midterm 2 fall 2002 solutions

Linear Algebra with Applications

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Unformatted text preview: Name: (4.0 {we/- k2,? Math 21b Midterm 2 Tuesday, November 19th, 2002 Please circle your section. Tom Judson Andy Engelward Andy Engelward Katherine Visnjic (CA) Jakub Topp (CA) Erin Aylward (CA) MWF 9-10 MWF 10-11 MWF11-12 —-— You have two hours to take this midterm Pace yourself by keeping track of how many problems you have left to go and how much time remains You don't have to answer the problems in any particular order. So move on to another problem if you find you 're stuck and that you are spending too much time on one problem To receive full credit on a problem, you will need to justify your answers carefully - unsubstantiated answers, even if correct, will receive little or no credit (except if the directions for that question specifically say no justification is necessary, such as the True/False). Please be sure to write neatly — illegible answers will also receive little or no credit. If more space is needed, use the back of the previous page to continue your work. Be sure to make a note of that so that the grader knows where to find your answers. You are allowed a half page of notes on it during the test, but you are not allowed to use any other references or calculators during this test. Good luck! Focus and do well! Question 1. (18 points total) True or False (3 points each) No justification is necessary, simply circle T or F for each statement. a F (a) If A is an invertible matrix then the kernels of A and A"1 must be equal. ‘3ch 55ng— 71C 4 if MMCLQ/ fim \Cer (A) ==§53 SD A" .3 Awe/fun. as well] Pug-(AT) :: 233 an) fa TM: Viz—arm]; are. 67an T @ (b) If A is a 5 x 8 matrix then it is possible for the dimension of kernel(A) to equal two. M0) 6‘ LT {‘5‘ka Lu—L— L3 Mmb’C-‘Autki‘j , 518% .. rabk-v-nV‘u: 2“ Y Lil! 0? claim»; a: 4)) Mn 2“: huafil-‘jrl (5'— JP‘ (WNJ(A’))=§. Them male wulc) kw; +6 ‘92. 6/ HL»:LA .3 «Matt; m2, énkm) s a news mnkCA/gflca/s ”,5. Mic gr T ® (c) If a subspace Vof SR3 contains two linearly independent vectors then Vmust contain at least one of the standard basis vectors as well. < MD MSQA kt X“ SLculJ - V MIBLT‘ L‘— G SULSPQCL Spénf‘u) L) P] “J l?) a—a - fulfil; New; Twi— Jfisn‘T— LMA— “:3 l ‘ ‘ ~--\ _.1 A . .0 4a,, g orea M 5t- T CID (d) If the product AB of two n x n matrices is the zero matrix, then BA must also equal the zero matrix. , , . . _ Monummfl‘l‘afiuaLn gJ'mk‘QS Gama - 4:..- a 51"va Cann‘l‘ervcamfiL cans-Jar A: [O 1;] B = Ll (Y o 1 O O F (e) If A is a (square) orthogonal matrix, and the product AB is also orthogonal, then the matrix B must be orthogonal as well. “I: 3 a: uanfm'l‘ on 1kg, 41d“ MT M PMJE‘Z op ‘I'wa / arm aA<l M‘t‘l‘u-LQS .5 «(So OF 0M1. No7“. 