference when defining coordinates. For example, if B = {| 4 ) , |} and B! = { ’ ; I} , then ne= (3 ]h [8] === [[s]le-L] Given the coordinates relative to one basis B = {v1,...,Vn} to, find the coordinates of the same vector relative to a second basis B’ = {v4,...,v,} a transition matrix can be used. To determine a transition matrix: e Find the coordinates of each vector in B relative to the basis B’, ¢ Form the transition matrix where the column vectors are the. coordinates found in the first step. That is, WB = [ (valor [vale [vn]ar } ¢ The coordinates of v relative to B’ given the coordinates relative B are given by the formula vie = (18 ve. The transition matrix that changes coordinates from B’ to B is given by . (NB = (1B). Let B= {| ; , 7 |} and B’ = {| ; a _ |} be two bases for R?, The steps for finding the transition matrix from B to B’ are: 78 : Chapter .3 Vector Spaces e Since 1-2/1 -1] sancestol 2 ~2/ 2 2 2 -t}1 1 [PME QO 3g fer 3? so the coordinates of the two vectors in B relative to B’ are I = 1/3 and “t = Lj ity —1/3 1 B [7] : vn'=| tis 3 | stool D)= [28 HILT [48 TLR] E58] Mi Solutions to Exercises 1. The coordinates of ; , Telative to the basis B are the scalars c; and cz such that C1 3 + C2 ~ = 8 . The vector equation yields the linear system Ser 2en, = 8 , which has the 1 2 0 : c+2ce, =0 unique solution c; = 2, and cg = —1. Hence, [v]g = a . . f -2 -l -2 apf 10° -1/2 : _f -1/2 2. Since 4 1 1 reduces tc 041 3 , then [v]z = 3 . 3. To find the coordinates we form and row reduce the matrix 1 38 1} 2 1 0 O} 2 2 -1 -1 0O;/-1}—]0 1 0]-1 |, sothat.jvjp= | -1 |]. 2 1 2) 9 0 0 1) 3 3 2 1 0] 0 2 1 oo 1/2 4, Since | 2 0 0] 1 | reducesto] 0 -1 0/1 |, then [vJg=] —-1 |. 1 2 1); 1/2 001 | 2 5. Since cy + cg(@ — 1) + cgu* = 3 + 2a — Qu? if and only if cy — cg = 3,c2 = 2, and cg = —2, we have that 5 VJe= |} 2 2 6. The equation cy (a? + 2a + 2) + c2(2a + 8) + ¢3(—a? + a + 1) = 8+ 60 — 32? gives the linear system, in matrix form, ‘ 2 3 1 | 8 1 0.0] -1/3 22 1 6 | , which reduces tg] 0 1 0 2 1 0 ~1]-38 0 0 14} 8/3 [| so [v]g = 2 : 8/3 . 1 -1 0 1 1 : 1 0 1 up vp 7. since «| 9 o|ta[t oftele frei a 0 |=[-2 2 | Hand only 3.4 Coordinates and Change of Basis 79 a +eg+eq =1 ~e1 + ee =8 2? and the linear system has the solution c, = —1,cyg = 2,cz = —2, and cq = 4, C2. —~Ch =o C3 = —14 2 . we have that [v]p = _9 |: 4 ot + 1-1 fo 1 1-1], f14 8. since e1| 9 i}+e] 9 rita t a |te[o 3|- [i 5 | leds to the tinar system 1 0 4 1] 2 10 0 Of] 2 2 -1 1 -1 Ij -2 0 10 OF 2 2 0 0 1 ol4 , which reduces to 0010 7 , then [v]p = 1 1 2 0 3); 8 000 1)-— -1 9. [vy] U4 (vie 1/2 10. [v] ly + (VJB, 4 2] + |ViB, = lv 1/8 1/2 : 1 1 ; 7 11. [v]B, = 2 ; lv]. = 1 12, [va = 1 sv] B, = Z —1 0 4 3, 13. The column vectors for the transition matrix from a basis By to a basis By are the coordinate vectors for the vectors in By relative to Bg. Hence, WR = [ : 7 = ; ; ‘ . Then this matrix Bz By transforms coordinates relative to B; to coordinates relative to Bg, that is, [v]s, = [I] 2? BV], = =| ; . 