Class Note for MATH 290 with Professor Mandal at KU
Class Note for MATH 290 with Professor Mandal at KU
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Chapter 4 Vector Spaces 41 Vectors in R Homework Textbook7 41 EX 157 217 237 277 317 33d7 457 477 497 557 57 p 189 We discuss vectors in plane in this section In physics and engineering7 a vector is represented as a directed segment It is determined by a length and a direction We give a short review of vectors in the plane De nition 411 A vector x in the plane is represented geomet rically by a directed line segment whose initial point is the origin and whose terminal point is a point xhzg as shown in in the textbook7 115 116 CHAPTER 4 VECTOR SPACES page 180 7 The bullet at the end of the arrow is the terminal point mag See the tea tbookmage 180 for a better diagram This vector is represented by the same ordered pair and we write x 1 2 1 We do this because other information is super uous Two vectors u abuz and V 01112 are equal if ul 01 and uz 112 2 Given two vectors u uhu2 and V 0102 we de ne vector addition uV u1 vlu212 See the diagram in the textbook7 page 180 for geometric interpre tation of vector addition 3 For a scalar c and a vector V 111112 de ne CV 0111 am See the diagram in the textbook7 page 181 for geometric interpre tation of scalar rnultiplication 4 Denote 7V 1V 41 VECTORS IN RN 117 Reading assignment Read Textbook7 Example 1 37 p 180 and study all the diagrams Obvioulsly7 these vectors behave like row matrices Following list of properties of vectors play a fundamental role in linear algebra ln fact7 in the next section these properties will be abstracted to de ne vector spaces Theorem 412 Let uVW be three vectors in the plane and let cd be two scalar 1 u V is a vector in the plane closure under addition 2 u V V u Commutative property of addition 3 u V W u V W Associate property of addition 4 u 0 u Additive identity 5 u 71u 0 Additive inverse 6 cu is a vector in the plane closure under scalar multiplication 7 Cu V cu cv Distributive propertyof scalar mult 8 c du cu du Distributive property of scalar mult 9 Cdu cdu Associate property of scalar mult 10 1u u Multiplicative identity property Proof Easy7 see the textbook7 papge 182 411 Vectors in R The discussion of vectors in plane can now be extended to a discussion of vectors in nispace A vector in nispace is represented by an ordered nituple x1z2zn The set of all ordered nituples is called the nispace and is denoted by R So7 1 R1 1 7 space set of all real numbers7 118 CHAPTER 4 VECTOR SPACES 2 R2 2 7 space set of all ordered pairs 1 2 of real numbers 3 R3 3 7 space set of all ordered triples 123 of real numbers 4 R4 4 7 space set of all ordered quadruples 1z2x3z4 of real numbers Think of space time 6 R n7space set of all ordered ordered n7tuples 1 2 xn of real numbers Remark We do not distinguish between points in the n7space R and vectors in n7space de ned similalry as in de nition 411 This is because both are describled by same data or information A vector in the n7space R is denoted by and determined by an n7tuples 12 xn of real numbers and same for a point in n7space R The ith7entry is called the ith7coordinate Also a point in n7space R can be thought of as row matrix Some how the textbook avoids saying this So the addition and scalar mul tiplications can be de ned is a similar way as follows De nition 413 Let u 111112 un and V o1o2on be vectors in R The the sum of these two vectors is de ned as the vector uvu1 o1u2o2unon For a scalar 0 de ne scalar multiplications as the vector cu cu1cu2cun Also we de ne negative of u as the vector 7u 71u1u2 un 7u1 7u2 71in and the di erence u7vu7vul7o1u27o2un7on 41 VECTORS IN RN 119 Theorem 414 All the properties of theorem 412 hold for any three vectors u V W in nispace R and salars c d Theorem 415 Let V be a vector in R and let c be a scalar Then 1 V 0 V Because of this property 0 is called the additive identity iri Rn Further the additive identitiy unique That means if V u V for all vectors V in R than u 0 2 Also V 7V 0 Because of this property 7V is called the additive irwerse ofv Further the additive inverse of V is unique This means that V u 0 for some vector u in R then u 7V 3 0V 0 Here the 0 on left side is the scalar zero and the bold 0 is the vector zero in R 4 c0 0 5 lfcv0thenc00rv0 6 77V V Proof To prove that additive identity is unique suppose V u V for all V in R Then taking V 0 we have 0 u 0 Therefore u 0 To prove that additive inverse is unique suppose vu 0 for some vector u Add 7V on both sides from left side So ivVu7v0 120 CHAPTER 4 VECTOR SPACES So So 0u7V So u7V We will also prove So suppose CV 0 If c 0 then there is nothing to prove So we assume that c 31 0 Multiply the equation by 0 1 we have C 1CV 0 10 Therefore by associativity we have c lcV 0 Therefore 1V 0 and so V 0 The other statements are easy to see The proof is complete I Remark We denote a vector u in R by a row u u1u2 un As I said before it can be thought of a row matrix uu1 uz In some other situation it may even be convenient to denote it by a column matrix U1 U2 u l u l Obviosly we cannot mix the two in fact three di erent ways Reading assignment Read Textbook Example 6 p 187 Exercise 416 Ex 46 p 189 Let u 00 78 1 and V 1 78 07 Find W such that 2u V 7 3W 0 Solution We have 