7347‘ 2 IS Orbs! M .7 ,. .1 , 1, /" Thea So :3 A =AT [93m {Len wjréqa’ééqyrfl lsA'A’ ‘f— .Ln, ‘5 Tl’ SAAJDLLR/ So A‘lor-Njomll AB 0"»le wit; ”40‘" 'N‘OGL‘cT’ quA'B) =3 “S “S “<1 T @ (f) If ATA =AAT for an nxn matrix A, thenA must be anorthogonal matrix. “0/ 1" Le, Dr‘lkojon-J AT/Q' =1) Lw—r JVJ?’ kaouu‘nv Mr Wfl=AN’A%fl’«H«e~ek«QJ/$ R:- a Ounbaflwpéa/ ConfiJer— A: [23%] / MT O’NJ‘"*/ (wa/urAM=gg=MT - Question 2 (14 points total) Suppose that A = . Find a basis for each of the following subspaces. l 2 3 o 3 3 l l 2 '2 o 2 (a) (6 points) Image(A). Net) +9 5‘) thCA’) +0 Pal pfuoT uhmng 63(- S.’:’\% Mtbta Ingz-u- ipcfléfici. CC m[uMA.S"’ duff? “AS ’1b~ “"L ("19,43 .AJePQAJMT/ 1s.“ Mtg, of. 3 = u/ ”A, 9/1: 0/ / - .x , 0/5: 1.91.)- 4"3‘347/ \xs—Iv—Mm} \O\lo -033'36+3’—°O(l"l ~3011—11 Itlol-(I) o-,-(|-; +1" 00000 3.0:. D. o ~1£1t) 0—7—17 —8’ +911) 000 0° ‘ 77‘ PNJoT—CDKS ‘R— PTA I («[$ "2 «nib-:2 A O > s, flab/24): slom< , J D. . 1 l “ a] H (b) (6 points) Kernel(A) [u o LL91.) min/x) v. Sn )(\+)(3+)(¢1=C> U x3=SJ XLft-t’ X§=W o—MIJ Xl+X3 ‘X‘t'fl‘r :0 -g-t De,“ Erna/[43:6 ':+C'1” {r ngeR 0 1 \ 3: H .g/\ l 2 r—_—_'\ oo-LL l...‘_2 ¥ l o 1 q 53 a has t3 _ I _ (‘9 ' g 4:!" WMA‘) “' it 9Q o ‘ (xv-6:513 a: O I :1) I 3 M TLra. KJd’WS’ b l (c) (2 points) What is the dimension of the kernel (AT)? We. knew 313t- W(AT\=C—va§e—QOC§>LI an) 5.6%. E~1eA,LImafleP\-)J‘ are Lolk kaspaccs at [RH] QJ Jm’n (Eu 0%)): k 9m pvt (a) Man 01.; (Ence)L)=‘/~A=l s. (lawman a? kiuan (4*) Is A. Question 3. (12 points total) (a) (6 points) Let {5], 172, ..., 17," be a set of vectors that span a subspace V. Suppose W is another vector _. in the subspace V. Show that the set of vectors W53, ,172,...,vm is linearly dependent. __L Sykcc- 5". Vi 37"" \/j‘ "linen-x {Chm N {S ix V/ __g .5 -—-‘ —-I . .2 $0M. 0C The, cong'l‘an‘i's (,1 ant. 00AM} So C3— C47: —- COS/7;.” . . . - QM-Ji,‘ = 8 I: c nm‘i‘r‘iw‘ai Nixhen A \ amp/‘3 M 5(1- V—J/ V‘ / O”, a / So I I: c: iuxgfimt‘D )QPQI‘JENT 891. E? d” :5 M \TJ an) an OM wt: 4 ES lmwj Jefenr/en'f' as Cd-fiOv nonzero consTmT <.. (b) (6 points) Suppose that 561, 562,. ..,5Ek is a set of k linearly independent vectors in SR" . Suppose A is an nxn invertible matrix, and thatji = Aic'i for i = 1 to k (i.e. 531 = Ailjz = A552,...jk = Aik ). Show that set of vectors 5?], 552,” .,?k is also linearly independent. \ SMPPOSC. ’tkc, $2.1, 3" flk’ "'J jK VJuSAI-z— Ingahij t‘nABPWxJEnTJ 11...,“ c‘g‘4. c1131... .. '+CK.S‘K =0 Gr- same, SJ 6: Cpnflcn‘rs c,‘ ) W'Cl‘k 41L lea“ anti. C; A0+ earn-[MAS game. l 8;? 1km Mui‘hp‘biaj L3 A“. 32‘,le Adzci 5-:4- -.-+C.4,<5‘K :- O V .. _' _ ‘ GDJ So CqA‘BJfl“ “ 4’C‘k EAR—O) Lui. Stat—L. _.