MATH 218 Problems Solution Manual Chapter 4.4, Fall 2011

44 Matrix Transformation of a Linear Transformation ; 101 37, Let v be a nonzero vector in R°, Then a line L through the origin in the direction of the vector v is given by all scalar multiples of the vector v. That is, L = {tv| t € R}, Now, let T : R? —> R® be an isomorphism, Since T is linear, then T'(tv) = tT(v). Also, by Theorem 8, T(v) is nonzero, Hence, the set L' = {tT(v)| t € R} is also a line in R® through the origin. A plane P is given by the span of-two linearly independent vectors u and v. That is, P = {su + tv| s,¢ € R}, Then T(su+tv) = sT(u)+tT(v), and since T is an isomorphism T(u) and T’(v) are linearly independent and hence, P! = T(P) = {sT(u)+tT(v)| s,t € R} is a plane. Exercise Set 4.4 If A is an m xX n matrix, a mapping T’; R” —+ R™ defined by the matrix product T(v) = Av is a linear transformation. In Section 4.4, it is shown how every linear transformation T : V —+ W can be described by a matrix product, The matrix representation is given relative to bases for the vector spaces V and W and is defined using coordinates relative to these bases. If B = {v1,..., Vn} is a basis for V and BY’ a basis for W, two results are essential in solving the exercises: ¢ The matrix representation of T relative to B and B’ is defined by (TB = ((va)}o (wa)har «.. (vn) J ¢ Coordinates of T(v) can be found using the formula [T(w))]a" = (1B Ie. To outline the steps required in finding and using a matrix representation of a linear transformation define a —& y = |] —-y | and let. z z 1 0 B= ol,/ 1], and B’= , , 0 0 : ; 0 0 Tr 0 =| 0 1 l e Apply T to each basis vector in B, ¢ Find the coordinates of each of the vectors found in the first step relative to B’. Since ren 2( ROO Ree EO, Orn two bases for R°, ll (i) EDL 11 2/-1 0 0 10 0] 1/2 -1 1/2 10 1/0 -1 0}/—5]1 0 1/-1f2 1 1/2 J, 1 10/0 01 1 1 0/-1/2 0 -1/2 Lela TE TL Cad ¢ The column vectors of the matrix representation relative to B and B’ are the coordinate vectors found in the previous step. , 1/2 -1 1/2 [T]B =|] -1/2 1 1/2 —1/2 0 -1/2 102 Chapter 4 Linear Transformations ¢ The coordinates of any vector T(v) can be found using the matrix product [T(v))] a: = (NE Wve. -4 rsDi Lee AAI] Since B is the standard basis the coordinates of a vector are just the components, so 1 1/2 -1 1/2 17 4 1/2 T —2 = —1/2 1 1/2 —2 |= | -9/2 |. PSD) [Se SBS} Le] This vector is not T(v), but the coordinates relative to the basis B’. Then (DESH sG- (3) Other useful formulas that involve combinations of linear transformations and the matrix representation are: 1. : e As an example, let v = —2 , then after applying the operator’T the coordinates relative to B’ is given by o(S+T]5 = (SIF +712 ° (KT)B = kITIZ «(Sor |B = [S]B IIR) (Ta =((T]a)" © (P™]e = ((T]e)™! W Solutions to Exercises 1. a. Let B = {e1,e2} be the standard basis. To find the matrix representation for A relative to B, the column vectors are the coordinates of T(e1) and T(eg) relative to B. Recall the’coordinates of a vector relative to the standard basis are just the components of the vector. Hence, [T]z = [-[T(e1]z [T(ee]e ] = 2 7 . b. The direct computation is T : = ’ and using part (a), the result is C2 IR) Pa-[3 Ue )-E) 3. a. Let. B = {e1,e2,e3} be the standard basis, Then (Tle -1 1 2 1 3 fs 0 8 41 . b. The direct computation is T —2 = —3 |, and using part (a) the result is b. The direct computation is T > = 4 and using part (a), the result is i ‘(Peale (P(eela (Teale ] = LQ -1 3 2 EEE 4.4 Matria Transformation of a Linear Transformation 103 0 2 2 : 0 b. The direct computation is T | —5 | = 5 | , and using part (a) the ~l 1 ~l ao -EEIELE) 5. a. The column vectors of the matrix representation relative to B and B’ are the coordinates relative to B’ of the images of the vectors in B by T. That is, [T]3’ = Ir ([ i i} [r (| * )) . Since B’ J/ JB Br is the standard basis, the coordinates are the components of the vectors T (| A }) and T (| ; \) » SO / ~3 2 . Bl 3 mig=[~3 eI]: —1 b, The direct computation is T’ 2 = 3 and using part (a) rl 2 ]ef-3 -2][-1] ~[-3 -2 2 -]_f-3 2/7 [ 3 6][-2],>[ 3 6]| -3/2}*{-3 |: ~3 2 4 1 1 —3/2 6a. (TR =| 2 1°2} by 1 fer} -1 | =[rye fb -1 > =e] -3 2 02 2 ‘ il, 5/2 7. a, The matrix representation is given by | DI, lr ([ DI | —3 | , BI m= (EDL P(e DEP We can find the coordinates of both vectors by considering 38 01-2 2 _,} 1 0 —2 -2/-3 2 0 1 b. The direct computation is T vi = ~2 . Using part (a) we can now find the coordinates of the 4,a. [T]B =a ORO ww —3 -4 image of a vector using the formula. [T'(v)]’] = [7]8 [v]g. and then use these coordinates to fine T(v. That is, cain us RD ° Ny cm L a i i rls, eee ss}e-sf 4] af 4] -[]} 1 a) 1 8. a. (T]B ‘| 301 b. The direct computation gives T >| 1 | = 4 | . Using the matrix in 8 ; : “3 4 —2 l 1 part (a) gives i =(T]2 |T] 1 =(T}2 | 1 | =] —3 |, so that 3 |i, 1} | ~4 —2 0 1 -1 : 1 1 }=|o}-3} 0 |-4}-1 fe fa]. 3 1 1 0 L 4 | | 104 Chopter 4 Linear Transformations 11 1) 9, a. Since B’ is the standard basis for Py, then [7]2° = | 0 -1l ~2 | . b. The direct computation is 0 0 1 T (a? — 84 +3) = 2? — 3x +3. To find the coordinates of the image, we have from part (a) that = [TB [e? - Se +3)e = (212 [T (a? — 3a + 3)] », 1 3 | = | -3 | , 80 D(a? — 8a +3) = 3 — 8a +2”, 1 1 ° 1 -1 -2 10. a. [T}B' = | -1 0 1 | b. The direct computation gives T(1— 2) = 4 (1-2) + (1-2) =~. -3 1 8 1 Fou 0 Using the matrix in part (a) gives [T(1—2)|e = (7) (1 —e]e = (1) 1 | = | 0 | , SO 1 T(L-«) =0(-1+2)+0(-1+a+a*)-2=~a, 11. First notice that if A= ‘ - , then T(A) = 0 2b . 2c 0 oe 0 0 2 1 0 ~2 a. [T]p= | 0 -2 O | b. The direct computation is T {| 5 = , Using part (a) 0 0 2 3-2 |} > 6 0 r([3 3 ))] =i 1 | so \ i | 3 by\_ Ba b+ 2c r,-{0 1 2 d|)=la+e sal) *MB=)o 9 | b. The direct computation gives T (| 4 ; \) = 12, First notice that T (| a a8 | 4.4 Matriz Transformation of a Linear Transformation : 105 0 0 , 0 0 ; Qo ol 016 ba (TB =| 1 0}b WB =] 0 lle WS =}] 0 Of] a [SZ = [° 0 | , 0 1/2 1/2 0 1/2 0 , 10 , 00 0 e. [S]Z(T]2 = 01 TIE (S]B, = | 0 1 0 | £ The function $07 is the identity map, that is, 00 1 (So T)(aa +b) = aw +b so S reverses the action of T, 2, 0 0 2 2 Q -2 2 q 0 -2 2 -2 0 0 -2 2 Bil Be 16a (Me =a] 3 4 1 9 | MIB=] yg 1 <1. -3 1 -1 0 0 0 2 2 4 0 0 0 2 0 0 2 1 0 -1 -1 7 -2 0 -6 2 7 0 2 2 0 Oo 1 1 +i Bid Bit Boo oe =a) 1 9 3 7) %Ule=a} yy 41} le =]o 1-1 4 -3 0 5 1 -l1 -1 1 1 10 1 1 17. [T]e = ; 3 . The transformation T 18. The transformation rotates a vector by @ ra- dians in the counterclockwise direction. reflects a vector across the w-axis, 19. [T]p = ol . 20. Since T(A) = A~ Al = [ = mar 0 0 0 0 ; _{ 0 4, -1 0 then [T]g = 0-1 1 0 00 0 0 B! 10 21. [T)Z =[1001 22. a, [-85]o = 815] =-3| | | [ 6 mL) 5 2 31 23. a. 2T+S)e=2[T]e+[Sln=| _1 7 24. a. (ToS]e=(TlalSja=| 5 3 ~4 —3 b. 33 bios 24 a 2 -2 -2 25. a [So T]z = (S]a[Z]B = [i ‘| b. 26. a, 2T]B = 2[T]s = 0 4 4 2 2 2 ~ [ —10 10 b. | 16 | 10 3 3° «4 2 0 -2 27. a, [-3T +25]n= | 2 -6 -6 28. a. [ToS|x=[Tja[Sja=| 2 0 2 3-3 -1 . ~2 0 2 3 [ -8 b. | —26 b. | 4 79 L 8 106 Chapter 4 Linear Transformations °0 00 6 0 —12 00 0 0 24 . —48 31. Since [T]p = | 0 0 0 0 O | and [T(p(@)jp= | 0 |, then T(p(x)) = p(x) = -12 — 482. 0000 0 o | 000 0 0 0 oor 0 32. Since B is the standard basis, then (T(1)]z = [l]z = | (T(«))p = alg = | 2 | , and [T(2?)]_ = 0 0 100 : (3e7]5=| 0|,s0(T]p=-|0 2 0]. 3 - 100 3 33. [S]2 = OoOrooe 0 0 2 | = 0 3 CoOrsa 34. The linear operator that reflects a vector through the line perpendicular to ; , that is reflects across me-((, [e]gleF2 a] -b a-d 0 b e 0 d-a@ ~c 0 c —b 0 the line y = —a, is given by T e = | md , So 35. If A= ‘ y , then the matrix representation for T is [T]g = 36. Since T(v) = v is the identity map, then 0 1 0 (TB = (va) le ((va)| er (P(vs)]B'] = [ [vale [vale [vale] = | 100 | . 00 1 a ; b : If [v]z = | 6 |, then (v]g- = | @ |. The matrix [T]3’ can be obtained from the identity matrix by c ¢ interchanging the first and second columns. 37. 1100 0 0110 0 . 0011... [Tle = [[P(va)]a Mva)la-.. (Cyn) Ja = [Ivale [vatvale... Wa-atvn|a}=] ois! : 00 0 0 1 0000 0 1 00 0 0 0 Exercise Set 4.5 . If: V —>» V is a linear operator the matrix representation of T relative to a basis B, denoted [T]a