2 1 2 1 1 8 16 W7uiv70707787171778707777 if 777 3 3 3 3 3 3 3 Exercise 417 Ex 50 p 189 Let ul 1321u2 2 727541 u3 2713 6 If V 25740 write V as a linear combination of u1u2u3 If it is not possible say so 41 VECTORS IN RN 121 Solution Let V aul bu2 cu3 We need to solve for a b 0 Writing the equation explicitly we have 25 740 a132 1 b2 72 754 C2 713 6 Therefore 25 740 a 2b 2036 7 2b 7 02a 7 5b 3ca 4b 6c Equating entry wise we have system of linear equation 1 21 20 2 3a 72b 70 5 2a 75b 30 74 1 41 60 0 We write the augmented matrix 1 2 2 2 3 72 71 5 2 75 3 74 1460 We use Tl to reduce this matrix to Gauss Jordan form 1 0 0 2 0 1 0 1 0 0 1 71 L 0 0 0 0 J So the system is consistent and a 2 b 10 71 Therefore V2u1u2u37 which can be checked directly 122 CHAPTER 4 VECTOR SPACES 42 Vector spaces Homework Textbook7 42 EX7 97 157 197 217 237 257 277 35 p197 The main pointin the section is to de ne vector spaces and talk about examples The following de nition is an abstruction of theorems 412 and theorem 414 De nition 421 Let V be a set on which two operations vector addition and scalar multiplication are de ned It the listed axioms are satis ed for every u vw in V and scalars c and d7 then V is called a vector space over the reals R 1 Addition a u V is a vector in V closure under addition b u v v u Commutative property of addition c d There is a zero vector 0 in V such that for every u in V we have u 0 u Additive identity uv w u vw Associative property of addition e For every u in V7 there is a vector in V denoted by 7u such that u iu 0 Additive inverse 2 Scalar multiplication a cu is in V closure under scalar multiplication0 42 VECTOR SPACES 123 b cu V cu cv Distributive propertyof scalar mult C d cdu cdu Associate property of scalar mult u c du cu du Distributive property of scalar mult e 1 u Scalar ideritity property Remark It is important to realize that a vector space consisits of four entities 1 A set V of vectors 2 A set of scalars In this class it will alawys be the set of real numbers R Later on this could be the set of complex numbers CC 3 A vector addition denoted by 4 A scalar multiplication Lemma 422 We use the notations as in de nition 421 First the zero vector 0 is unique satisfying the property 1d of de nition 421 Further for any u in V the additive inverse 7u is unique Proof Suppose there is another element 6 that satisfy the property 1d Since 0 satisfy 1d we have 600000 The last equality follows because 6 satisfes the property1d The proof that additive iriverse of u uriique is similar the proof of theorem 232 regarding matrices Suppose V is another additive inverse of u uv0 and u7u0 124 CHAPTER 4 VECTOR SPACES So iu07u uv7u Vu7u V0V So the proof is complete I Reading assignment Read Textbook Example 1 5 p 192 These examples lead to the following list of important examples of vector spaces Example 423 Here is a collection examples of vector spaces 1 The set R of real numbers R is a vector space over R 2 The set R2 of all ordered pairs of real numers is a vector space over R 3 The set R of all ordered nituples of real numersis a vector space over R 4 The set CR of all continuous functions de ned on the real num ber line is a vector space over R 5 The set Cab of all continuous functions de ned on interval 11 is a vector space over R 6 The set ll of all polynomials with real coef cients is a vector space over R 7 The set l of all polynomials of degree 3 n with real coef cients is a vector space over R 8 The set Mm of all m gtlt n matrices with real entries is a vector space over R 42 VECTOR SPACES 125 Reading assignment Read Textbook Examples 6 6 Theorem 424 Let V be vector space over the reals R and V be an element in V Also let 0 be a scalar Then 1 0V 0 2 CO 0 3 If CV 0 then either 0 0 or V 0 4 1V 7V Proof We have to prove this theorem using the de nition 421 Other than that the proof will be similar to theorem 415 To prove 1 write W 0V We have W 0V 0 0V 0V 0V W W by distributz39mtyProp20 Add 7W to both sides W W WW W By 1e of 421 we have 0WWW W0W So 1 is proved The proof of 2 will be exactly similar To prove 3 suppose CV 0 lf 0 0 then there is nothing to prove So we assume that c 31 0 Multiply the equation by 0 1 we have C 1CV 0 10 Therefore by associativity we have c lcV 0 Therefore 1V 0 and so V 0 To prove 4 we have V 1V 1V 1V 1 1V 0V 0 This completes the proof I 126 CHAPTER 4 VECTOR SPACES Exercise 425 Ex 16 p 197 Let V be the set of all fth degree polynomials with standared operations ls it a vector space Justify your answer Solution In fact V is not a vector space Because V is not closed under additionaXiorn 1a of de nition 421 fails f 5 z 7 1 and 97z5 areinVbutfg5z717z5z71 is not inV Exercise 426 Ex 20 p 197 Let V z 2 0y 2 0 with standared operations ls it a vector space Justify your answer Solution In fact V is not a vector space Not every element in V has an addditive inverse axiorn i1e of 421 fails 71 1 71 71 is not in V Exercise 427 Ex 22 p 197 Let V xrealnumber with standared operations ls it a vector space Justify your answer Solution Yes V is a vector space We check all the properties in 421 one by one 1 Addition a For real numbers Ly We have x zyzy So V is closed under addition b Clearly addition is closed under addition c Clearly addition is associative d The element 0 00 satis es the property