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# Linear Algebra & Matrix Theory MATH 225

Yale

GPA 3.81

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Date Created: 10/29/15

26 Let f be a map f U a V We say f is onto or injective if for each 1 6 V7 there exists u E U such that fu 1 We say f is oneto one or injective if7 for any uhuz 6 U7 fu1 fu2 implies ul u2 le if ul 31 uz then fu1 31 fu2 We say the map f is bijective if it is both injective and surjective Let U7V7W be three sets and f U a V and g V a lV7 then the composition function g o f is a function that maps U to W and is de ned by g 0 MW 9fu Note that the composition of functions is associative However7 it may not be commuta tive lndeed7 let U7 V7W7X be sets and fU VgValV andhWaX 7 Then the composition function h o g o f is de ned as h 0 g 0 f U a X by h 0 g 0 MW h9fu Then h o g o f h o g o f h 09 o f On the other hand7 if the two set U and W are different7 then it does not make sense to have f o 9 Suppose f is one to one and onto le fU V Then for each 1 E V there exists a unique u E U such that fu 1 310 Now7 consider the map 9 V a U by 91 u7 where u and 1 are related by 310 Note that fgv fu 1 Thus f o g is an identity on V7 denoted by IV Similarly7 we have gfu 91 u7 which shows that g o f IU Thus the following proposition holds Proposition 8 Let f U a V be a bijective map Then there em39sts an inverse off denoted by f l such that fquotlV U7 andf lofIU7 andfof 1IV Here 5 denotes the identity map on S Ie 55 s for all s E S In particular lfU V then f lofIUfOf 1 Example 16 Let U 077T and V 7171 Then fx cosx is a bijective map from U to V De nition 11 Linear maps on vector spaces Let U7V be vector spaces over F7 and let f U a V We say f is a linear map or homomorphism if the following conditions hold let u7u E U and a 6 F7 LM1 fu w M M LM2 fau afu 27 Example 17 Examples of linear maps 1 Let U be a vector space The identity I U a U de ned by u u is a linear map 2 Let U be a vector space with an inner product lt3 Fix an element 1 E V The map L U a F de ned by Lu ltuvgt To show LM1 Lu u ltu u vgt ltu71igt ltu vgt Lu Lu To show LM2 Lau ltau711gt altu711gt aLu Note that the map Lu lt07 gt is also linear However7 the map Lu lt07ugt is not linear if F C 9 Let U The projection 7T1 Fn F de ned by mltlt17 7397 1 is a linear rnap 4 Let W be the vector space of polynomials p of degree 3 n with coef cients in F7 then the map D W a W given by dpm D pm dz is a linear rnap Note that Da1a1x anx a1 2a2znanz 1 Sirnilarly7 the map Dk7 k 2 17 given by Dkpz 13 is also linear Properties of linear maps Let f be a linear map from the vector space U to the vector space V We have 1 f0 0 To clarify the algebra7 we denote by 6 the zero element in U and V and by 0 the zero element in F We have f0 f0 0 0f0 0 2 fu fu7 and u 7 7J u 7 fv Let f be a linear map from the vector space U to the vector space V The kernel of f7 denoted by Kerf7 is a non ernpty subset of U which is de ned as Kerf u E U fu 0 E V Note that 0 E Kerf The image of f7 denoted by Imf7 is a non ernpty subset of V which is de ned as Imf UE Vfu 1 for sorneuEU Note also that 0 E Imf dimKer is called the nullality of f and d m1mf is called the rank of f 28 Proposition 9 Let f be a linear map from the vector space U to the vector space V Then Kerf and Imf are subspaces of U and V respectively Proof Homework D Proposition 10 Suppose f U a V is linear map Then f is injective if and only if Karo 0 Proof To show i Suppose f is injective We know that 0 E Kerf lf 0 31 u 6 U7 then by de nition of being injective7 fu 31 07 which shows that u is not in Kerf Hence Karo 0 To show Suppose Kerf Let U71 6 U such that fu fv we need to show that u v lndeed7 uiv fu fW 07 which shows that u 7v E Kerf But Kerf This shows that u 7v 07 and hence D 111 Proposition 11 Let U and V be vector spaces with U being nite dimensional and f is a linear map from U to V Then dimKerf dimImf dimU Proof Let ij7 be