__, ‘4 , n T A'uc'JX'. (3:0er A“ " 1». “3° QRJ+."+<'KX\‘L:O/ GAJ Sl'ncz. 11':— 357— >441“ “/xl‘t ’5 imwb "‘\ £91“; (MT 11‘6“ C4 =- (‘35: ‘hsi S UhTt‘aal‘Qi'S M ‘Fil—Sul~ GSS uM‘D‘l‘Aan ’lluj’ 'i'hfi. 3" , lsgtar'ib thgnden‘r 50 19*. 52]“ 3|: ‘ .J 3 K WT Lgaiclm‘zcienr ‘=C‘K=Q/ L\A1— Then WK Question 4. (16 points total) . . , 1 1 1 1 (a) (10 points) Flnd an orthonormal ba51s for the kernel of the matrlx [I 2 3 4 [i i i o _( ‘1 ‘ C-‘ .o. l = (In? r'r-b 0‘ A 3 0‘ D. 3 Sb x, X314.“ / *L="lX3-3x~1 ‘ 1 basis ~9- W.{:;]l{:;1 Lxerc. we. 3°! G—tNm-SJWQJT' +oiwt. i o \ \ 0 l 5. L ._> ,__._, ‘1 \IJ\ = MT gas ‘3 .— ‘ [‘3] J: L] 2 [f Pm“ i, i J. '4, l ( 373/ ’ l S. cm w‘l‘xoM/‘fiu( S» M unTt' («walk audio? «5 WE‘VE]: 173-: :ll / L‘SLS gr 15*— lCer-MJ Jill: 3 y | D. 3— 'X J— “ '5 Fe ( J 5'; —*t o 3 (b) (6 points) Suppose thatA = BC where B is a 4 by 3 matrix and C is a 3 by 4 matrix. Is it possible for A to be an invertible matrix? If so, give an example (write down matrices A, B and C). If not, explain why not. M0 7]— iSA‘T— P°S§~blb\ wait. “*1- W (C) 72‘ :33 . Mam r‘aAk (C3 S 3 (£255 vi 7F "0ij/ 50 L 3 mi: in ullib=7 am A “11:5 2 4.; (lewd (c)) = L/ - ml: (9) >/ i NexT 00+:— TLC!— aflik-ml :« MM] CC.) :5 C~(So :n ICCI‘M/l (84/ S‘ACL TC C7 :6/ MA clam-[Q BQ—J‘ :A/ 5“ 2p vék‘f'i/ICC)’ V: 13 raise: in ice/we! (BC). 30 kennel (Cl #5 £03 implies kermi (3(4-37E 333/ s. A: g; is Mr ieruo‘fiblt. Question 5. (16 points total) (a) (6 points) Write down a basis for the linear space of all skew-symmetric3 x 3 matrices (recall that A is skew-symmetric if AT = -A), and thus determine the dimension of this space. 5179i” £3 wrii'inj sou-m A jeagr-xl 3x3 Sictd-Sjmm‘l‘N—L m‘i‘rtk. “ look iii: 0 Q E “Q; JL-‘DONJ Lu“ in £70.. BN/. B- M d, e 136‘ O C- ag cscL e“+'13 67"“(5 muffin-E. Vin/K3) —E ‘g o a“) a; L, C. Can be 4.43 AHMEe‘S. So 4 anwt‘Vi 5kg.) -— SUMMET‘PIL waft-ix C-'| '\ L71 kl H‘H’LK “-S ‘- linem— leaika .;a Si: 3 Sde-Sjmmtt‘ic M-d'm'LLS', ('43 rte—In ”\J) o o ( o o o . . 6‘ E3 '02] +- L 00 O] + C..[O O I ‘ Ru! 4 Legus ConxasTS dc oco "00 o—IO + . (3 l O o o l .J a Q o ‘ ' ‘ -l o E O ( he. Tim-<4. Law; “(224 SJmM‘i‘nL MA bag 0 (a 3 l E), 02 0-04 Q a“) so The, JORLASL-cq ,C ’6“: inkcm— Sr“. .5 3 (b) (6 points) Find the dimension of the linear space of all symmetricn x n matrices (recall that A is symmetric if AT = A) Mow which 0(on ’i‘lae. tjPI.CA( WLV‘ 5C 'hAQ. 5 (2.. g Stain/:44. Hi; (\Kh ms‘i'm'us ‘. o‘u‘j‘u ' " qua Moi-L cwnfklxl in, w ’1“; Mam IQ 3:; Ga ‘ A-‘Agonsi ts Jus‘l' a beggar!" a: a age 1 \.¢I .‘3 y. gen-{span “:3 gn‘f‘ q ou-Q. m& can nnmLet‘S “I." am‘ _ J-LuaoA-«i lace-MAR. 6i: Janm‘h‘udij So a. memiru- as: M, (\KA 53mm+ht.c Mel-Ax spam is éefer-mfi‘gc} i) ‘13“. (Jew—2.3 'Qr' a" q‘k' ' ' an > n W';°i:ic5 Q.I,CH~," ' chm j CiIA +A~l v-u-i-LQJ a“) H. qi" . - “1‘" + l ‘. V‘H“;LL¢_ QM . Sb ai'i‘afichwr‘ 'hnu—c «r-L. n+(n-a) +... +[ = A; A...