of the zero elernent 43 SUBSPACES OF VECTOR SPACES 127 e We have 7 z iz7z 807 every element in V has an additive inverse 2 Scalar multiplication a For a scalar c we have 1 1 c cxpm 2 2 So7 V is closed under scalar multiplication b The distributivity Cu V cu CV works for u V in V c The distributivity c du cu du works7 for u in V and scalars c d d The associativity Cdu cdu works e Also 1u u 43 Subspaces of Vector spaces We will skip this section7 after we just mention the following De nition 431 A nonernpty subset W of a vector space V is called a subspace of V if W is a vector space under the operations addition and scalar rnultiplication de ned in V Example 432 Here are some obvious examples 1 Let W x0 z is real number Then W Q R2 The notation Q reads as subset of It is easy to check that W is a subspace of R2 128 CHAPTER 4 VECTOR SPACES 2 Let W be the set of all points on any given line y ms through the origin in the plane R2 Then7 W is a subspace of R2 3 Let P2 P3 Pn be vector space of polynornials7 respectively7 of de gree less or equal to 2371 See example 423 Then P2 is a subspace of P3 and Pn is a subspace of Pn1 Theorem 433 Suppose V is a vector space over R and W Q V is a nonempty subset of V Then W is a subspace of V if and only if the following two closure conditions hold 1 If uV are in W then uv is in W 2 If u is in W and c is a scalar7 then cu is in W Reading assignment Read Textbook7 Examples 1 5 44 SPANNING SETS AND LINEAR INDIPENDENCE 129 44 Spanning sets and linear indipendence Homework Textbook7 447 EX 277 297 31 p 219 The main point here is to write a vector as linear combination of a give set of vectors De nition 441 A vector V in a vector space V is called a linear combination of vectors u1u2 uk in V if V can be written in the form V C1111 62112 Ckuk where 0102 ck are scalars De nition 442 Let V be a vector space over R and S V1V2Vk be a subset of V We say that S is a spanning set of V if every vector V of V can be written as a liner combination of vectors in S In such cases7 we say that S spans V De nition 443 Let V be a vector space over R and S V1V2Vk be a subset of V Then the span of S is the set of all linear combinations of vectors in S spanS clvl CgVZ ckvk 0102 ck are scalars 1 The span of S is denoted by spanS as above or spanv1 V2 Vk 2 If V spanS7 then say V is spanned by S or S spans V 130 CHAPTER 4 VECTOR SPACES Theorem 444 Let V be a vector space over R and S V1V2Vk be a subset of V Then spanS is a subspace of V Further spanS is the smallest subspace of V that contains S This means if W is a subspace of V and W contains S then spanS is contained in W Proof By theorem 433 to prove that spanS is a subspace of V we only need to show that spanS is closed under addition and scalar multiplication So let uV be two elements in spanS We can write uCiV102V239Cka and Vd1V1d2V2 dek where 0102 ckd1d2 dk are scalars It follows 11 V 01 d1V1 02 d2V2 39 Ck dkVk and for a scalar c we have cu 001V1 002V2 cckvk So both u V and cu are in spanS because the are linear combina tion of elements in S So spanS is closed under addition and scalar multiplication hence a subspace of V To prove that spanS is smallest in the sense stated above let W be subspace of V that contains S We want to show spanS is contained in W Let u be an element in spanS Then 1101V162V2Cka for some scalars Ci Since S Q W we have 1 E W Since W is closed under addition and scalar multiplication u is in W So spanS is contained in W The proof is complete I Reading assignment Read Textbook Examples 1 6 p 207 44 SPANNING SETS AND LINEAR INDIPENDENCE 131 441 Linear dependence and independence De nition 445 Let V be a vector space A set of elements vectors S V1 V2 Vk is said to be linearly independent if the equation CiVi 62V2 Cka0 has only trivial solution 010020ck0 We say S is linearly dependent7 if S in not linearly independent This means that S is said to be linearly dependent if there is at least one nontrivial ie nonzero solutions to the above equation Testing for linear independence Suppose V is a subspace of the nispace R Let S V1V2 Vk be a set of elements ie vectors in V To test whether S is linearly independent or not7 we do the following 1 From the equation CiVi 62V2 Cka 07 write a homogeneous system of equations in variabled 0102 ck 2 Use Gaussian elemination with the help of Tl to determine whether the system has a unique solutions 3 If the system has only the trivial solution 51 0020 Ck 0 then S is linearly independent Otherwise7 S is linearly depen dent Reading assignment Read Textbook7 Eamples 9 127 p 214 216 132 CHAPTER 4 VECTOR SPACES Exercise 446 Ex 28 P 219 Let S 6217132 De termine7 if S is linearly independent or dependent Solution Let c62 1 di13 2 0 0 0 lfthis equation has only trivial solutions7 then it is linealry independent This equaton gives the following system of linear equations 60 7d 0 20 3d 0 c 2d 0 The augmented matrix for this system is 6 71 0 1 0 0 l2 3 0l itsgaussiJordanform l0 1 0 l 1 2 0 0 0 0 J So7 c 0d 0 The system has only trivial ie zero solution We conclude that S is linearly independent Exercise 447 Ex 30 P 219 Let S3231332363 Determine7 if S is linearly independent or dependent Solution Let 353 7 3 7 Z9347 7762000 277 77 lfthis equation has only trivial solutions7 then it is linealry independent This