any basis for U7 then x 7 spans Imf Thus Imf is a nite dimensional subspace of V Now7 let v17 7vm be a basis for Imf7 and let uL7 be a basis for Kerf Let mi 6 U such that rm vi Claim ul77umw177wm is a basis for U 1 To show linear independence Let ahbj 6 F7 i 17 7n7j 17 quot77717 such that alul anun blwl bmwm 0 We have 0 fa1u1 anun b1w1 bmwm fb1w1 bmwm blul bmum Since v17 7vm is linearly independent7 bj 0 for all j 17 7m Thus 0 alul anun blwl bmwm alul anun Again7 since ul77un is linealy independent7 we have al 0 for alli 177n Therefore7 ul77umw177wm is linearly independent 29 2 To show that u17umw17wm spans U Pick any u 6 U7 we have fu E Imf Since 11177om is a basis for Imf7 there exist unique element b17bm in F such that fu blol bmum Let it blwl bmwm7 then fw b1fw1 bmfwm blol bmum Now let u u 7 w7 then fu fu 7 w fu 7 fw 07 which shows that u E Kerf This implies there exist a17an in F such that 1 alul anon Therefore7 u u w alul anun blwl bmwm7 which shows that u17umw177wm spans U D Let U and V be vector space and f U 7 V be a linear map We say f is an isomor phism if it is a bijective map Two vector spaces U and V are isomorphic if there exists an isomorphism f U 7 V Proposition 12 Let U rid V be vector spaces arid f U 7 V be an isomorphism Theri there epists an isomorphism f l V 7 U such that filofIU7 andf0f71IV In particular ifU V theri f 1 of IU fof l Proof For each i 6 V7 there exists a unique u E U st fu 1 De ne f l V 7 U by fo u By proposition 87 f 1 is bijective It remains only to show that f 1 is linear lndleedl7 Let iii 6 U and 111 E V such that oi for i 12 We have film1 12 f71fu1 fu2 f71fu1 111 U1 U2 f71111 film2 Similarly7 using the linearity of f7 we get f 1ao af 1o for all i E V and a E F It is clear from proposition 8 that f 1 ofIU and fof l IV D Example 18 Let U C and V R2 be vector spaces over R The map f C 7 R2 given by fz iy Ly is linear and bijective This shows that f is an isomorphism Hence C and R2 are isomorphic ln general7 C is isomorphic to R2 30 The algebra of linear maps Let U and V be vector spaces over F Denote by HomU7 V the set of all hornornorphisrns from U to V In other words7 HomU7 V f U a V f is linear We can de ne and scalar on HomU7 V by Let f and 9 be linear maps in HomU7 V7 and a 6 F7 f 995 f96 99 and af9 af97 for all 9 e U Claim HomU7 V is a vector space over F 1 F 9 7 To show HomU7 V is closed under Let f9 6 HomU7 V By de nition7 f 9 maps U to V It remains to show that f 9 is a linear rnap lndeed7 let 979 E U and b 6 F7 we have f 99 9 f9 9 99 9 1 f9 99 99 WC 99 f9 99 f 99 f 99 Also7 f 9b9 fan 9099 bf9 599 WW 99 W 9X9 Thus7 f 9 is linear To show HomU7 V is closed under scalar Follow the above steps All the associative7 commutative and distributive laws are preserved We will show next that there is a unique neutral element in HomU7 V Let 0 be the map from U to V such that 09 0 for all 9 E U Clearly 0 is a linear map from U to V To show uniqueness Suppose there exists 9 E HomU7 V such that f g f7 for all f e HomU7V In other words7 f 99 f9 for all 9 E U and f E HomU7 V We need to show 9 0 Pick any 9 6 U7 we have f9 99 f 9X9 f9 f9 and 99 are elements in V In V7 0 is the unique neutral element such that f9 0 This shows that 99 0 Since 9 is arbitrary7 we must have 99 0 for all 9 E U which shows that 9 0 31 5 for each f E HomU V there exists a unique addive inverse denoted by if E HomU V such that f if 0 To show existence For each f de ne if to be ifx Clearly if is a linear map from U to V We have f f ifz f 7 f 0 for all z E U This shows that f if 0 To show uniqueness For each f E HomU V let 9 be an element in HomU V such that fg0 Thus for any x E U f95 996 f 9X96 0 311 In V 7fx is the unique additive inverse of Thus 9a 7fz Since z is arbitrary we must have 9a 7fx for all z E U which shows that g if Let fg E HomU V If U 31 V then it does not make sense in general to cosider the composition 9 o f or f o 9 However if U V then the composition between