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# LinearModelingforEngineers MATH290

Drexel

GPA 3.63

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This 242 page Class Notes was uploaded by Moshe Swift III on Wednesday September 23, 2015. The Class Notes belongs to MATH290 at Drexel University taught by Staff in Fall. Since its upload, it has received 8 views. For similar materials see /class/212294/math290-drexel-university in Mathematics (M) at Drexel University.

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Date Created: 09/23/15

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J 2quot WK SandiRS CSXLJB F A Ma y rWMQJ39R Cream a QUL Gun A I aL QLI 01 039 am a 39 4499 0 am 4 Earn Mk 00 O 0039o at 0 A EB I Aamp B an TL 4w Wee WM 9 ABBA GD AJ CBec3Argc CD A4 05A 69 rCMm grAHB CchMA YAesA EL 69 mm CmA A LY Lo L A 3A L1 gw 1x 5 39LVDI393 MW jg A U m 2 8Com fame gfgtquotjf Him ms LN A15quot AM sup rg RB 51 w l 1 23 ii A i L B L g 96quot c 3X C M s we26m ca gm ab M L s a A L E M ML A E 3 33 193 AM L WU L 21 em a 2 g r I A m5i gt1 4 sx3 W A mxu anp 1335 v GERquot w r5 MCyre 6 a n quot 4h 42 QC 39 39 3 62 3 133 auegmxgw 61 C CMLMS Recall A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix ie if A PDPl where P is invertible and D is a diagonal matrix In this case Al O PV1 VnD E 5 0 An V1 39 39 39 7 Vn are the eigenveotors ofA corresponding to the eigenvalues A1 An 56 Discrete Dynamical Systems Xk 1 AXk X given initial vector Assume A is diagonalizable Let V1 39 39 39 Vn be 71 linearly independent eigenvectors of A corresponding to the eigenvalues A1 7 An respectively Then 8 2 V1 Vn a basis for R LGt C1 X013 39 Then C X0 C1V1 39 CnVn Cl1V1 I I CnAnVn 01A12V1 39 39 39 Cnn2Vn 3 gt4 I gt4 M II In general xk amnm o Cami6v Note If Aj lt 1 then Agk z 0 for large values of k Example Discuss the longterm behavior of the system Xk1 AXk where 9 0 39 2 39 A o 6 X0 339 Solution eigenvalues 1 971 2 ei envectors 9 V1 0 7 V2 1 C1 In eneral let X g 0 C2 Then X0 01V1 02V2 and Xk Cl9kV1 C26kV2 Thus xk a 0 as k a oo We say that the origin is an attractor of the dynamical system and the direction of the greatest attraction is the line through the origin and the vector V2 4 8 E L t 39 39 xamp e e A 16 28 A PDP l where p IMI I I IH I Discuss the longterm behavior of the solutions to x ii i Note that A1 A2 gt 1 Xk 6112kV1 022kV2 Since V2 corresponds to the larger eigenvalue if we take X0 2 V2 then Xk tends away from the origin the fastest We say that the origin is a repellor of the dynamical system and the direction of greatest repulsion is the line through the origin and the vector V2 What if some eigenvalues have absolute value greater than 1 and the others less than 1 Answer The origin is a saddle point of the system That is some solutions tend toward the origin and some solutions tend away from the origin Example Let A be a 2X2 matrix with eigenvalues 2 and Z and corresponding eigenvectors V1 2 respectively Consider the difference equation Xk1 Axk XOB 3 V1 V2 39 1 Find a formula for Kit in terms of V1 V2 2 Find an X0 for which a Xk tends to the origin b Xk tends away from the origin 1 1 and V2 1 2 1O Trajectories of Solutions For the dynamical system Xk1 Axle the graph of X0 X1 X2 is called a trajectory of the system 11 Examples Use MATLAB Sketch some trajectories for the following dynamical systems 1 Xkl Z Axk A 06 A A 4 8 A Xk1AXk A 16 3 A 1LX 11 39 61 I 12 8 J 12 4 Xk1 AXk A L 56 3 11 1 J 13 Considerthe system Xk1 Axk with 3 4 3 11 repellor or a saddle point of the system A Is the origin an attractor a 1 attractor 2 repellor 3 saddle point 14 57 Applications to Differential Equations For a system of differential equations 3 a111 m392 a211 37 a7115171 m all a1nn 127151771 l arminn we have an equivalent matrix system a1 Easy Case When A is diagonal O 5 O 7 AU u This is a decoupled system 16 For a decoupled system X AX 1 