equaton gives the following system of linear equations 3 3 1a 3b 750 0 3a 41 6c 0 3 7 7 Ea b 20 70 44 SPANNING SETS AND LINEAR INDIPENDENCE 133 The augmented matrix for this system is 3 3 2 0 1 0 0 0 4 6 0 its Gaus 7 Jordan form 0 1 0 0 g 0 0 0 1 0 mm man No So7 a 0b 00 0 The system has only trivial ie zero solution We conclude that S is linearly independent Exercise 448 Ex 32 P 219 Let S 07 0 0747 0 07 07 76 17 3957 Determine7 if S is linearly independent or dependent Solution Let C11 07 0 020747 0 63010776 641a5a 73 07 07 lfthis equation has only trivial solutions7 then it is linealry independent This equaton gives the following system of linear equations 01 04 0 462 564 0 7603 7304 0 The augmented matrix for this system is 1 0 0 1 0 0 1 0 0 4 0 5 0 its GausiJordan form 0 1 0 125 0 0 0 i6 i3 0 0 0 1 5 0 Correspondingly 01 C4 0 02 12504 0 Cg 504 134 CHAPTER 4 VECTOR SPACES With 04 t as parameter we have 01 it 02 Cg C4 The equation above has nontrivial ie nonzero solutions So S is linearly dependent Theorem 449 Let V be a vector space and S V1 V2 Vk k 2 2 a set of elements vectors in V Then S is linearly dependent if and only if one of the vectors 1 can be written as a linear combination of the other vectors in S Proof Assume S is linearly dependent So the equation CiVi 62V2 Cka0 has a nonzero solution This means at least one of the c is nonzero Let c is the last one with c 31 0 So 01V102V267Vr0 and 01 02 0771 Vr7ivliivzii c c 0 So Vr is a linear combination of other vectors and this irnplication isproved Vril i to prove the other irnplication we assume that Vr is linear corn bination of other vectors So Vr CiVi Csz Crilvril 671Vr1 Cka So CiVi Csz Crilvril Vr 671Vr1 Cka 0 The left hand side is a nontrivial ie nozero linear combination because Vr has coef cient 71 Therefore S is linearly dependent This completes the proof I 45 BASIS AND DIMENSION 135 45 Basis and Dimension Homework Textbook7 45 EX 17 37 77 117157 197 217 237 257 287 357 377 397 417457 477 497 537 597 637 657 717 737 757 777 page 231 The main point of the section is 1 To de ne basis of a vectoiquot space 2 To de ne dimension of a vectoiquot space These are probably the two most fundamental concepts regarding vectoiquot spaces 136 CHAPTER 4 VECTOR SPACES De nition 451 Let V be a vector space and S V1v2 Vk be a set of elements vectorsin V We say that S is a basis of V if 1 S spans V and 2 S is linearly independent Remark Here are some some comments about nite and in nite basis of a vector space V 1 We avoided discussing in nite spanning set S and when an in nite S is linearly independent We will continue to avoid to do so An in nite set S is said span V if each element V E V is a linear combination of nitely many elements in V An in nite set S is said to be linearly independent if any nitely subset ofS is linearly independent 2 We say that a vector space V is nite dimensional if V has a basis consisting of nitely many elements Otherwise7 we say that V is in nite dimensional 3 The vector space P of all polynomials with real coef cients has in nite dimension Example 452 example 1 p 221 Most standard example of ba sis is the standard basis of R 1 Consider the vector space R2 Write e1 170762 07 Then7 e1e2 form a basis of R2 45 BASIS AND DIMENSION 137 2 Consider the vector space R3 Write el 17 07 0 eZ 071707 2 07 07 Then7 e1e2e3 form a basis of R3 Proof First7 for any vector V 17273 E R3 we have V 161 262 363 3 i So7 R is spanned by e1e2e3 Now7 we prove that e1e2e3 are linearly independent So7 sup pose 0161 6262 6363 0 610203 0 So7 cl 02 03 0 Therefore7 e1e2e3 are linearly indepen dent Hence e1e2e3 forms a basis of R3 The proof is complete I 3 More generally7 consider vector space R Write e1 100e2 010en 001 Then7 e1e2e3en form a basis of R The proof will be similar to the above proof This basis is called the standard basis of R Example 453 Consider v1 111v2 1711v3 1171 m R3 Then V1V2V3 form a basis for R3 138 CHAPTER 4 VECTOR SPACES Proof First7 we prove that V1V2V3 are linearly independent Let 01V102V203V3 0111102171103111 00 We have to prove cl 02 03 0 The equations give the following system of linear equations 01 62 63 0 01 02 63 1 l o 01 02 03 0 The augmented matrix is 1 0 0 0 1 71 1 0 its Gauss 7 Jordan form 0 1 0 0 1 1 i1 0 0 0 1 0 So7 cl 02 03 0 and this estblishes that V1V2V3 are linearly independent Now to show that V1V2 V3 spans R3 let V 1 2 3 be a vector in R3 We have to show that7 we can nd 0102053 such that 1 2 3 CiVi Csz 63V3 OR 17273 01111 021 711 031171 This gives the system of linear equations 01 02 03 1 1 1 1 01 1 01 02 03 2 1 1 1 02 2 01 02 03 3 1 1 1 03 3 45 BASIS AND DIMENSION 139 The coef cient matrix 1 1 1 l 0 5 5 l A 1 71 1 has inverse A71 5 75 0 1 1 71 l5 0 75l So7 the above system has tha solution C1 1 0 5 5 1 02 A71 2 5 75 0 2 03 3 0 3 So7 each vector 17273 is in the span of V1V2V3 So7 they form a basis of R3 The proof is complete I Reading assignment Read Textbook7 Examples 1 57 p 221 224 Theorem 454 Let V be a vector space and S V1V2 Vn be a basis of V Then every vector V in V can be written in one and only one way as a linear combination of vectors in S In other words V can be written as a unique linear combination of vectors in S Proof Since S spans V we can write V as a linear combination V01V1 62V2 quot Cnvn for scalars 