“ "\‘Je—Pv‘w V4¢NALL2J \ . - ~ _ k ~ ~ JUL‘MW‘E- ‘ SWSL‘ “V“ YDMMV“ ”‘fi‘x \a 3. 11-1: equals 11». Jimenhén. (c) (4 points) If A is an n x n symmetric matrix, then is A2 necessarily symmetnc as well? Explain why or why not. SMFC. - - - we, (.Arx (Acck Rifle M1. US$13 CtMpoM/‘TS : Question 6. (10 points total) Use the method of least squares to find the linear function y = mx + b that best fits the following data: y-1216 £74133 TR. 477: “gr J93 ‘lb 71:. equdtln j=MX+L 3mg u: £9!" qum‘lxoms‘. _(=M(—6)+lb ~é l “I L=-(“~)*‘L °’ 0" [I]: l {:M<l) +5 I l l é=m(¥) +L 1* ' E A :2 =5 an hams-idea?“ spilt/w. . q T mrmx/ ever-slum ATfl—X : A la , .. 7:41-17- :6‘_oc3 #51 [(14) Rig-Z: I1] 1 1 AT = [76 1‘ I UH {1;} é o M LIS- Jan. Dita ”a; 5:: now S°lv¢ (q: ‘1][5] : [g] / r ‘ ‘ a ‘A slug... +0 C‘xlcVJc‘l‘L (Arfijd' GS 71V:- Sash/v1 LS 5 8-353 S -_L 7:: fiOM:‘7/5v «J ‘74:?3 85 Mga/ A 6") +1"; lage-V‘ 7411\974‘44 75m?" 4337‘ {'6’ +52. Jfilcct - _._L_ ,5 3... B~X+§x Question 7. (14 points total) (a) (10 points) Let T: V——> Vbe a linear transformation, where Vis a linear space. Suppose that kemel(Tz) = kemel(T). Show that it must also be the case that kemel(T3) = kemel(Tz). As before, a good strategy is to first show that kemel(Tz) g kemel(T3), and then show that kemel(T3) g kemel(T2). (Note, Vis a general linear space, so don't assume that Vis equal to ER“ in your answer). pc’f‘ 34‘0“) ku- (T‘A‘) é ku— (Te) . I? X é It?» (TA) ‘l'l-tvx TVxFO, La— The“ TKCX) = 7(Txlx» = TCO'FO J So X 6 ker’ (T3) <5 we,“ . So kw” (T\) $- LCU" (1—3) (\k‘d- S‘L‘W‘ ker‘éTBJ éku‘éfl‘), This .1: LthcP/ Slaw-1- u-‘fk an element“ x .5— kc" LT"), So Ts @‘lzo. L—Cf‘ TBKKB szZ'l/CKX- Here‘s wiser; TM. «ss«mp‘l';o« 0L0“? ker- L1") = leer (1“) am in sham TKO-04)) :0). M m) e lemme), w wwkkmum, a .HC chnxb=o fken So Joe: TCFCX» =0/ ya T\(X)‘—=O. M f xeww) So mm a“ m) =6, 50 xe kernel (Th) / So kemldfij g. kerne( 0.x). 71“ WW] 04} :- Puma! 0-3) (b) (4 points) Give an example of a nonzero linear transformation T from ‘33 to SR3 such that kemel(T2) is equal to kemel(T). E453 “ASWU—u- TCX‘) = E 32 / 6.5 lith-scl (:3) 2: £53 AAJ S'RCR— (6E3),L : I3 ‘t‘wvx kv (TA) 2 £33 «S wall Hfir-th' +° 523—.” 'I'F T :IA'T' 1&2. :‘Jen‘l‘nu T‘Pqflfpbl‘fi‘hén 11cm "lake, :2 look «if (a Ohfhaaan-s/ pNJ‘QKJLlo/z/ SkcL a5 o-n+D ‘hat. X5 CflO(‘JH\\-\‘('{. plfi-AQ.«' kw = 1“”(‘3/ bur 1': jOU :39 M orbxoaamx! pMch‘A;- “fwlkm, 112, 513mg}. mpuT dc flag, Tmarporwm‘lkc‘m has M cRS—T- ~ Intense! 43 g1...” Z‘AX:S. agclc H‘sh‘ M+Ptezj I TZE):E)OI 00:12)...
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