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ppt on Vector spaces (VCLA) by dhrumil patel and harshid panchal

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This document outlines the concepts of vector spaces and related topics in linear algebra, including subspaces, linear independence, and basis. It provides definitions, properties, examples, and theorems regarding vector spaces, demonstrating how to identify and work with these mathematical structures. The content is organized as an active learning assignment for students in the chemical division at L.D. College of Engineering, Ahmedabad, for the academic year 2014-15.

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Guided by: Prof. L.D. College Of Engineering Ahmedabad TOPIC : VECTOR SPACES BRANCH : CHEMICAL DIVISION : F SEM : 2ND ACADEMIC YEAR : 2014-15(EVEN) Active Learning assignment LINEAR ALIGEBRA AND VECTOR CALCULAS ACTIVE LEARNING ASSIGNMENT 1
NAME ENROLLMENT NO. PATEL DHRUMIL P. 140280105035 SHAIKH AZIZUL AMINUL HAQUE 140280105048 SHAIKH MAHOMMED BILAL 140280105049 PANCHAL SAHIL C. 140280105031 SONI SWARUP J. 140280105056 RAMANI AJAY M. 140280105040 VAKIL JAYADEEP A. 140280105060 UMARALIYA KETAN B. 140280105057 MODI KATHAN J. 140280105027 SONI KEVIN R. 140280105054 PANCHAL SARTHAK G. 140280105032 2
CONTENTS 1) Real Vector Spaces 2) Sub Spaces 3) Linear combination 4) Linear independence 5) Span Of Set Of Vectors 6) Basis 7) Dimension 8) Row Space, Column Space, Null Space 9) Rank And Nullity 10)Coordinate and change of basis 3
4 Vector Space Axioms Real Vector Spaces Let V be an arbitrary non empty set of objects on which two operations are defined, addition and multiplication by scalar (number). If the following axioms are satisfied by all objects u, v, w in V and all scalars k and l, then we call V a vector space and we call the objects in V vectors. 1) If u and v are objects in V, then u + v is in V 2) u + v = v + u 3) u + (v + w) = (u + v) + w 4) There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u for all u in V 5) For each u in V, there is an object –u in V, called a negative of u, such that u + (-u) = (-u) + u = 0 6) If k is any scalar and u is any object in V then ku is in V 7) k(u+v) = ku + kv 8) (k+l)(u) = ku + lu 9) k(lu) = (kl)u 10) 1u = u
Example: Rn is a vector space Let, V=Rn u+v = (u1,u2,u3,. . . . . un)+(v1,v2,v3,. . . . vn) = (u1+v1,u2+v2,u3+v3, . . . .un+vn ) Ku =(ku1,ku2,ku3,. . . . . kun) Step :1- identify the set V of objects that will become vectors. Step :2- identify the edition and scalar multiplication operations on V. Step :3- verify axioms 1 & 6, adding two vectors in a V produces a vector in a V , and multiplying a vector in a V by scalar produces a vector in a V. Step :4- confirm that axioms 2,3,4,5,7,8,9 and 10 hold. 5
Example: Set Is Not A Vector 6
Examples of vector spaces (1) n-tuple space: Rn ),,,(),,,(),,,( 22112121 nnnn vuvuvuvvvuuu   ),,,(),,,( 2121 nn kukukuuuuk   (2) Matrix space: (the set of all m×n matrices with real values) nmMV  Ex: ：(m = n = 2)                     22222121 12121111 2221 1211 2221 1211 vuvu vuvu vv vv uu uu             2221 1211 2221 1211 kuku kuku uu uu k vector addition scalar multiplication vector addition scalar multiplication 7
(3) n-th degree polynomial space: (the set of all real polynomials of degree n or less) n nn xbaxbabaxqxp )()()()()( 1100   n n xkaxkakaxkp  10)( (4) Function space:(the set of all real-valued continuous functions defined on the entire real line.) )()())(( xgxfxgf  ),(  cV )())(( xkfxkf  vector addition scalar multiplication vector addition scalar multiplication 8
9 PROPERTIES OF VECTORS Let v be any element of a vector space V, and let c be any scalar. Then the following properties are true. vv 0v0v 00 0v     )1((4) or0then,If(3) (2) 0(1) cc c
10 Definition: ),,( V : a vector space      VW W  : a non empty subset ),,( W ：a vector space (under the operations of addition and scalar multiplication defined in V)  W is a subspace of V Subspaces If W is a set of one or more vectors in a vector space V, then W is a sub space of V if and only if the following condition hold; a)If u,v are vectors in a W then u+v is in a W. b)If k is any scalar and u is any vector In a W then ku is in W.