two maps in HomU U is de ned In other words the composition is again an element of HomU U Thus we can de ne the product between two f g E HomU U as f996 f996 Note that is associative ie f g h f g However may not be commutative Consider Let U R2 and fm 0721 and 9m 1796 Cleally both f and g are linear maps from R2 to R2 Let my 6 R2 we have f 9m f9m ow 0796 and 9 fm 9006711 9079 170 which shows that f g 31 9 f However for special types of linear maps the commutative law for holds Let f U a U be a linear map from U to itself We denote by f the composition of f n times and we write f0 to mean the identity map I Note that f o f f t Proposition 13 Let f E HomU U then f is also an element of HomU U In partie ulaiquot any linear combination 2 alf l is also an element of HomU U Proof Let us induct on n Suppose n 0 then f0 I which is clearly linear Now suppose fk is linear we need to show that flitl is also linear lndeed We y ffk9c 9 ffk96 fky7 The last equality uses the assumption that fk is linear Now since f is linear ffk96 fky ffk96 ffky kaW fk1y Following the same steps we get fk1ax afk1 By induction f is linear for all n 2 0 D 32 Theorem 14 Let U be a vector space and f U a U be a linear map Let Sf Emfl for some n 2 04116 F C H0mU7U i0 Then S is a subspace of HomU7 U andfor my 97h E Sf g h h 9 Proof To show S is a subspace7 ie Sf is closed under 7 and 0 E S Exercise Before showing the general case7 let us show that arm W W arm for any mm 2 o 312 lndeed7 for any x 6 U7 afm 39 bf afmbf 95 abfm 95 bf afm bf 39 afm7 which shows 312 To show 9 h h 97 for any 97h E Sf Let g ELI aifi and h lbkfk Fix m and lnduct on 71 Suppose n 07 then clearly for any x 6 U7 9 71W aofo 71W 0010106 a01196 h0096 haoI9 haof096 h 9W Henceghhg Suppose true for n k7 to show true for n k 1 lndeed7 k1 k k g h 2 an h Zaifl we h Zuni h ltak1fk1gtn i0 i0 i0 We have awfw h m1ka bofo bmfm aknfk befo akufk bmfm befo 39 ak1fk1 bmfm 39 ak1fk1 bofo amfm 39 ak1fk1 h 39 ak1fk1 The third equality uses 312 By the inductive assurnption7 we have k k Zaifih hZaifi Thus k k 9 39 h 2010 h ltak1fk1gth h 2 hak1fk1 h g i0 i0 By induction7 for any xed m7 9 h h g for all 71 Since m is arbitrary7 this must also hold for all m D 33 Remark 11 We have 2 mix bjfj aoltbo M bmf quot i0 j0 a1fbo M bmf anf bo M bmf quot aobo aobl a1b0f anbmfmm nm Z Ckfk7 k0 where ck Zai j aibj Example 19 Examples of linear maps on the vector space HomU7 U Fix 107 7an C F Consider P HomU7 U a HomU7 U by Pfa0a1fanf Chapter 4 Matrix Algebra 41 Matrix Algebra De nition 12 An m gtlt 71 matrix of coef cients in F7 denoted by ahan or simply calj is a rectangular array with m row and 71 columns le a11 a12 aln a21 a22 0 1an 7 aml 1mg amn The entry 239 indicates the row and j indicates the column that aij belongs to If m 717 then the matrix is called the square matrix Denote by ManF the set of all m gtlt n matrices with coef cients in F If we dont need to distinguish the eld F7 we simply write Mmm for short notation For each matrix A calj E mem denote the 2 row of A by 0i an at am 6 M1gtltm and the jth column of A by 11739 12739 a7 7 E mel am 35 Thus A can be rewritten as On mem and scalar multiplication can be de ned as follows Let AB 6 mem and c E F all b11 112 blg am b1 A B avan byan all b21 an by a2 b2 aij bigMM aml bml 1mg bmg amn bmn and call 0012 cam cA 401739an can can 002 caljan camL camg cam Clearly7 Mmm is closed under and scalar multiplication Proposition 14 With the above arid scalar multiplication Mmm becomes a vector space over F The zero matricc is given by 0 aw lm where aij 07 for all i7j For each A awmxm the additive iriverse is given by 7A emam Proposition 15 dimMmm mn Proof Let EM ahan E Mmm such that ani 0 ifiy k70rj7 l 7 1 ifihandjl Then clearly7 Mmm SpanEklk 1m7l 1n Moreover7 Ed is linearly independent This implies Ed is a basis Hence7 dimMmm mm which is the number of elements in Ekhh 1m7l 17 De nition 13 Let A ahan E Mmm and B bik gtltp E MW77 then