0102 0 To prove uniqueness7 also let Vd1V1 d2V2 ngn for some other scalars d1 d2 dn Subtracting7 we have 01 d1V1 02 d2V2 6n dnVn 0 Since7 V1V2 Vn are also linearly independent7 we have Clidl0627d2076n7dn0 140 CHAPTER 4 VECTOR SPACES OR 01 d162 d2Cn This completes the proof I Theorem 455 Let V be a vector space and S V1V2 Vn be a basis of V Then every set of vectors in V containing more than n vectors in V is linearly dependent Proof Suppose Sl u1u2 um ne a set of m vectors in V with m gt n We are requaired to prove that the zero vector 0 is a nontrivial ie nonzero linear combination of elements in SI Since S is a basis7 we have 111 C11V1 012V2 CmVn 112 C21V1 022V2 62nVn 11m CmiVi Cm2V2 CmnVn Consider the system of linear equations C11 0222 Cm1m 0 C12 0222 Cm2m 0 Cin i 62n2 Cmnm 0 which is C11 022 39 39 39 Cmi 1 0 C12 022 39 39 39 sz 2 0 Cln CZn Cmnj rm L Since m gt n this hornegeneous system of linear equations has fewer equations than number of variables So7 the system has a nonzero solution see Textbook7 theorern 117 p 25 It follows that 1111 2u2 mum 0 45 BASIS AND DIMENSION 141 We justify it as follows First7 C11 022 cm 111 uz umv1 V2 Vn 012 622 0mg lcln CZ Cmnj and then 1 2 1u12u2mumu1 u2 um m which is C11 022 le x1 V1 V2 Vn 612 622 sz x2 Cln 62 cm zm which is Alternately at your level the proof will be written more explicitly as follows zlul zguz xmum m m V L V L m V L ji j1 i1 i1 j1 i1 The proof is complete I Theorem 456 Suppose V is a vector space and V has a basis with n vectors Then7 every basis has n vectors 142 CHAPTER 4 VECTOR SPACES Proof Let SV17V27quot397Vn and 1 ulaUZaquot39aum be two bases of V Since S is a basis and S1 is linearly independent7 by theorem 4557 we have m S n Similarly7 n S m So7 m n The proof is complete I De nition 457 If a vector space V has a basis consisting ofn vectors7 then we say that dimension of V is n We also write dimV n If V 0 is the zero vector space7 then the dimension of V is de ned as zero We say that the dimension of V is equal to the eardinalz39ty of any basis ofV The word eardz39nalz39ty is used to mean the number of elements in a set Theorem 458 Suppose V is a vector space of dimension n 1 Suppose S V1V2 Vn is a set of n linearly independent vectors Then S is basis of V 2 Suppose S V1V2 Vn is a set of n vectors If S spans V then S is basis of V Remark The theorem 458 means that7 if dimension of V matches with the number of ie 7cardinality7 of S then to check if S is a basis of V or not7 you have check only one of the two required prperties 1 indpendece or 2 spannning Example 459 Here are some standard examples 1 We have dimR 1 This is because 1 forms a basis for R 45 BASIS AND DIMENSION 143 2 We have dimR2 2 This is because the standard basis e1 10e2 01 consist of two elements 3 We have dimR3 3 This is because the standard basis e1 10 0e2 010e3 001 consist of three elements 4 Mor generally dimR n This is because the standard basis e1 1000e2 01 0en 001 consist of n elements 5 The dimension of the vector space Mm of all m gtlt n matrices is mn Notationally dimMm mn To see this let eh be the m gtlt 71 matrix whose z jth7entry is 1 and all the rest of the entries are zero Then Seu z3912mj12n forms a basis of Mm and S has mn elements 6 Also recall if a vector space V does not have a nite basis we say V is ini nite dimensional a The vector space ll of all polynomials with real coef cients has in nite dimension b The vector space CR of all continuous real valued functions on real line R has in nite dimension 144 CHAPTER 4 VECTOR SPACES Exercise 4510 Ex 4 changed p 230 Write down the stan dard basis of the vector space M33 of all 3 gtlt 27matrices with real entires Solution Let eh be the 3 gtlt 27matrix whose z jth7entry is 1 and all other entries are zero Then 6117912792179227 393317 932 forms a basis of M32 More explicitly 1 0 0 1 0 0 611 0 0 12 0 0 21 1 0 0 0 0 0 0 0 and 0 0 0 0 0 0 622 0 1 31 0 0 33 0 0 0 0 1 0 0 1 It is easy to verify that these vectors in M32 spans M32 and are linearly independent So they form a basis Exercise 4511 Ex 8 p 230 Explain why the set S 71 2 1 72 24 is not a basis of R2 Solution Note 71 2 1172 0274 So these three vectors are not linearly independent So S is not a basis of R2 Alternate argument We have dirn R2 2 and S has 3 elements So by theorern 456 above S cannot be a basis 45 BASIS AND DIMENSION 145 Exercise 4512 Ex 16 p 230 Explain7 why the set S 271772a727 7172 4727 74 is not a basis of R3 Solution Note 472774 271772 7 7217112 OR 27 17 72 7 727 717 2 7 4a 2a 74 0 07 So7 these three vectors are linearly dependent So7 S is not a basis of R3 Exercise 4513 Ex 24 p 230 Explain7 why the set S 6x 733952 7 2x 7 952 is not a basis of P2 Solution Note 17 2x 7 z 73695 i 3 i 13952 3 3 OR 1 1 1 7 2x 7 x2 6z i 3 gm 0 So7 these three vectors are linearly dependent So7 S is not a basis of P2 Exercise 4514 Ex 36p231 Deterrnine7 whether S 17271171 146 CHAPTER 4 VECTOR SPACES is a basis of R2 or not Solution We will show that S is linearly independent Let a12 b1 71 0 0 Then ab0 and 2a7b0 Solving7 we get a 0 b 0 So7 these two vectors are linearly indepen dent We have dirn R2 2 Therefore7 by theorem 4587 S is a basis of R2 Exercise 4515 Ex 40 p231 Deterrnine7 whether S 07 0 17 57 67 67 27 is a basis of R3 or not Solution We have 10 0 0 0156 062 1 00 0 