math294_hw6fa06 - 52 16. Yen may be able te find the...

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Unformatted text preview: 52 16. Yen may be able te find the eelntiene by educated gueeeing. Here is the systematic eppreeeh: we firet find all vectere :E that are erthegenei te 171., fig, and '33, then we identify the unit vectere eineng thein. Finding the vectere if with :if - "'1 = :E * {£2 = f - i323 : 0 eineunte te eelving the system (we can emit all the ceefficiente 1:; t ' H —f The eelntiene ere ef the ferin :c = = :53 -f 134 t Since “if” = 2|t|, we have 3. unit 1eecter if t = a: er t = —-%. Thus there are twe peeeihle eheicee fer ‘54: genel, but tllfly are net unit vectere: 26. The twe given vectere spanning the enhepece ere erthe the euhepece, we diwde heth have length 7. Te ebtein en erthenerinel heeie £1,112 ef by 7: 2 3 31:; 3 ,e;:% #6 6 2 49 New we can use Feet 5.1.5,r with 1i," 2 49 49 2 3 1i! preij = (“11.1 - f)fi] + (“l-£2 * f)fi2 = 11 3 “A *6 = 39 6 2 64 28. Since the three given vectere in the subspace ere erthegenel, we have the erthenerinel basis 1 1 1 e 1 1 e 1 1 i. _ l —1 “1:5 1 rug—i _1 13F2 F1 1 —1 1 New we can nee Feet 5 1 5, with if = El prejvzif = (nil - f)fl'1 + (112 * flfig + (1—53 5W3 = 3 l 1 4 #1 —1 -.L .. .. - _ U _ ”2_[u1'fi2}E1 1 O H ——- +_. —— _. _. _‘ __ = 2 {I112 H t’s-(“rflelfll 7'5 1 U In Exercises 15—28, we will use the results ef Exercises 1—14 (nete. that Exercise A1, where A: = 1, . . . , 14, gives the QR fscterizeticn cf Exercise (k: + 14)). We can set Q = [11' 1 .. . Em]; the entries cf R are T11 = "‘51" T22 = W?“ = “‘52 — (1-51 *fiflifilli r33 = “173+” = “173 r (131 @9111 — (E2 ' 53%“ n}.- = ii} #13}, where r' «C j. 4.....- 1 72' U __1._ vi 0 34. rref(r=1)=[1 D —l _2] 0 1 2 3 1 2 . . _. —2 _. '—3 A beers cf ker(A) is 111 = 1 , v2 = O 0 1 We apply the Grain-Schmidt precess and ebtain l 11' — 1 s — 1 ‘2 1 — Tell: 1 _ 75 1 U 2 —u-J,_ —: -I- -I- -I- —- _. __F vi __, ve—{'fll-v2}fll _ 1 1 “2 _ H132 _ IIfiE-{firfis “Ell _ :30 —4 513 28. Write L(:i:') = ALE; by Definitien 5.3.1, A is an erthogenal fl )5: in matrix, segiat 141114;}... by Fact 5.3.7. New L(fi) - LEE) = (At—f) ' (A15) = (Afi)TAtEi = fiTA Aw = an Inn: 2 e at = 17 * iii, as claimed. Note that we have used Facts 5.3.6 and 5.3.9a. 1 0 32. a. Ne! As a ceunterexaniple. censider A : U 1 (see Exercise 30). {J 0 I). Yes! Mere generally, if A and B are n X n matrices such that BA 2 I“. then AB = I”, by Fact 2.4.9c. 1 1 s s 75 7e 36. Let the third celunln be the cress precinct ef the first twe: A 2 fi: — 12 7% 1 fl _ 4 3 Is There is anether selntien1 with the signs in the last cehnnn reversed. R 4‘2. a. Snppese we are prejecting ente a subspace W ef IR”. Since .437? is in “Bairearly. the erthegenal prejectien ef ALF ente IV is just .45? itself: A(A:r) = ALE, er A :r : Arr. Since this eeuatien helds fer all :ir', we have A2 : A. b. A = QQT. fer seine matrix Q with erthenerinal cehnnns 1T 1 , . .'.‘, EMF}; Nete that QT :2 1.”, since the ijth entry ef QTQ is 11‘. - it... Then A2 : )QIQQ = Q(Q Q)Q = (21....0'1" = QC.)T = A. 60. Using Fact 4.3.2, we find the matrix scan—- Czar-It: GHQ: was: ...
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math294_hw6fa06 - 52 16. Yen may be able te find the...

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