0 0 39 mm A A2 I 7 X 2 General Solution 1 X CleAlt 5 Cneknt O 17 Decoupling a Dynamical System X AX APDP 1P v1vn D change of variable system becomes decoupled General solution Going back to X I General solution of orig system 18 Example Solve the system az l Cl 2 21 1Ox2 21 72 19 For a dynamical system X AX with real eigenvalues A1 o An 1 If all the eigenvalues are negative then the origin is called an attractor or sink of the system Direction of greatest attraction 2 If all the eigenvalues are positive then the origin is called a repellor or source of the system Direction of greatest repulsion 3 lfA has some positive and some negative eigenvalues then the origin is called a saddle point of the system 21 Example Solve X AX A of the system X AX 2 0 0 1 3 o 2 2 1 Is the origin an attractor a repellor or a saddle point 7 XO l C 1 22 What if the eigenvalues are complex nonreal Example Solve XI AX 7 10 A 4 5 Eigenvalues 1 273 1 2239 Eigenveotors 3 z 3 H 2 2 The real part of each eigenvalue is negative The solutions spiral toward the origin 56 Discrete Dynamical Systems Xk 1 AXk X0 given initial vector A1 O Write 0 A X0 ClVl I I CnVn The X1 Cl1V1 I CnAnVn X2 01A12V1 39 39 39 C nn2Vn etC39 In general Xk Cl1kV1 quotIquot 39 39 CnnkVn If IAJl lt 1 for allj then the origin is an attractor of the system If Aj gt 1 for allj then the origin is a regellor of the system If Aj lt 1 for some eigenvalues and lAJl gt 1 for others then the origin is a saddle point of the system 1 Note For complex eigenvalues A a z b a2 b2 57 Applications to Differential Equations For a system of differential equations 3 a111 m392 a211 37 a7115171 m all a1nn 127151771 l arminn we have an equivalent matrix system a1 3 For a decoupled system X AX 1 0 0 39 mm A A2 I 7 X 2 General Solution 1 X CleAlt 5 Cneknt O For the general case X AX APDP 1P v1vn D General Solution If an initial condition is given ie if xO is specified then we can solve for the constants Cl 39 39 7 On to find a particular solution Example Consider the system d3 d tl 1 42 d3 1 Rewrite this system in matrix form X AX 2 Write down the general solution of the system 3 Is the origin an attractor a repellor or a saddle point of the system 4 Solve the system if 1O 2 2320 2 1 For a dynamical system X AX with real eigenvalues A1 o An 1 If all the eigenvalues are negative then the origin is called an attractor or sink of the system Direction of greatest attraction 2 If all the eigenvalues are positive then the origin is called a repellor or source of the system Direction of greatest repulsion 3 lfA has some positive and some negative eigenvalues then the origin is called a saddle point of the system Example Solve X AX 2 0 0 0 A 1 3 O XO 2 2 2 1 1 Is the origin an attractor a repellor or a sagidleboint of the system X AX 10 What if the eigenvalues are complex nonreal Answer Look at the real parts of the eigenvalues If all the eigenvalues have negative real parts then the origin is an attractor of the system If all the eigenvalues have positive real parts then the origin is a repellor of the system If some eigenvalues have positive real part and others have negative real part then the origin is a saddle point of the system 11 Example Solve XI AX 7 10 A 4 5 Eigenvalues 1 273 1 2239 Eigenveotors 3 z 3 H 2 2 The real part of each eigenvalue is negative The solutions spiral toward the origin Questions about the Review Problems or any general questions about the exam 13 IfA is 25x25 and rank A 25 then 1 O is an eigenvalue ofA 2 O is not an eigenvalue of A 14 lfA is 25x32 then the largest possible value of rank A is 1 25 2 32 15 IfA and B are 4x4 and rank A 2 can rank AB be greater than 2 1 Yes 2 No 16 For a dynamical system Xk1 AXk where A 1 1 1 1 Is 2x2 With eigenvalues i 7L 2 22 1 the origin is an attractor of the system 2 the origin is a repellor of the system 3 the origin is a saddle point of the system For a