the product A B is the matrix C clQMgtltp E mep where Cik 0i 39 bk ailblk aijbjh ambnk 36 Remark 12 By the above de nition7 we have Cik a blk aijbjk ambnk 11quot bk7 where the row vector al 6 Mlm and the column vector bEMnxl Erereise 1 Let A E mem B E Mnxp and C E Mpxq Check that ABOABO Let D A B C7 then D E meq For short notation7 we7ll write AB for A B Remark 13 Matrices as linear maps Let A E Mmm with coef cients on F7 then A can 1 be made into a map from F to F le for each x 2 6 F72 the map TA F a F 1271 given by TAx Ax E F is a linear map It is easy to check that for any my 6 F and e 6 F7 TA y Az y Ax Ay TAx TAy7 and TAC Aez eAz CTAx Thus7 for each matrix A7 we can associate to it a linear map TA Proposition 16 Let A E Mmm arid B E Mnxp Let TA F a F arid TB F17 a F be the liriear maps associated to A arid B respectively Theri the map T F17 a F giveri by Tm TATB96 is liriear Moreover Tz TAB ABx for all z E F Proof Clearly T is linear because it is a composition of linear maps TA and TB Morevoer7 for all x7 Tz TATB TABx AB ABx The last equality uses the fact that matrix multiplication is associative Hence T TAB D For each A ahan E mem the transpose of A7 denoted by A2 is the matrix in Mnxm which is de ned as 111 121 am a a a t 12 22 m2 A ajirigtltm am 12 amn Proposition 17 Let A E Mmm arth 6 Mnxp Theri ABt B At 37 Proof Homework 1 When there is multiplication7 we can talk about the multiplicative dentity and existence of inverse elements The identity matrix in Mnm7 denoted by m is the matrix caljmm such that aij 0 if z39y jandaij1isz39j Remark 14 Let A E mem then ImA A AI Let A 6 MW We say A is invertible or non sigular if there exists a matrix B E Mmm such that AB In BA The matrix B is called the inverse of A7 which is denoted by A l Note that A 1 is the unique matrix such that AA l A lA I lndeed7 let B be a matrix such that AB BA m then B BI BAA 1 BAA 1 IA l A l Example 20 Computing the inverse of a 2 gtlt 2 matrix Let A If A 1 exists7 then i 1 1 0 i 1 2 111 112 I i gt 1 i 3 7 121 122 011 2021 1 which show that 012 2022 0 3011 7021 1 3012 1 7022 0 This is a linear system of 4 equations and 4 unknowns Solving this this system7 one obtains that 7 72 1 i A 7 73 1 This is the most intuitive and naive way of solving the inverse of a matrix In general7 if A is an n gtlt n matrix7 then we will need to solve a system of n2 equations with n2 unknowns7 which can be slow when n is large We will discuss other methods for solving A l 42 Matrix Coordinates Recall from remark 13 that for each matrix A E mem we can associate to it a linear transformation T that maps F to F Now given any nite vector spaces U and V7 and a linear transformation T U a V7 we want to make the reverse connection 38 Suppose U and V are nite dimensional vector spaces Let B uL7 be a basis for U For any u 6 U7 we know that there exist unique b177bn in F such that u blul bnun7 and Tu b1Tu1 bnTun Note that bj ltu7ujgtHujll27 for j 177n7 is the jth coordinate of u with respect to B Thus7 Tu is completely determined by Tu7 j 17 quot771 Next7 let 8 0177vm be a basis for V Then7 for each TUj 6 V7 Let 11739 ltTUjwlgtHvlll27 7amj ltTuj7vmgtvall2 Then 11739U1 am lm For each j7 a1j77amj are the coordinates of Tu with respect to 8 Now let A ahan be a matrix with the z j entry equals to llj In other words7 The column vector aj of A7 j 17 quot7717 is the vector 11739 aj 02739 am The matrix A7 denoted by MfgT7 is called the matrix coordinates of T with respect to B and 8 Note that MfgT is uniquely determined by T Recall for each u E U and a basis 8 uL7 7un7 we have u blul bnun The column vector 51 bn is the coordinates column vector of u E U with respect to the basis 8 Thus 1 0 0 0 0 ul 7 1 7 Us 0 0 1 Then 111 112 11 1 111 Aug 7 021 022 am 0 i 021 1 7 7 am 1mg amn 0 am 39 In general7 111 112 am 0 11739 B i 7 Au 7 all Jig am 1 i a aml amZ amn 0 amj With the new notation7 we have Thus7 Aug TuB for all j 17 7m And for any UK 6 U7 