11 Every vector space V has at least two subspaces (1)Zero vector space {0} is a subspace of V. (2) V is a subspace of V.  Ex: Subspace of R2    00,(1) 00 originhethrough tLines(2) 2 (3) R • Ex: Subspace of R3 originhethrough tPlanes(3) 3 (4) R    00,0,(1) 00 originhethrough tLines(2) If w1,w2,. . .. wr subspaces of vector space V then the intersection is this subspaces is also subspace of V.
Let W be the set of all 2×2 symmetric matrices. Show that W is a subspace of the vector space M2×2, with the standard operations of matrix addition and scalar multiplication. sapcesvector:2222  MMW Sol: )(Let 221121 AA,AAWA,A TT  )( 21212121 AAAAAAWAW,A TTT  )( kAkAkAWA,Rk TT  22ofsubspaceais  MW )( 21 WAA  )( WkA Ex : (A subspace of M2×2) 12
WBA        10 01 222 ofsubspaceanotis  MW Let W be the set of singular matrices of order 2 Show that W is not a subspace of M2×2 with the standard operations. WB,WA              10 00 00 01 Sol: Ex : (The set of singular matrices is not a subspace of M2×2) 13
14 kkccc uuuv  2211 formin thewrittenbecanifinvectorsthe ofncombinatiolinearacalledisspacevectorainA vector 21 vuuu v V,,, V k Definition scalars:21 k,c,,cc  Linear combination If S = (w1,w2,w3, . . . . , wr) is a nonempty set of vectors in a vector space V, then: (a) The set W of all possible linear combinations of the vectors in S is a subspace of V. (b) The set W in part (a) is the “smallest” subspace of V that contains all of the vectors in S in the sense that any other subspace that contains those vectors contains W.
15              2123 2012 1101    nEliminatioJordanGuass             7000 4210 1101 )70(solutionnohassystemthis   332211 vvvw ccc  V1=(1,2,3) v2=(0,0,2) v3= (-1,0,1) Prove W=(1,-2,2) is not a linear combination of v1,v2,v3 w = c1v1 + c2v2 + c3v3 Finding a linear combination Sol :
dependent.linearlycalledisthen zeros),allnot(i.e.,solutionnontrivialahasequationtheIf(2) t.independenlinearlycalledisthen )0(solutiontrivialonly thehasequationtheIf(1) 21 S S ccc k     0vvv vvv   kk k ccc S   2211 21 ,,, : a set of vectors in a vector space V Linear Independent (L.I.) and Linear Dependent (L.D.): Definition: 16 Theorem A set S with two or more vectors is (a) Linearly dependent if and only if at least one of the vectors in S is expressible as a linear combination of the other vectors in S. (b) Linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S.
17 tindependenlinearlyis(1)  dependent.linearlyis(2) SS 0   tindependenlinearlyis(3) v0v  21(4) SS  dependentlinearlyisdependentlinearlyis 21 SS  tindependenlinearlyistindependenlinearlyis 12 SS  Notes
      10,2,,21,0,,32,1, S 023 02 02 321 21 31    ccc cc cc  0vvv 332211 ccc Sol: Determine whether the following set of vectors in R 3 is L.I. or L.D.             0123 0012 0201   nEliminatioJordan-Gauss           0100 0010 0001  solutiontrivialonly the0321  ccc tindependenlinearlyisS v1 v2 v3 Ex : Testing for linearly independent 18
19 Span of set of vectors If S={v1, v2,…, vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S. )(Sspan   )invectorsofnscombinatiolinearallofset(the 2211 S Rcccc ikk  vvv  If every vector in a given vector space can be written as a linear combination of vectors in a given set S, then S is called a spanning set of the vector space. Definition: If S={v1, v2,…, vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S.