system X AX where A is 2x2 with 1 10 1 1n eigenvalues i 7 T 1 the origin is an attractor of the system 2 the origin is a repellor of the system 3 the origin is a saddle point of the system 18 annqu r 1 2 WNW Jx r15 I WU 3 m j ida H WKS 3 M 3 rhw W 71HMW Van WNW L max a W Jr Wm W 01 3430 Fv I N at Wm 0297 0 7 raw11 4 Aqviuv w r as RL blunw gm E m w g m qlsxsgx rk x53 mgt Re CPS 35h ww ck lt4 v mFANMeNux O iLraHngTiif b 1 4 Egg asbprux g r iggtgt5j Vyv v mp Tw wa WV vmvmw J WEE L KW W m 1913 Alb H 11b dzi7ltg1 I 11 LE 4 1AL IA 17 71i7 bqAEA V l 3 Ult H 711L1 Lan 3392 5 1 39L 1 11 z L 1quot1 1 l L UAW FA m m c E WARM JL L J U73ch SWLMH 511 l J L V L ITL ll m 39z LL 7 3W1 QC J4wSTM mm 62 g a QWth my g A755 5 miwk HIEw H A J 32 14R AkmwB task ELM Mr w L3 a V m i rm Ffj nitu m b u 9 5L akirrra gt5 gt3 my ii g Egtya zf asxmyi gp r A L Mariam Kmgtgt QJ W TWIP O I L m TOTd 436 PKvacW N m if mt m6t 95qu F E 3 N9 3 wiry W am my P xyvxvvuh m a L1 3 o 3 2 gt 7 Lo 3 39 3 fji w 39L 1q 56M g 9 Ll fOBQ L A r X H F 391 v J r a n m 0 0 00 2 ac F V i J 7 JJ L J we NM 5 3 E if L h auguno E f L r C2 5 PEEK 25 13on C DANHll 9 j f mv ax 2 A Em CAI nxrfvul R JD P by Ta ruu no DX TE f f 3 Er 1 Ar Frfmf s E L Ax 7 x e e W VA mgvm3 u 6G W q n CVJ 3 h m A R d 36 0U hIN 4 Tu c 9 All rHVJUoU u e g 3 H Q nmfwy jgi CJV a wasp 44L 3 it 6 i 3 3 36 Jn XJ6 Rm si ke f in v q 7 Lk m m H Um 3 t LEWK 9WNquotWLgt FIE MbSquLR 5 gs4 9er ATA ll1 lv3x I Lil LL GMAQ 9amp6 FLA h r WM wg lt quot7J 392 Why H vhf Co CCMx Vquot g Ca 439 CI X4quot 4 CMXM grL lcoclkzch l I We WT MI9TM Era 4D m l39 39l Ca Y K C L Yh 3quot Cam I H h CD Ctr arrmm lIA A inf 839 L1 L 439 I O 7 m 7 s 4 quot 2 a l VLV h 7 h 0 O 3 I l h K5 2 K 1 7f TWLs wq 39 S 0J W WELL gig 5393 rxLDf 9 5 m w m ms 711gf71 u 70 and MW 9lhx3 q a q 3 L 0 O 7031C Co 6 w u 1quot rv a L 0 01 C w L qg SW 0 1 0944 qM mL o WW W J u L 621939 chfx 4 D39Lsx Mm 7010 Led vb lb Note Title 11162006 A No 9ch Twig Em 2 Miami 4mm 5 PmLJ Mam 5 s Vrb ffamf Iftkwo Exam Dec15J 39bVV1 5 AdaL w rulf I 139in Ejoeigul r Side f 95 kMB a mm qume KL M a u 590 MM Xn I san xl7t f 39thqc WW xtm m J m4 TM Mix aw 39 Xlllk QM 0m QM Ki 391 9 aw 3 x49 cam QM fUc A ilk b x K E ugtxE inf XS 3 mix can gt S OmrmL x v 3 m 20 o r E O y i x H D M 4 xv have m T15 J x for 4 MN u Luis NAWHLM X E u v x C K E f er E u mm 57 5 9W gt9 Ct GAFR39 2 QCAlf 3C7JC3 Cl 3 1 COA Ct 39 n 3006 5 Lu phi Me m avwwsem W Mt G sO ampo 29 962 ffAV7Q A 1 o a f AL 3739 L39Y v quot 17 4k 8 q a XGELCle ave b Jrcae HT gm 39W WW1 1 cm UP 39Fv CIJC LC 5 me if F83 nm a w Pm MWW 9 pf gm 3 4 grip fax lvHQLM v garLP M M L u w a kiss ME P a r P 954 T is wag nl IJl u Ha mfg WV mvoim Qixo m 0 0 9 mu 3 3 n E U 3 5 Rag L 5 A ummlmxv H GARbug n yevanSmy mM N mfg 8 A main WEW m KP 1 A xumx8xv fk u 53 3 x U m N O V g r h N E king SE V ngyxwx n PM Hmy n0yltuxyi V E 1o f y gtng u K 39 L E SO A7IL I 3A Lgt Ani V quot0 LVD 13 gt9th 0 d 56771 4 0 Seq o 8 ek mfwao Rn 197 m 9 fig 93 A53 in am 540 6 Vrv f 1 A HOFQVZ ltVF1T DSNVIXI W 91 U a CT 5 29 TS KP mmmix7 q 3130 also xvi s ghda Hm gtnu Pb of gtm f R X63Wx r gm 43 10 I E A CV 03 io Z J r SE35 ltf7e gt2 W 8M3 194 OYijM do 1 I VQ U bd Tit gt 7 GM 91m CL U 439 CL Gift7 05 77 7 7 A 11 f 4 r 3 L V1 A531vi 01 l 1 L73 0 cc A 9 6 GU 12quot c737 vai T g I W TyeV2 5 CU C C3 he SEGA W93 w 961 am am m CL 0 CL C1 0 MLgum y th c4436th i ok 3 M 6 or 28 Subspaces of R A subspace of R is any subset H of R satisfying 1 The zero vector of R is in H 2 Foreach 11V in H uv is in H 3 For each u in H and scalar C 11 is in H Example Let V17 V29 39 39Vp be vectors in R Let H Spanv1V2 Vp Then H is a subspace of R Example Let S xanyrealno a 1 Is S a subspace of R2 Column Space and Null Space of a Matrix A m X n The column space of A denoted ColAis the set of all linear combinations of the columns of A That is if A 211 an then The null space of A denoted NulA is the set of all solutions