So7 S is linearly dependent and hence is not a basis of R3 Remark In fact any subset S of a vector space V that contains 0 is linearly dependent Exercise 4516 Ex 46 p231 Deterrnine7 whether S 4t 7 5 t33t 5 2P 7 3 is a basis of P3 or not Solution Note the standard basis 1tt2t3 45 BASIS AND DIMENSION 147 of lP g has four elements So7 dim P3 4 Because of theorem 4587 we will try to check7 if S is linearly independent or not So7 let 014t 7 t2 025 t3 033t 5 042t3 i 3t2 0 for some scalars 010203C4 If we simplify7 we get 502 503 401 303t 701 7 304t2 02 204t3 0 Recall7 a polynomial is zero if and only if all the coef cients are zero So7 we have 562 503 0 461 303 0 701 7304 0 02 204 0 The augmented matrix is 0 5 5 0 0 1 0 0 0 0 4 0 3 0 0 0 1 0 0 0 its Gaussi Jordan form 71 0 0 73 0 0 0 1 0 0 0 1 0 2 0 0 0 0 1 0 Therefore7 cl 02 03 c4 0 Hence S is linearely independent So by theorem 4587 S is a basis of P3 Exercise 4517 Ex 60 p231 Determine the dimension of P4 Solution Recall7 P4 is the vector space of all polynomials of degree 3 4 We claim that that S 17t is a basis of P4 Clearly7 any polynomial in P4 is a linear combination of elements in S So7 S spans P4 Now7 we prove that S is linearly 148 CHAPTER 4 VECTOR SPACES independent So let 001 Clt Cgtz Cgts C4t4 Since a nonzero polynomial of degree 4 can have at most four roots it follows 00 cl 02 03 c4 0 So S is a basis of P4 and Exercise 4518 Ex 62 p231 Determine the dimension of M32 Solution In exercise 4510 we established that S 1179127921792279317932 is a basis of M32 So dirnM32 6 Exercise 4519 Ex 72 p231 Let W tst st E R Give a geometric description of W nd a basis of W and determine the dimension of W Solution First note that W is closed under addition and scalar rnulti plication So W is a subspace of R3 Notice there are two parameters st in the description of W So W can be described by z 2 Therefore W represents the plane z z in R3 1 suggest guess that u 1707 will form a basis of W To see that they are mutually linearly indepen dent let au bv 000 OR aba 000 45 BASIS AND DIMENSION 149 So a 0b 0 and hence they are linearly independent To see that they span W we have t s t tu 8V So uv form a basis of W and dirnW 2 Exercise 4520 Ex 74 p232 Let W 5t73ttt te R Fnd a basis of W and determine the dimension of W Solution First note that W is closed under addition and scalar rnul tiplication So W is a subspace of R4 Notice there is only parameters It in the description of W So I expect that dirnW 1 I suggest guess e 57311 is a basis of W This is easy to check So dirnW 1 150 CHAPTER 4 VECTOR SPACES 46 Rank of a matrix and SOLE Homework Textbook7 46 EX 77 97 157 177 197 277 297 337 357 377 417 437 477 497 577 63 Main topics in this section are to de ne Z We de ne row space of d mutants A and the column space of d mutants A 2 We de ne the mnh of d mdtmd 3 We de ne nullspace NA of d homoheneous system Ax 0 0f linedquot equations We also de ne the nullity of d mutants A 46 RANK OF A MATRIX AND SOLE 151 De nition 461 Let A calj be an m gtlt 71 matrix 1 The nituples corresponding to the rows of A are called row vectors of A 2 Similarly the mituples corresponding to the columns of A are called column vectors of A 3 The row space of A is the subspace of R spanned by row vectors of A 4 The column space of A is the subspace of R spanned by column vectors of A Theorem 462 Suppose A B are two m gtlt n matrices If A is row equivalent of B then row space of A is equal to the row space of B Proof This follows from the way row equivalence is de ned Since B is rwoequivalent to A rows of B are obtained by a series of scalar multiplication and addition of rows of A So it follows that row vectors of B are in the row space of A Therefore the subspace spanned by row vectors of B is contained in the row space of A So the row space of B is contained in the row space of A Since A is row equivalent of B it also follows the B is row equivalent of A We say that the relatz39onshz39p of being mw equtvalent is reflective Therefore by the same argumen the row space of A is contained in the row space of B So they are equal The proof is complete I Theorem 463 Suppose A is an m gtlt 71 matrix and B is row equivalent to A and B is in row echelon form Then the nonzero rows of B form a basis of the row space of A Proof From theorem 462 it follows that row space of A and B are some Also a basis of the row space of B is given by the nonzero rows of B The proof is complete I 152 CHAPTER 4 VECTOR SPACES Theorem 464 Suppose A is an m gtlt 71 matrix Then the row space and column space of A have same dimension Proof You can skip it I will not ask you to prove this Write 011 012 013 39 39 39 aln 021 022 023 39 39 39 am A 031 032 033 39 39 39 13m aml amZ am 39 39 39 amn Let V1 V2 Vm denote the row vectors of A and u1u2 un de note the column vectors of A Suppose that the row space of A has dimension 7 and Sb1b2br is a basis of the row space of A Also7 write bi bilabi27 1 We have V1 Ciibl 012b2 Cl39rbr V2 C21b1 022b2 Czrbr Vm Cmibi Cm2b2 Cm7 br Looking at the rst entry of each of these in equations7 we have 011 C11511 012521 quot39 Cl39rb39rl 021 C21511 022521 39 39 0239rb39rl 031 Cslbll 032521 39 39 Csrbri aml Cmibii Cm2b21 39 39 Cm39rb39rl Let c1 denote the ith column of the matrix C clj So7 it follows from these in equations