MATH 218 Problems Solution Chapter 6.1, Spring 2012

138 Chapter 6 Inner Product Spaces WM Solutions to Exercises L. we v = (0)(1) + (1)(—1) + (8)(2) = 5 3. 3 ortaman| Lo | =0+1-12=-11 -4 5. || u l= V1? +5? = V26 7. Divide each component of the vector by the 1 . . 1 is ' norm of the vector, so that Te 5 is a unit vector in the direction of u. wo | 2 | 0. | LL. | a = (3)? + (2)? +8 = 29 17. Since two vectors in R? are orthogonal if and 8 1 = 0, gives -c+6 = 0, that is, c= 6. only if their dot product is zero, solving e ‘ 2 19. The pairs of vectors with dot product equal- ing 0 are vz-Lv2, Vilva, Vi-LVs, V2-Lv3, Vsiva, and vglvs5. 21, Since vg = —vj1, the vectors vy and vg are in opposite directions. Q, wy — O-146 =& vv [+1+4 ~~ 6 1 1 . — O+1-9 — 8 4. aw = Theo 1 |=-i| 1 3 —3 -1 ~1 : vm-n-lfa lal 8. Since cosé = pHing = Jae then the vec- tors are not orthogonal, 10. The vector w is orthogonal to u and v if and only if w+ 5we = 0 and 2w, + we = 0, that is, wy =0= wa. 12. ju ~v|| = /(u—v)+(u—v) 14, Since cos@ = ~aive # 0, then the vectors are not orthogonal. 16. A vector w is orthogonal to both vectors if and only if Sur = 2u2 + 3g = 0 —W1 — We — 8s = 0 wy = 9w3, We = —12ws. So all vectors in 9 span —12 are orthogonal to the two vec- 1 tors. -1 0 18. e : 2 =0+2c-2=0e6c=1 2 -1 20. The vectors v2 and vy are in the same direc- tion. 1 i 22. |Ivall = (4 +44 4 i i i 1 i | 6.1 The Dot Product on IR" 139 26. w = u. v 29. Let u be a vector in span{u1,Ug,**+ ,Un}. Then there exist scalars ¢1,c2,-++ ,¢p such that u = CU + CQUg +11 + Cnn. Using the distributive property of the dot product gives veu=v-(cyuy + cola +... + Cnn) = e1(v- uz) + ¢o(v+ ua) + +++ + en(v + Un) = ¢1(0)+ €2(0) Foe + Cn (0) = 0, 30. If u and w are in 9 and cis a scalar, then (utew):v=uivtc(w:v)=0+0=0 and hence, S is a subspace. 140 . Chapter 6 Inner Product Spaces 31. Consider the equation e1vi + cava tet + enVa = 0. We need to show that the only solution to this equation is the trivial solution 0) = Cg = 1+ = Cn = 0, Since vi (civ + c2v2 + +++ + CnVn) = v2 ‘0, we have that er(va V1) + ¢2(va V2) to0 + env 'Vn) = 0. Since § is an orthogonal set of vectors, vi-vj = 0, whenever # # j, so this last equation reduces to ex||val|? = 0. Now since the vectors are nonzero their lengths are positive, so ||v1|| #0 and hence, ¢, = 0, In a similar. way we have that co = cg = +++ = Cn = 0. Hence, S is linearly independent, 32. Since AA7! = I, then hy AKO, =O fort # j. 33. For any vector w, the square of the norm and the dot product are related by the equation ||w||? = wew. Then applying this to the vectors u + v and u — v gives : \lu-+ vil? + Ila — vi? = (atv) - (atv) + (uv) (av) =u-ut2u:v+vivturu-2u:vtvev = 2|[ul|? + 2I[v|)?. 1 34, a. The normal vector to the plane, n = 2 | , is orthogonal to every vector in the plane. -1 12 -1 bA=|0 0 0 00 0 35, If the column vectors of A form an orthogonal set, then the row vectors of A’ are orthogonal to the column vectors of A. Consequently, (A‘A)ij = 0 if i 4 j. If i= 7, then (AtA)iy = || Au]|?. Thus, Aa? 0 oe 0 AtA = 0 |jAai[?_ 0 ; : 0 a 0 0 a 0 Anil? 36. First. notice that u: (Av) = u'(Av). So u: (Av) = u'(Av) = (u'A)v = (Abu)*v = (Atu) -v. - 87. Suppose that (Au): v = u- (Av) for all u and v in R”. By Exercise 36, u+ (Av) = (A’u) -v. Thus, (Atu) -v = (Au) +v for all u and v in R®, Let u =e and v = ej, so (A*);; = Ajj. Hence A! == A, so A is symmetric. For the converse, suppose that A = A’, Then by Exercise 36, u: (Ay) = (A‘u)-v = (Au) -v. Exercise Set 6.2 Inner products are generalizations of the dot product on the Euclidean spaces. An inner product on the vector space V is a mapping from V x V, that is the input is a pair of vectors, to R, so the output is a number. An inner product must satisfy all the same properties of the dot product discussed in Section 6.1. So to