we have Tu5 b1Tu1B bjTu7B bnTunB allbl alibi alnbn 111 a12 aln b1 ailbl aijbj ainbn an 012 am 5 amlbl amjbj ambn am am am bn Aug Thus7 we have the following proposition Proposition 18 Let U and V be vector spaces with basis 8 and 8 respectively Let T be a linear transformation from U to V then for each u E U Mg T uB Tu5 Example 21 Let T R2 a R2 be the rotation of vectors in R2 by an angle 0 This implies 1 7 cosl9 Tltogt W and Tltlt gtgtltzigtgt Thus the matrix 7 5 7 cos 6 7 sin 6 A 7 MB T 7 ltsinl9 cosl9 7 where B 8 17 07 07 40 Theorem 15 Let U V and W be nite dimensional vector spaces with basis A B andC respectively Let S U a V and T V a W be linear maps Then Mgr Mffs MfTS Proof Let A u1um B v1vn and C w1wp For each j 1 m let aljvl li ll39 amvn Similarly For each i 1 n let bliwl binmp Next for each j 1 m we have T11 1 Si li amp 1le11 aijTv amTvn a1jb11w1 bklwk bplwp ajb1w1 bm wp anb1nw1 bknwk bmwp bnalj bliaij bylaw1 bk1a1j bkiaij bknanjwk bzmam bmam bzmammp cljwl ckjwk cmwp where ckj bklalj bkiaij bknam for h 1 p and j 1 m Thus M TS ijlpm On the other hand we have Mfg ahan and Mgr MW Thus MgT Mgs dkilpxrm where dkj bk1a1jbkaj bmaM But ck dkj for all h 1 p andj 1 m Therefore Mgr Mffs M ltTS D Theorem 16 Let U and V be vector space with dimU n and dimV m respectively Let B and 8 be cced basis for U and V respectively Then the map L HomU V a Mmm given by LT MET is an isomorphism 41 Proof Clearly7 L is linear and injective To show L is surjective Let B uL7 and 8 o177om Pick any A ahan E Mmm We want to show that there exists T e HomU7 V st MB A Indeed de ne Z li ll39 i1 Then T e HomU7 V and Mg T A D By the above Theorem and proposition 157 we have the following result Lemma 2 Let U and V be nite dimensional vector spaces Then dimHomU7 V dimU dimV Let U be a vector space and B a basis for U We know that for each T E HomU7 U7 we can associate to it uniquely the matrix MfgT We want to say that T 1 exists if and only if Mngl exists Note that the matrix associated to T 1 is MgT l Moreover7 I MfgTT l Mg T MfgT4 Thus it must be that MS TW Mg rl lndleedl7 we have the following result Proposition 19 Let A E MnmF Suppose B E MnmF is a matiia such that AB In Then BA In which shows that A 1 eaists and is equal to B Proof Let U be a vector space over F of dimension n7 and B be a basis for U Let TA and TB be elements in HomU7 U that corespond to A and B respectively Since AB In7 we have M TATB MgTA MgTB AB I which shows that TATE U is the identity map from U to U We claim that both TA and T3 are isomorphisms7 which show that both TA and TB have inverses see proposition 12 Hence TB IUTB TngATB T TATB TglrU Tgl This implies B MgT MgS l Moreover7 AB META Mngl M TAT1 M T1TA Mngl MgTA BA I To show that both TA and T3 are isomorphisms First let us note that KerTATB 0 and ImTATB U Hence ranhTATB ii Let x E KerTB7 then TBx 0 This implies 0 TATB TATB7 42 which shows that z E KerTATB Thus kerTB 0 But 71 d mU d mKerTB d mImTB 0 rankTB rankTB n This implies that ImTB U7 which shows that TB is surjective Combining the fact that KerTB 0 and ImTB U7 we see that TB is an isomorphism As sets7 we have U TATBW TATBU TAU7 which shows that TA has full rank le rankTA d mU n This implies d mKerTA 0 Therefore7 TA is also an isomorphism The proof for TB being an isomorphism is the same D 43 Change of Basis Given any transformation T U a V7 the matrix associated to T7 MgT7 is completely determined by the basis A and B for U and V7 respectively Let A u177un and A be bases for U and let 8 11177vn and 8 017 quot7143 be bases for V What is the relation between MgT ahan and MET agamxn We have T IVTIU Thus7 MET Mg IV MffT MQIU Note MgT aijmxm where for each j 17 quot7717 Tuj 117111 ai h39 amjvm MQIU pgkWm where for each k 17 quot7717 IUu u plkul pjkuj pnkun And MfgU 0me where for each 239 17 7m7 IVvl Di qua pm1 gm1 ln particular7 if U V and A A and B 8 7 then Mg IU MgIUrl

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