to AX 0 The null space of A denoted NulA is the set of all solutions to AX 0 Remark N ulA is a subspace of Proof Basis Let H be a subspace of R A set 8 2 b1 bp is called a basis forH if 1 l3 spans H 2 I3 is a linearly independent set Basis Let H be a subspace of R A set 8 2 b1 bp is called a basis forH if 1 l3 spans H 2 I3 is a linearly independent set Example H R 3 91 en is a basis for R 10 L C39C t D I x ltd39 PCD39ltD O 4 44 2 EC CDtltCD GD TU N C O C r39 C CD C3 C3quot GD Ill 2 cm L 2 4 Y I unquotrq CU C quot r4 LC ltT39lth GE r39 rl 1rI 8 LC LC 0 LO m I lt1 J lt1quotCVI39ltD GD 3 r39 C l rll Y X I lt1 Find bases for NulA and ColA 11 29 Dimension and Rank Fact Any two bases for a subspace has the same number of vectors The dimension of a subspace H denoted dimH is the number of vectors in any basis for H 12 29 Dimension and Rank Fact Any two bases for a subspace has the same number of vectors The dimension of a subspace H denoted dimH is the number of vectors in any basis for H Examples 1 dimR n CL 217 O ab any real nos dimH b 13 The rank of a matrix A denoted mnkA is the dimension of the column space of A 14 LC 00 CH C m ltl39 Q C O 4 I C LC 0 C C CD 5 l C C C r39 C C C C39 CD Ill 2 cm L 2 d v I unt ru CU C 4quot r4 LC ltT39ltn GE r39 rl 1rI 5 LC m 0 L0 m lt N lt139 N 390 d E r39 C l rll Y gtlt LIJ lt1 16 Math 39qu am 2 NW6 Tnle 5 Fwt m Wei IRJ39Mquot wmoub 51 Mj num MK 5th K Y1 Y3 YR KY K H 5 a O t to quotl Yl xLL HL 0 O IM h39o Wldles 1 XX3 Mwuu EU I gr Wquot equot Y I x 1m V x q SxL LYL Yb 3914 V L v1 YL39 39 Y Y k1quot Y nxr 33 Um L4 Mam is 04 um 54 a wait a Wm MM 7 6 M 39f39Nerxqu QM gm L31 ML 71 97114 C WWW Q1 1 A17 I MAL4 44 9b 7 GD w wabm Wave LuxELQ 39 7 m2 Ck 39 395 quot x 3 60 Mfm j L599 Iva542 WI39nLlc 14 WAqu wit a wrach we Bum KL z 1 393 X3 4 l L I H L Y3 Said ijPJI N J x 0 l l 9 3 t 39L 3 I 7 l L 3 l I N it t l o 1 4 q 4 7 1 q LL bow 5 Q E 1 3 lR tEa 3 L O l O q 0 0 1 3 P v G 0 39 3943 LR L i I 0 E L o q C C E 3 5 5 1 3 q W V l 0 o c 39L 23ktae 0m 0 394 q YL 2 4K 0 0 l L 3 393 7 I 8 fl ii ltL an 2 x LCI1 q M K39s P301 3 Ya quotQt M Llc 1 VA II Wig A Kr I K J wxx K K IKJ E p 032 i9 21 K JANITwa u N K NKN T Kw U K wKN 63 r p23 r fw f TWP 55 mm 1 is sz nu SEEM rv w wait Ix ix N E F W at W 4P r U P Jr 1Vr P51 NVL N rfLwPuLPu u N 0 a 11fo Q 0 l I ovPhA 3 I z no L rTlLO Q nigm mu mP PHMInIJp VHF PU rfgmn s 5 mar T tnnruo oh Li fa Drum kn ruT vv 93 air a lmP To Pk rw 5L1 grwm lt2 L 5 we goiimsnm u EEK mltw lf Hg gax Rd Tm x du Tm V V be wlt H r FE t r u ESL n in msi 2er m i akin 36 4ampvaQE gpk www W4 H v VJ39u awe 5 uto9 w quotquot amp A 6 LL M C b IA gig1quotquot 37pr v 2 J J39ULI S LU CF KL vacHr 55 MM gt s DO 5 Cl V 39k 39 fquot A va did a Lima mum r6 vi 6r wit wdijg C J J Cf gt quot r as 2 mm wa v 5 73 L a L ml 5 VIVw J VE MR wath 0J0 J 7 0L 5L1ALWG m wh V65 CS LV I 39L Q Wake d r x LANA ma a 191 iv 7 39L m it 1m 1 2 3911 a 7 Yl tLVL 7 lyl LVLT39I KL zxLo i 35 LL 9 IV2 U Q gym 4m KL Q N quot K 91 OK 5 3 5quot gawk 0 E i Algal 0 E3 7 Y1 m w ax 111mg W 16 Dial M v7 0 Kwrgmtm39l MltJIJ P H cF7fltn ch MvQ GM W W av w ems E 4 1 5 q 3 L 1 3 if Wm WM st OJ C 1 4 I 2 CM womaam C2 0 6 em MU my M 9479 7 le4 XLE L 15 f M 4sz a 1 tquot I I I 3 m g j 0 uskw3haj Z 1 1 LLLLAtl q l 4 L L M O L0L 3a Z 0 La WRWII Q 9 a a v swagaha 1 SWJJ j a 9 J ltgt Htwa W w W 4 0 J A YLQIYLQL K7L ltgt aquot f K gt f P Uh 1 Fl H4 x 06 a baskMd SulsJ c CQ J4 Kw wech MW WWW cw 9104 Malt OJ ampw MLA We 7 1 2 Let u v 4Forwhichvalueofkis 2 4 the vector u orthogonal to the vector v k2 k4 k52 No such k exists spews 62 Orthogonal Sets A set of vectors 111 u2 up in R is called an orthogonal set if u7 uj 0 whenever 239 7amp j Theorem Suppose 5 11171127 yup is an orthogonal set of nonzero vectors in R and W Spanu1 u2 up Then S is a linearly independent set and is therefore a basis for W is an 1 Example8 1 n u CDl l OO t orthogonal basis for R3 Theorem