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math103hw7solns

math103hw7solns - ~2ch Ax x0:9 V i o grjdli(W las‘r WNW...

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Unformatted text preview: ~2ch§ Ax x0 :9 V: i o grjdli (W las‘r WNW MM 9% 110 M La. ’6 mhlx WWW M4“) BUT V1 x= 0 93¢ MU: v1:1,)“ @ $6 sgva-LBJ’ ' 59/ Ian/0r ab)”; j ”Wits; HM“, LanQ 2. Letia ERandeetlm lwpeplal amydfiledb P={:1'ER | +---+£znrn=d}. 8110 011-1 that ‘th32 h01:it.9stdsta.11€:00b tween Pa 11121 the 1:0igi11is 71—. ‘Hmr: Firstchoosen=2, 31+...a3‘ ' anddrauiapcture Then twa angd alize Your argument; "J “1' “VI/d— 321(3me 1 5"“:(«1 I“ ievee‘ifi’zol. Val/6949 Mnlwmiyfl WI’W (MG/:17 ’5 I“; §:_CL,Z7/ 5mg 3. {a} Let B = {611. “16”} be an orthonormal basis of R”. Let 13;; 6 JR”: and 0:11. “foam 9311- - H.311 be such that :E' = Zia—'1'?!- and 33' = 21. 81-171}. Show that :E"- 33' = 21-91-93}. :No- tide that {5:13 = 5', and {m3 = 3. This problem says that the dot product E- 3', where E and y" are in standard coordinates is the same as (5)3 - (5’33, where coordinates are in basis B {b) Give an example to show that E'- Q‘% {EJB- {3)}; if B is not orthonormal. @gy lOI'L—mrlly of WI. 060* erM) film 71% C? W) EZWG‘ " “3‘ J:\ gk/l' V2” . VJ : i” — 6:) yNWO i“£~?\: . O ‘E 5:1 M (me; (We; ‘5 (M): o. 4. {21) Le tAbeaI11mxn. nat1x 31112156137”. LetV=i1 111A][111etet11atVisasubspace BERT"): andsuppeseLker Ar: '{[1}. Show that PVHJJI= fish” A)_ “A5. HINI: Leas squares.._: {bjLMit{1_.W}beabassofson subspaceVCRm_aI11dAbet1emxn11313131131111t1 vector rs L"- s as colu111111s. Fi1 1d 31 111atr rix M [i11ter11111s of the 111131; rix A) such that Py—[bj = M’b. :This tells you how to find the matrix of an orthogonal projection when you have a basis which is not orthogonal. In reality however, it is better to use Gram-Schmidt to convert your basis to an orthonormal basis, and then use the formula from class— @109, New 1’29"?va 6 VL‘: MAY. 3 WAT/57 Ami—11117)) : 5’ a» ATM”): AT‘E’ Swag PVC-C") é V=ivV\/°Y/ we, knew RICE»): A)? “1:9’1 7’3 ® SW‘CJL V: 47me:75: i‘M/‘i/ 110» Jove. @9ij M: A (AMY/1" g; 4-. Not orthogonal, the first and third oolnmn vectorh fail to he perpendicular to each other. 1 U _ 32. a. No! As acounterexample, considerA = [0 I] (see Exercise 30}. 0 Ill b. Yes! More generally. if A and B are n x n. matrices such that BA :— Ifi, then AB 2-1., by Fact 2.4.9e. ' -' ' d per triangular, then A” is both lower . _ upper triangular {hein the inverse of an triangular matrix; compare with Exerdse 2.3.35c], g upper Therefore, A'1 2 AT ' with pesitive diagonal entries, all the diag b. Using the terminology Suggested'in the hint We ohse , r th t Fact 5.3.4) and Rgflfl ' ' “5 3- Q the ifltttl‘bc leo, = R23? is it... so that Q1 = Q2 and n, s. a. By Fact 5.4.6, hm?) = {ATAJ-lATtt The transformation 45+ is linear sinoe it is “given by a matrix,” by Definition 2.1.1. b. If L {and therefore A} is invertible, then Lfig’) = n-lmTrIAT" = 44—14 = L-Ig; so that I.“ = If]. e. L+(L{fi]) = ( the unique leastsquares solution it of Ht?) 2 MED = f. d. L twin 2 A ATA 4.4%}: proj ,s, Where V = im .4), by the 5.4.7. 17 1 0 1 0 o ' e.HereA= 0 1 .ThenL+[gT}={ATA]_1AT"=[O 1 OFF. 0 0 . . 10. a. If :E is an arbitrary solution of the system A3? = 3, let it}, = projvii, where V = kerfiA), and :50 = 53'» projvi". Note that 5: A5:- Afiit + 554]} 2 A55 + Air}; = Ania, sinoe :3}, is in lint-(A). b. If '30 and :31 are two solutions of the system As? = 3, both from (kerAJ-L, then :31 _— 50 is in the subspace {ken-13H- as well. Also, Afil — £0) = A3; —- Afin = 5 — 3: 5, so that 5:} —- in. is in kerfA). By Fact 5.1.813, it follows that it} — £1] = ll, or E1 = is, as claimed. (3. Write 573', = 3'2}. + in as in part a; note that f}, is orthogonal to Ed. The claim now follows from the Pythagorean Theorem [Fact 5.1.9). . —1 2D. Usmg Fact 5.4.5, we find if" = [3] and 5‘. At? t [ i ' ' 1 Note that ,5, As?“ is perpendicular to the two columns of A. ...
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