7th - EXERCISE SET 5.2 [I ti El- 0

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Unformatted text preview: EXERCISE SET 5.2 [I ti El- 0 F]—f[?]—irl]+[—l]=reg1ud63=[gld[fl—h[fl:|—b|::]=TE3;thlfi{T-r]= I) 0 Li H l —l —1 — —1CI I'J [0—1 III]. 11' lIii—1 lfi. {a} r" 26. The matrix A is a rotation rnatrhr since it is orthogonal and det{A] = l. The axis of rotation is —1 I) .1: found by solving the system [I - A}: = ti, which is [ fl 1 —1:| [a] = Fl. A general solution -1 {I I. s o r of this system is x = [a] = t [I]; thus the axis of rotation is the line passing through the origin 1 and the point (111.1}. The plane passing through the origin that is perpendicular to this line has equation : +y + z = fl, and w = {4,1113} is a vector in this plane. Writing w in eolurnn vector form, we have -1 [I 1 {l —'l l l w = 1 1 Aw = I} {l l l = IJ , w x Aw = l [i 1 Li I] [l' —1 1 Thus the rotation angle 5 relative to the orientation u = w x Aw = [11 1,1} is determined by eosfi' = fi‘fi” = JEL, and so 9: “T'ilflflfi. D3. The two columns of this matrix are orthogonal for any values of a and b. Thus1 for the matrix to he I::-rthop_1:onal_.2 all that is required is that the column vectors he of length 1. Thus a and 5 must eatrefy [a + tr} + (a — M2 = l, or [eqtuvalentlfl 2n2 + 2.53 = I. Dr. From the polarization identity. we have x . y = HM}: + y"? — “x — gin”) = the m o = s. EXERCISE SET 6.3 . . 1 r x E. The kernel of T Is equal to the solution spaoe of the system [I “I [y] = [3]. The augmented ii'..' . . 1 a i '0 matrix of this system can he reduced to [0 l D E 0]; thus keriT} consists of all vectors of the form —] $=t[ 0] where—meteor). 1 12. The vector w is in the range of the linear operator T if and on];r if the linear system 3—3; = i I+y+z= 2 5..“ +2z=—1 is consistent. The augmented matrix of this system can he reduoed to long i 010:; stung Thusthe.stemhasauniue [' =1g_§ _ T q 50L].th H: agar all! and “'2' have Ell-L .—s = 19. {a} The linear transformation T; : R: —> RE ha oneto-one if and only if the linear system An = II} has only the trivial solution. The augmented matrix of Ar: = {I is 1-1:e s sin s-Me and the redunei rot-tr echelon form of this matrix is 1 1] I] l} I] I} all-l: From this we ooneluele that A: = D has onl}r the trivial solutionT and so T1 is one-to-one. {In} The augmented matrix of the system Ax = I} is [ 1 2 s 1 I] —1 l} -4 i [I and the reduced row echelon form of this matrix is 1 o 4 : o o 1 —% i o "-4 From this are eonelude that Ax = [I has the general solution x = t i . In particular, the 1 spatem Ax = D has nontrivial solutions and so the transformation TA : R3 -r R2 is not one- to—nne. D1. {a} True. If T is oneto-one, then T): = a if and oni},r if x = I]; thus Tin — vi = 0 implies n — tr 2 fl and u = o. (h) True. If T -. R“ —t R" is onto1 then {from Theorem $3.14} it is one-to—one and so the argu- ment givmi in part {a} applies. {c} True. See Theorem 6.3.1.5. (:1) True. If TA is not oneto—one. then the heterogeneous linear system Air = D has infinitelyr many nontrivial solutions. e True. The standard matrix of a shear operator T is of the form A = l k or A = J u . In e 1 k 1 either case, we hate det{A} = 1 of El and so ’1' = TA is one-to—one. P1. If B}: = ID. then [213]): = Elfin} = At] = I}; thus x is in the nullspaoe of AB. ...
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7th - EXERCISE SET 5.2 [I ti El- 0

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