Let 111 ug up be an orthogonal basis for a subspace W of R Then if y in W is given by y C1111 39l39 C2112 39 39 39 39I39 Cpup we have yuj C39 O o An Orthogonal Projection Problem For a fixed nonzero vector u in R write y in R as y 2 011 Z where C is a scalar Z is a vector orthogonal to u An Orthogonal Projection For a given nonzero vector u in R we can decompose y in R as y z where y u orthogonal projection of y onto u 37 11 Z y ll component ofyorthogcme to 8 u 3 7 and 3991 Example Let y 4 7 1 L Spanu L is the line through 0 and 11 Find the point in L that is closest to y What is the distance from y to L Orthonormal Sets A set of vectors u1 u2 up in R is called an orthonormal set if it is an orthogonal set and ui 1 for all 7 If W Spanu17 1127 39 39 39 711p7 then 11171127 39 7 up is an orthonormal basis for W I isan l J orthogonal basis for R3 An orthonormal basis is Example 5 OI I x on xr x l CDO 5 LetSI 1 1 i 1 2 7 NIH NIH 1 I orthonormal basis for R2 1 True 2 False Then S is an Theorem An mxn matrix Uhas orthonormal columns if and only if UTUI Theorem Let U be an mxn matrix with orthonormal columns and let x and y be in Rquot Then a 39JEXU ile 0 lUX 39 WY X39y C UXUy 0 ltgt XyzO 11 63 Orthogonal Projections Example Suppose 111112113 is an orthogonal basis for R3 and let W Spanu1 112 Write y in R3 as a sum of a vector y in Wand a vector 2 in Wi Solution 12 Theorem The Orthogonal Decomposition Theorem Let Wbe a subspace of Rquot Then each y in Rquot can be uniquely represented in the form y z where y is in Wand z is in Wi In fact if 111 112 o up is any orthogonal basis for W then yu yu A 1 19 y 11139111u1 up up up Z A and y y 13 Theorem The Orthogonal Decomposition Theorem Let Wbe a subspace of Rquot Then each y in Rquot can be uniquely represented in the form y z where y is in Wand z is in Wi In fact if 111 112 o up is any orthogonal basis for W then 37 m ul 3243 up and Z y 3957 The vector y is called the orthogonal projection of y onto W 14 Example Let u1 7112 n sooo DI 0 and y o 3 Observe that u1 112 is an orthogonal basis for Chanf114 UVUIIIUl in Wand a vector orthogonal to W UV VV 112 Write y as a sum of a vector 15 Geometric Picture 16 Let y 6 3 4 3 u1 4 u2 3 Find the 2 o o orthogonal projection ofy onto Spanul 112 a b 39 1 C 7 M000 000C Theorem The Best Approximation Theorem Let W be a subspace of Rquot y any vector in Rquot and 3957 the orthogonal projection of y onto W Then 37 is the point in Wclosest to y in the sense that My ill lt ly V for all V in Wdistinct from 7 Outline of Proof 18 Example Find the closest point to y in Span ll 112 where y andu2 I I CDO OOl H QCbk Solution 19 61 Inner Product Length Orthogonality Not all linear systems have solutions 1 2 Example No solution to exists Why m A I Ll 331 5172 l 00 Instead find 3 so that ASE is closest to b Segmentjoining A and b is perpendicular or orthogonal to the set of solutions to AX b Need to develop fundamental ideas of length orthogonality and orthogonal projections The Inner Product The inner product or dot product of u and V2 If ul v1 U U u 7 V 2 7 71 Du Then 39 39 39 39 Notethat uVzvu Theorem Let 11 V W be vectors in R and let 0 be any scalar Then a ll39VZV39ll b uVWuWVW C CuVCuVuCV d uuZO and uuOltgtuO Combining b and c we have 31111 quot39Cpup 39 W q I I c1u1 W1 I I pkup39 Length of a Vector 01 For v a the length or norm of V is the Un nonnegative scalar defined by V V V U 717 NcgtteIVI2 V 39 V 9 Example In R2 v V b 7 For any scalar c cv Distance in R The distance between 11 and V in R distu V 2 Mn Example In R2 Example In R3 Orthogonal Vectors Two vectors u and V are said to be orthogonal to each other if u V O Theorem The Pythagorean Theorem Two vectors u and V are orthogonal if and only if Ul V2 Ul2 V2o Orthogonal Complement If a vector Z is orthogonal to every vector in the subspaceWof R then Z is said to be orthogonal to W Orthogonal Complement of