that 111 bllcl 521 b39rlcr 46 RANK OF A MATRIX AND SOLE 153 Similarly7 looking at the jth entry of the above set ofelt11uations7 we have uj 51701 52702 b39rjcr So7 all the columns uj of A are in spanc1c2 cr Therefore7 the column space of A is contained in spanc1c2 cr It follows from this that the rank of the column space of A has dimension 3 r rank of the row space of A So7 dimcolumn space of A S dimr0w space of A Similarly7 dimr0w space of A S dimcolumn space of A So7 they are equal The proof is complete I De nition 465 Suppose A is an m gtlt 71 matrix The dimension ofthe row space equivalently7 of the column space of A is called the rank of A and is denoted by rankA Reading assignment Read Textbook7 Examples 2 57 p 234 461 The Nullspace of a matrix Theorem 466 Suppose A is an m gtlt 71 matrix Let NA denote the set of solutions of the homogeneous system Ax 0 Notationally NAX R Ax0 Then NA is a a subspace of R and is called the nullspace of A The dimension of NA is called the nullity of A Notationally nullityA dimNA 154 CHAPTER 4 VECTOR SPACES Proof First NA is nonempty because 0 E NA By theorem 433 we need only to check that NA is closed under addition and scalar multiplication Suppose Xy E NA and c is a scalar Then Ax0 Ay0 so AXy AxAy000 So X y E NA and NA is closed under addition Also Acx CAx CO 0 Therefore ex 6 NA and NA is closed under scalar multiplication Theorem 467 Suppose A is an m gtlt 71 matrix Then rankA nullityA 71 That means dimNA n 7 rankA ProofLet r rankA Let B be a matrix row equivalent to A and B is in Gauss Jordan form So only the rst 7 rows of B are nonzero Let B be the matrix formed by top 7 ie nonzero rows of B Now rankA rankB rankB nullityA nullityB nullityB So we need to prove rankB nullityB n Switching columns of B would only mean re labeling the variables like 1 gt gt zhxz gt gt 3 gt gt 2 In this way we can write B INC where C is a r gtlt n 7 7 matrix and corresponds to the variables x71n The homogeneous system corresponding to B is given by 1 quot39 611r1 012r2 Cln739rn 0 2 39 39 621r1 022r2 02m77n 0 7 071r1 Cr r2 Crn739rn 0 The solution space NB has n 7 r papameters A basis of NB is given by S Er1Er2 En 46 RANK OF A MATRIX AND SOLE 155 where Er1 01151 62152 Crler 6H1 50 071 and e1 6 R is the vector with 1 at the ith place and 0 elsewhere So7 nullityB cardinalityS n 7 r The proof is complete I Reading assignment Read Textbook7 Examples 67 77 p 241 242 462 Solutionf of SOLE Given a system of linear equations Ax b where A is an m gtlt n matrix7 we have the following 1 Corresponding to such a system Ax b there is a homogeneous system AX 0 2 The set of solutions NA of the homogeneous system Ax 0 is a subspace of R 3 ln contrast7 if b 31 0 the set of solutions of Ax b is not a subspace This is because 0 is not a solution of Ax b 4 The system Ax b may have many solution Let Xp denote a PARTICULAR one such solutions of Ax b 5 The we have Theorem 468 Every solution of the system Ax b can be written as xxpxh where Xh is a solution of the homogeneous system Ax 0 Proof Suppose x is any solution of Ax b We have Ax b and Axp b 156 CHAPTER 4 VECTOR SPACES Write xh x 7 Xp Then AXhAXXp AxiAxpbib0 So7 X is a solution of the homogeneoud system Ax 0 and x Xp Xh The proof is complete I Theorem 469 A system Ax bis consistent if and only if b is in the column space of A Proof Easy lt is7 in fact7 interpretation of the matrix multiplication Ax b Reading assignment Read Textbook7 Examples 897 p 244 245 Theorem 4610 Suppose A is a square matrix of size n gtlt n Then the following conditions are equivalent A is invertible Ax b has unique solution for every m gtlt 1 matrix b Ax 0 has only the trivial solution A is row equivalent to the identity matrix In detA 31 0 RankA n The n row vectors of A are linearly independent The 71 column vectors of A are linearly independent 46 RANK OF A MATRIX AND SOLE 157 Exercise 4611 Ex 8 p 246 Let 2731 A 5106 8775 a Find the rank of the matrix A b Find a basis of the row space of A c Find a basis of the column space of A Solution First7 the following is the row Echelon form of this matrix use Tl 1 7875 625 B 0 1 2 0 ool The rank of A is equal to the number of nonzero rows of B So7 rankA 2 A basis of the row space of A is given by the nonzero rwos of B So7 V1 17875625 and V2 01 2 form a basis of the row space of A The column space of A is same as the row space of the transpose AT We have 2 5 8 AT 73 10 i7 1 6 5 The following is the row Echelon form of this matrix use Tl 10 1 7 Z 3 C 0 1 02857 0 0 0 158 CHAPTER 4 VECTOR SPACES A basis of the column space of A is given by the nonzero rows of C to be written as column 1 0 111 1370 112 1 3 02857 Exercise 4612 Ex 16 p 246 Let S 17 2a 2 71a 07 0 17 17 g RS39 Find a basis of of the subspace spanned by S Solution We write these rows as a matrix Now the row space of A will be the same as the subspace spanned by S So7 we will nd a basis of the row space of A Use Tl and we get the row Echelon form of A is given by 122 B011 000 So7 a basis is u1172727 u20717139 Remark The answers regrading bases would not be unique The following will also be a basis of this space V1172727 V21707039 46 RANK OF A MATRIX AND SOLE 159 Exercise 4613 Ex 20 p 246 Let S 25 73 i2 i2 i3 2 75 13 72 2 1 7535 Q R4 Find a basis of of the subspace spanned by S Solution We