14. n= [18 8 ] solos = nBeotes | 2, |= [ 48 | 3 2 1 ~1 -1 15, [12 = | -1 -2/3 | v= a | 0 |-| 1 | 0 -1/3 0 2 0 —2 4/3 —4/3 2° —4 16. [p= | 1 1/3 2/8 | i = tj=] 8 0 1/38 ~1/3 1 0 17. Notice that the only difference in the bases B, and Bo is the order in which the polynomials 1, a, and a? are given, As a result the column vectors of the transition matrix are the coordinate vectors only permuted. 00 1 5 That is, [Ug = | 1 9 0 |. Then [vja, = [Z]p%Ivla, = | 2 | - 010 ; 3 18. Since [v]a, = [Z| 5) = [ (2? — 1a, (22? ++ 1]a [-2 +1] | = 1 13/8 (elvis, | 1 = | ~1/2 | ; 2 11/8 | 1/4 5/8 3/8 | -1 ~-1/2 1/2 |, then [v}p, = 3/4 11/8 -3/8 80 Chapter 3 Vector Spaces ee -1 1 -lia 1.0 O0|-a-—b+e a 1 0 t]/b}—}]0 1 0 a+b , we have that | b 1 1 Ofe 0 0 1) a+2b-c¢ e 20. Since 1 0 0 -1l a 100 0 2a+b-—c-2d 0 -1 -1 0 0b jl ‘ 0 10 0 -a-b+e+d the 11 -1 0c |S) 9 0 10 aed » en 0 -1 0 -1 ad 000 1 a+b-c-2d 0 0 1 2 ant [! 0 ° | ose 2] =[] aaa. B=[1 ~o] > W=[_2 {| since [i olf of=[o 2] me cng = ing. oomt-(2 3 [],-EHEEL “Uh hel elio[s! c d, —sind cos 6 cos 6 24, a. [v]p = cos _ xcosd — ysin#@ asind + ycosé Iti —a-b+e = a+b B a+22%b—-—e a 2a+b-—c—2d b _ | -a-b+e+d c ~ a-c-d d at+b—c~2d 3.5 Application: Differential Equations . 81 25. a. Since uy = ~v4 + 2v2, Ug = ~v1 + 2v2 — vg, and ug = —v2 + vs, the coordinates of uz, ug, and ug relative to Bg are -1 f 1 0 -1 -1 0 luije,=| 2 | lula, =| 2 |,[usle =| -1 |, so (f= | 2 2 1 0 -1 1 0 -1 1 2 1 b, [Quy — 3ug+ ug] By = (3? -3 |=] -3 1 4 Exercise Set 3.5 1. a, Let y =e", so that y/ = re" and y” = r2e"®, Substituting these into the differential equations gives the auxiliary equation r? — 5r + 6 = 0. Factoring, we have that (r — 3)(r — 2) = 0 and hence, two distinct solutions are y; = e?” and yg = e°*, ert et b. Since W[y1, ye] (a) = get” Zee | = e®* > 0 for all x, the two solutions are linear independent. ¢. The general solution is the linear combination y(x) = Cye?® + Cye**, where Cy and Co are arbitrary constants. 2. a. Let y = e"®, so that y! = re’ and y"” = r?e", Then the auxiliary equation is r? + 3r+2=0= (r + 1)(r + 2) = 0 and hence, two distinct solutions are y, = e~* and yo = e7?", en? en 2 3 —e7® = —Qe72# c. The general solution is the linear combination y(x) = Cye™* + Cge7", where C; and Cy are arbitrary constants, : 3. a. Let y = e”, so that y/ = re" and y” = re", Substituting these into the differential equation gives the auxiliary equation r? + 4r +4 = 0, Factoring, we have that (7 +2)? = 0. Since the auxiliary equation has only one root of multiplicity 2, two distinct solutions are y;, = e7?* and yo = we7?", b. Since W[y1, yo} (a) = = ~e~* #0 for all x, the two solutions are linear independent. . + ew de wen 2 b. Since Wly1, y2](2) = —2e72% er 28 _. Omen 2e independent. c. The general solution is the linear combination y(a) = Ce7?" + Coxe7?", where Ci and C2 are arbitrary constants, 4. a. Let y = e", so that y’ = re" and y= re", Then the auxiliary equation is r? — 4r +5 = 0 = (r +1)(r — 5) = 0 and hence, two distinct solutions are y, = e~* and yg =: e5*, . ene edt b. Since W[y1, ya](z) = —en® Bede general solution is the linear combination y(z) = Cie~* + Ce", where C1 and C2 are arbitrary constants. = e~4@ > 0 for all x; the two solutions are linearly Mi =e > 0 for all a, the two solutions are linear independent. c, The | |