W Wi 2 2w 0 for all w in W Example Let W 2 Find Wi a any real no 10 Theorem LetA be an mxn matrix Then Row AH Nul A Col AH Nul AT Why 11 1 0 1 39 Example Let A 2 O 2 1 O Basis for Nul A o 7 1 and therefore Nul A 1 O is a in R3 1 Basis for Row A O and therefore Row A 1 is a in R3 12 1 0 139 Example Let A 2 O 2 Basis for Col A 2 soCoIAisa Basis for Nul AT i so Nul AT is a 13 62 Orthogonal Sets A set of vectors 111 ug up in Rquot is called an orthogonal set if uz 113 0 whenever 7 7amp j 14 62 Orthogonal Sets A set of vectors u1 u2 up in R is called an orthogonal set if ui uj 0 whenever 239 7amp j Example ls set i l 1 O I I o I b l I I I 0 CD an orthogonal 15 Theorem Suppose S 1111123911p is an orthogonal set of nonzero vectors in Rquot and W Spanu1 u2 up Then S is a linearly independent set and is therefore a basis for W 16 Orthogonal Basis An orthogonal basis for a subspace W of R is a basis for W that is also an orthogonal set 17 Theorem Let 111 ug up be an orthogonal basis for a subspace W of R Then if y in W is given by y C1111 39l39 C2112 39 39 39 39I39 Cpup we have y uj 7 J u Cj 18 Example Let y 7 3 4 ands 1 1 O an orthogonal basis for R3 Find LYS l X I OO 19 An Orthogonal Projection Problem For a fixed nonzero vector u in R write y in R as y 2 011 Z where C is a scalar Z is a vector orthogonal to u 20 8 u 3 7 and 3991 Example Let y 4 7 1 L Spanu L is the line through 0 and 11 Find the point in L that is closest to y What is the distance from y to L Mm 7 10 Lao6M0 l3 Scum Exm TE A 9 8 SD AM W 08 Com 39 23quot LZJ 3112 5 1 Si3J 353 62444 H L M SKIr57 Hak new 4 4 5 1 33 3 is x 3335 y g 1 x d3 u lt ijN 343330 K 3 E g Jbux nkdi gnaw 33 e I e 3qu nd w A i 3Lrw LA Quaisb 19 mp8 39 5 m 36 g C brat i i W93 i mN New 3510 JP 4N0 ii max 2315 n va i o 3x s kc if is SSE c On mfw VS n6 355 r x Zigv kw m 31 E o H DELLEF 94 Wu H ru b Frag 53 9 meow 3e 9 V mmccrrxa nwgtv b 9 3 ma 9 u mu Alwyn 1nan a x Le 1t S ULMLN I 3 g ag nz a H gm an Ti 31 44 N ufs wCPpSaaxa a m if FF JOOV wig 7 ARK 1 0 L rfW Exmgtwarr LV CSHP We 2 has amp an r g an a WYEL4H N 9 c8 F7 Q lmyoo quupm mm Lox 6 DA H m Low 40 DYHJ h oa 6 m a uA oy a m cydm i 08 u fmmmyAXH ooW R MQX YL L L2 E gt 3 X W A st 0 1gg l l 503 39L0 All N 2 Los o 0 O 2 1 KL 3 11139 Lf K1 mi 403ij m 113 a 03ng ML vx H gtANSTLv vv H ifmnbVC 91x J H 31083 m H TM Lx nw qKQbr mama W L00 F A mqvbmv her8m l if min 1 w x g r mien J Y 0 Lo 2 a 30 c O IQ 320 r o C7 fj 5 1quot355 13 00 20 CI M h 3 Q VD 7 quot9 7m I 7 7g V9 70 156va 0 WV gt9va q dOd LV ag 279W 3 W0 Wvog 610amp I MTIM H 1 J l d 0quotdchslv I MOdZV 93 Q5 time i A Le QMJM ALPDP I 2 7amp0 ALDO PWquot 9337119433 L P L I 2 4 K 7 ll 7 0 av1amp1 KB f it 13 if 2 s A MM APDPquot 1 AmosleooP4 H H d 7 DUDO G uoo 0 0 7100 a k 0 L9 LL00 L00 L00 A A 00 6 5 MK 03 x w 01 A k f quotLL00 L l L Xllwo l 47100 1 XV wk 1 gtL iixxyfibfb Emz ip M as y Pig u up 29 a 3 gtmo 7f DAH gt H Q qm l q Onhooirmv O K SQ F is SHE EL on Quit d chSO l v amp v aw a who Hm km x 2 offiksf 52 H a v 5 3 bnqwumrkd vrv 9 c a ltw 991 YVSWWL PW V ox gfsm Q 03 Tmm k P i 933 sum Rhino Rm O Rm wacaoxmm A D 3 LLK D p Rxiffmxpm gt3F rampr 2 6 pi Lifu 90 mxm Du 3 vii wr0 TEL av C X 2 MA r gykwmn KW m MAW3 1 6 WISHIQ A z Z O o f I l 0 l 91km 49 game If A O MAMA w 3 j 0 gt ka GLIM 0 1A CLAKL MO39MO C3 l MW A F mi w P7356 Pm1mufxeri lt1 ltVMv rJ g1 VHN V m0 gt n m mglw nit 0 AT in if p at Qqh FU MJ 0 wit ffsk o 1 AP CAEQ fws fu lowwww Q Q AU AU 4 4 O V p N 0 v lx Q h 0 Q r O W 0ampqu 3 NEW a i a 9 gm T9 Similarity For square matrices A and B we say that A is similar to B if there is an invertible matrix P such that A BBB 1 Theorem If A PBP 1 then detA det B Consequently detA AI detB AI so Definition A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix ie if A PDPl where P is invertible and D is a diagonal matrix Theorem A is diagonalizable if and only if A has n linearly independent eigenveotors