write these rows as a matrix 2 5 73 72 72 73 2 75 A 1 3 72 2 7175 3 SJ Now the row space of A will be the same as the subspace spanned by S So we will nd a basis of the row space of A Use TI and we get the row Echelon form of A is given by 1 25 715 71 0 1 706 716 B 0 0 1 719 0 0 0 0 So a basis is u1 12571571 u2 01706716 u3 0 0 1 719 777 Exercise 4614 Ex 28 p 247 Let 3 76 21 A 72 4 714 1 72 7 Find the dimension of the solution space of Ax 0 160 CHAPTER 4 VECTOR SPACES Solution Step Z Find rank ofA Use T1 the row Echelon form of Ais 172 7 B 0 00 0001 So the number of nonzero rows of B is rankA 1 Step 2 By theorem 467 we have rankA nullityA n 3 so nullityA 3 71 2 That means that the solution space has dimension 2 Exercise 4615 Ex 32 p 247 Let 1 4 2 1 2 71 1 1 A 4 1 1 0 4 2 0 Find the dimension of the solution space of Ax 0 Solution Step Z Find rank ofA Use T1 the row Echelon form of A is 1 5 25 25 0 1 5 0 B 1 0 0 1 g 0 0 0 1 So the number of nonzero rows of B is rankA 4 Step 2 By theorem 467 we have rankA nullityA n 4 so nullityA 4 7 4 0 That means that the solution space has dimension 0 This also means that the the homogeneous system Ax 0 has only the trivial solution 46 RANK OF A MATRIX AND SOLE 161 Exercise 4616 Ex 38 edited p 247 Consider the homoge neous system 2x1 2z2 4z3 7224 0 961 2962 963 2964 0 1 2 43 M 0 Find the dimension of the solution space and give a basis of the same Solution We follow the following steps 1 First7 we write down the coef cient matrix 2 2 4 72 A 1 2 1 2 71 1 4 71 2 Use Tl7 the Gauss Jordan for of the matrix is 1 0 0 71 B 0 1 0 2 0 0 1 71 3 The rank of A is number of nonzero rows of B So7 rankA 3 by thm 467 nullityA nirankA 473 1 So7 the solution space has dimension 1 4 To nd the solution space7 we write down the homogeneous sys tem corresponding to the coee cient matrix B So7 we have 1 74 0 2 24 0 3 74 0 162 CHAPTER 4 VECTOR SPACES 5 Use 4 t as parameter and we have 1 It 2 721 3 It 4 6 So the solution space is given by t72ttt te R 7 A basis is obtained by substituting t 1 So u 17 727 17 forms a basis of the solution space Exercise 4617 Ex 39 p 247 Consider the homogeneous sys tem 9x1 74 72 720x41 0 12L 76m 74 729x41 0 3x1 72 77m 0 31 722 fig 784 0 Find the dimension of the solution space and give a basis of the same Solution We follow the following steps 1 First7 we write down the coef cient matrix 9 74 72 720 12 76 74 729 3 72 0 77 3 72 71 78 46 RANK OF A MATRIX AND SOLE 163 Use Tl7 the Gauss Jordan for of the matrix is 1007E 01015 3 0011 000 0 The rank of A is number of nonzero rows of B So7 rankA 3 by thm 467 nullityA nirankA 473 1 So7 the solution space has dimension 1 To nd the solution space7 we write down the homogeneous sys tem corresponding to the coee cient matrix B So7 we have 1 ig l 0 2 0 3 4 0 0 0 Use 4 t as parameter and we have 4 1 it 7151 3 z2 7 3 it 4 So the solution space is given by 4 gt715t 7m te R A basis is obtained by substituting t 1 So 4 7 715 711 u 37 7 7 forms a basis of the solution space 164 CHAPTER 4 VECTOR SPACES Exercise 4618 Ex 42 p 247 Consider the system of equations 3x1 7822 423 19 7622 223 424 5 5x1 22 z4 29 1 7222 223 8 Determine if this system is consistent If yes write the solution in the form x Xh xp where X is a solution of the corresponding homogeneous system Ax 0 and Xp is a particular solution Solution We follow the following steps 1 To nd a particular solution we write the augmented matrix of the nonhomogeneous system 3 i8 4 0 19 0 i6 2 4 5 5 0 22 1 29 l 1 72 2 0 8 J The Gauss Jordan form of the matrix is 1 0 0 i2 0 0 1 0 75 0 0 0 1 5 0 0 0 0 0 1 The last row suggests 0 1 So the system is not consistents Exercise 4619 Ex 44 p 247 Consider the system of equations 21 742 53 8 771 43 31 763 3 46 RANK OF A MATRIX AND SOLE 165 Determine7 if this system is consistentlf yes7 write the solution in the form x X xp where Xh is a solution of the corresponding homoge neous system Ax 0 and Xp is a particular solution Solution We follow the following steps 1 First7 the augmented matrix of the system is 2 74 5 8 77 14 4 728 3 76 1 12 Its Gauss Jordan form is 1 72 0 4 0 0 1 0 0 0 0 0 This corresponds to they system 1 722 4 3 7 0 The last row indicates that the system is consistent We use x2 t as a paramater and we have 1 42t zgt zg0 Thaking t 0 a particular solutions is xp 4 0 0 2 Now7 we proceed to nd the solution of the homogeneous system 21 742 53 0 771 43 0 31 763 3 0 166 CHAPTER 4 VECTOR SPACES a The coef cient matrix 245 A 77144 3 76 1 b Its Gauss Jordan form is 1 72 0 B 0 1 0 C The homogeneous system corresponding to B is 1 722 0 3 0 0 0 d We use x2 t as a paramater and we have zl2t zgt x30 e So7 in parametrix form Xh 2tt0 3 Final answer is Witht as parameter7 any solutions can be written as x Xh xp 2tt0 400 Exercise 4620 Ex 50 p 247 Let 1 3 2 1 A 71 1 2 and b 1 0 1 1 0 46 RANK OF A MATRIX AND SOLE 167 Determine7 if b is in the column space of A Solution The question means7 whether the system Ax b has a solutions ie is consistent Accordingly7 the augmented matrix of this system Ax b is 1 H HHOO 2 1 2 1 0 1 0 The Gauss Jordan form of this matrix is i 1 0 71 0 l 0 1 1 0 0 0 0 1 The last row indicates that the system is not consistent So7 b is not in the column space of A 168 CHAPTER 4 VECTOR SPACES
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