Let A PDP 1 wherex What is A5 39 32 33 1 39 32 31 39 32 Example A 391 0 Is A diagonalizable 1 1 Example Diagonalize the following matrix if possible 2 0 039 A 1 2 1 1 0 1 Step 1 Find the eigenvalues of A Step 2 Find linearly independent eigenvectors of A A21 Step 3 Construct Pfrom the vectors in Step 2 Step 4 Construct D from the corresponding eigenvalues Step 5 Check that AP PD Example Diagonalize the following matrix if possible 39 2 4 6 39 A O 2 2 O O 4 Example Why is A diagonalizable comm M60 I OO Theorem An 17 X 17 matrix with n distinct eigenvalues is diagonalizable Warning The converse is not true 10 IfA is a 5X5 matrix with three distinct eigenvalues and the eigenspace of one eigenvalue is three dimensional Is A is diagonalizable 1 Yes 2 No 11 55 Complex Eigenvalues o 1 A 1 0 Find all eigenvalues ofA and a loasis for each eigenspace Example rotation matrix by in clockwise direction 12 Example A What are the eigenvalues of A 1 2 0 TI 13 Example If C a b where a b are real I a numbers and not both zero then the transformation X I gt OX is a rotation followed by a scaling The eigenvalues of C are A a E ib Let 7 IAI a2527 and let 90 O 3 90 lt 27f satisfy a b C0590 Slngp 7 Then 90 is the angle of rotation and 7 is the scaling factor 14 Example l l I I I 15 2 3 39 1 39 Let A 3 2 Then 2 IS an eigenvector corresponding to which eigenvalue a 2 37 b 2 3239 c 3 2239 d 3 2239 56 Discrete Dynamical Systems Xk1 AXk X0 given initial vector Assume A is diagonalizable We can find an easy way to compute the iterates Xk Let V1 Vn be n linearly independent eigenveetors of Aeorresponding to the eigenvalues A1 An respectively Then X0 01V1 39 CnVn for some constants C17 7 Cn X1 2 X2 17 In general Xk Z Note If Aj lt 1then Agk z 0 for large values of k 18 Example Discuss the longterm behavior of the system Xk1 AXk where 9 0 39 2 39 A o 6 X0 339 Solution eigenvalues 1 971 2 ei envectors 9 V1 0 7 V2 1 C1 In eneral let X g 0 C2 Then X0 01V1 02V2 and Xk Cl9kV1 C26kV2 Thus xk a 0 as k a oo We say that the origin is an attractor of the dynamical system and the direction of the greatest attraction is the line through the origin and the vector V2 20 4 8 E L t 39 39 xamp e e A 16 28 A PDP l where p IMI I I IH I Discuss the longterm behavior of the solutions to x ii i 21 Note that A1 A2 gt 1 Xk 6112kV1 022kV2 Since V2 corresponds to the larger eigenvalue if we take X0 2 V2 then Xk tends away from the origin the fastest We say that the origin is a repellor of the dynamical system and the direction of greatest repulsion is the line through the origin and the vector V2 22 What if some eigenvalues have absolute value greater than 1 and the others less than 1 Answer The origin is a saddle point of the system That is some solutions tend toward the origin and some solutions tend away from the origin 23 Example Let A be a 2X2 matrix with eigenvalues 2 and Z and corresponding eigenvectors V1 2 respectively Consider the difference equation xk1 Axk XOB 3 V1 V2 39 1 Find a formula for Kit in terms of V1 V2 2 Find an X0 for which a Xk tends to the origin b Xk tends away from the origin 1 1 and V2 1 2 24 Recall For an m X 17 matrix A ColA span of the columns ofA NulA all solutions X to Ax0 rankA dim Col A no of pivots dim NulA no of free variables rankA dim NulA n The row space of A denoted RowA is the set of all linear combinations of the rows of A That is RowA ColAT The row space of A denoted RowA is the set of all linear combinations of the rows of A That is RowA ColAT Theorem Tan43A frankqu L C39C t D I x ltd39 PCD39ltD O 4 I 39 C LC CD C C CD TU N C O C gt 39 S r39 C CD C Cquot CD Ill 53 t 2 Z d v I unt ru CU C quot r4 LC ltT39ltn GE r39 rl 1rI 8 LC LC 0 LO a lt N lt139 N 390 d E r39 C l rll Y gtlt LIJ lt1 Tan 14 dimNulA ranAT dimNulAT What is dimNulAT 1234 1234 1234 1234 1234 Let A 1 2 3 4 1 2 3 4 Coordinate Vectors Let B 2 b1 bp be a basis for a subspace H For each vector X in Hthe coordinates of X relative to the basis 3 are the weights C19 39 39 39 70 Such that X Clbl I Cpbp and the vector in RP 01 XB C P is called the coordinate vector of X relative to the basis B or the Bcoordinate vector of X Example LGt I3 b17 b27 where bl Then 8 is a basis for R2 Let x 2 Find x B 3 11 39 ll Example LGt 73 2 b1 b2where bl 3 Then 8 is a basis for H Spanb1 b2 9 Let X 10 6 1 Is x in H If 80 find X13 3 D Let B What is X A g ltBgt 12 ltcgt 13 1 1 00M 1 X15 J gt 31 Introduction to Determinants Recall 2 X 2 case F a b quotI HA d J then det A 2 ad bc 10 31 Introduction to Determinants Recall 2 X 2 case r I CLO C d L J Furthermore A is invertible if and only if IfA then det A 2 ad bc 1 X 1 case 11 General Formula for jet A7 A n x n n 2 2 For n 2 2 the determinant of an n X n matrix A 2 adj is defined by d tA all detA11 Q12 detA12 I I 11 a1n detAln where Aij the matrix obtained from A by deleting its z th row and jth column Example 39 1 2 3 4 A 46 47 48 A0 9 1U 11 12 13 14 15 16 12 Common Notation for det A 2 Example det 1 5 42 gtI DU39I IAI 13 Example Compute the determinant of Solution A 1 3 2 2 039 1 2 0 1 14 The 7j cofactor of A denoted Cij is the number defined by CM 2 1ij det 15 The 7j cofactor of A denoted Cij is the number defined by CM 2 1ij det Thus 1011 2012 0013 MOOH l OI M l MO 16 In general 111 C112 G21 122 azn anl an2 arm cofactor expansion across row 1 17 Theorem The determinant of an n X n matrix A can be computed by a oofaotor expansion across any row or down any column detA 6171021 042072 39 39 39 aan D l m aljclj i a2j02j i39 39 39 i ananj 18 Theorem The determinant of an n X 71 matrix A can be computed by a oofactor expansion across any row or down any column Get A ailcz39l 042072 39 39 amCm det A alelj I anCQj I ananj Use a matrix of signs to determine 173j1 39 19 Example Compute the determinant of A 1 3 2 2 O 1 2 O 1 down column 3 Solution by a cofactor expansion 20 Example Compute Solution OOOH COMM cowl 00 Q39lI Q39ll 21 Note Method of cofactor expansion is not practical for large matrices see Numerical Note on page 190 22 Triangular Matrices 000096 000 Theorem IfA is a triangular matrix then detAz O O gtllt gtxlt gt1lt gtxlt O 96 96 96 96 96 96 96 96 96 96 96 96 96 C 3900 96 96 DOC 96 OOOC 23 Example 4515 3120 2200 1000 24 32 Properties of Determinants Theorem Let A be a square matrix 1 If a multiple of one row of A is added to another row of A to produce a new matrix B then detB detA 2 If two rows of A are interchanged to produce B then det B det A 3 If one row of A is multiplied by k to produce B then det B k det A 25 Examples 26 O l Qamp D CDG S KHO and detA 2 then det 3A 2 27 4 O 10 2 9 7 11 of row reduction and oofaotor expansion Example Compute using a combination MOH JU39IM 000 28 11 gtilt gtilt gtilt O u22 gtilt Suppose A has been reduced to U 0 0 U33 0 0 0 O O O O um by row replacements and interchanges Then if A is invertible detA if A not invertible 30 Theorem A square matrixA is invertible if and only if detA O 31 More Properties of the Determinant Theorem det A det AT Theorem Multiplicative Property detAB det A det B 32 More Properties of the Determinant Theorem det A det AT Theorem Multiplicative Property detAB det A det B If A is square what is det Ak in terms of det A 33 Example If B2 Iwhat are the possible values of det B 34 21 Matrix Operations Matrix Notation Two ways to denote m X n matrix A In terms of the columns of A al 32 A In terms of the entries of A 3 A Main diagonal entries an Zero Matrix 0 O O O O O THEOREM Let A B C be matrices of the same size and let 7 8 be scalars Then a Al BzBlA d TABTATB b Al BICAI Bl C e TSATA8A C AlO 2A f 7 3A 719M Matrix Multiplication Suppose A is m X n and B is n X pwhere Bb1 b2 bp39 Then A Abl Ab2 Abp Example Compute AB where A Solution 4 3 O 239 5 1 and B Oh N l 00

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