20 (a)span (S) is a subspace of V. (b)span (S) is the smallest subspace of V that contains S. (Every other subspace of V that contains S must contain span (S). If S={v1, v2,…, vk} is a set of vectors in a vector space V, then  0)((1) span )((2) SspanS  )()( ,(3) 2121 21 SspanSspanSS VSS    Notes: VS SV VS VS ofsetspanningais by)(generatedspannedis )(generatesspans )(span  
21 Ex: A spanning set for R3   sapns)1,0,2(),2,1,0(),3,2,1(setthat theShow 3 RS  .and,,ofncombinatiolinearaasbecanin ),,(vectorarbitraryanwhetherdeterminemustWe 321 3 321 vvv u R uuu Sol: 332211 3 vvvuu cccR  3321 221 131 23 2 2 uccc ucc ucc    .and,,ofvaluesallforconsistentis systemthiswhethergdeterminintoreducesthusproblemThe 321 uuu 0 123 012 201   A u.everyforsolutiononeexactlyhasbx  A 3 )( RSspan 
22 Basis • Definition: V：a vector space Generating Sets Bases Linearly Independent Sets  S is called a basis for V S ={v1, v2, …, vn}V • S spans V (i.e., span(S) = V ) • S is linearly independent (1) Ø is a basis for {0} (2) the standard basis for R3: {i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)  Notes:
23 (3) the standard basis for R n : {e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1) Ex: R4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} Ex: matrix space:                               10 00 , 01 00 , 00 10 , 00 01 22 (4) the standard basis for mn matrix space: { Eij | 1im , 1jn } (5) the standard basis for Pn(x): {1, x, x2, …, xn} Ex: P3(x) {1, x, x2, x3}
24 THEOREMS Uniqueness of basis representation If S= {v1,v2,…,vn} is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of vectors in S. If S= {v1,v2,…,vn} is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent. Bases and linear dependence
25 If a vector space V has one basis with n vectors, then every basis for V has n vectors. (All bases for a finite-dimensional vector space has the same number of vectors.) Number of vectors in a basis An INDEPENDENT set of vectors that SPANS a vector space V is called a BASIS for V.
26 Dimension  Definition: The dimension of a finite dimensional vector space V is defined to be the number of vectors in a basis for V. V: a vector space S: a basis for V Finite dimensional A vector space V is called finite dimensional, if it has a basis consisting of a finite number of elements Infinite dimensional If a vector space V is not finite dimensional,then it is called infinite dimensional. • Dimension of vector space V is denoted by dim(V).
27 Theorems for dimention THEOREM 1 All bases for a finite-dimensional vector space have the same number of vectors. THEOREM 2 Let V be a finite-dimensional vector space, and let be any basis. (a) If a set has more than n vectors, then it is linearly dependent. (b) If a set has fewer than n vectors, then it does not span V.
28 Dimensions of Some Familiar Vector Spaces (1) Vector space Rn  basis {e1 , e2 ,  , en} (2) Vector space Mm  basis {Eij | 1im , 1jn} (3) Vector space Pn(x)  basis {1, x, x2,  , xn} (4) Vector space P(x)  basis {1, x, x2, }  dim(Rn) = n  dim(Mmn)=mn  dim(Pn(x)) = n+1  dim(P(x)) = 
29 Dimension of a Solution Space EXAMPLE
30  Row space: The row space of A is the subspace of Rn spanned by the row vectors of A.  Column space: The column space of A is the subspace of Rm spanned by the column vectors of A.   },,{ 21 )((2) 2 (1) 1 RAAAACS n n n    }|{)( 0xx  ARANS n  Null space: The null space of A is the set of all solutions of Ax=0 and it is a subspace of Rn. Let A be an m×n matrix Raw space ,column space and null space The null space of A is also called the solution space of the homogeneous system Ax = 0.
                                 mmnmm n n A A A aaa aaa aaa A      2 1 21 22221 11211 )( (2) (1) ],,,[ ],,,[ ],,,[ nmnm2m1 2n2221 1n1211 Aaaa Aaaa Aaaa        Row vectors of A Row Vectors:        n mnmm n n AAA aaa aaa aaa A      21 21 22221 11211                                                   mn n n mm a a a a a a a a a    2 1 2 22 12 1 21 11 Column vectors of A Column Vectors: || || || A (1) A (2) A (n)
32 THEOREM 1 Elementary row operations do not change the null space of a matrix. THEOREM 2 The row space of a matrix is not changed by elementary row operations. RS(r(A)) = RS(A) r: elementary row operations THEOREM 3 If a matrix R is in row echelon form, then the row vectors with the leading 1′s (the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1′s of the row vectors form a basis for the column space of R.
33 Find a basis of row space of A =                    2402 1243 1603 0110 3131 Sol:                    2402 1243 1603 0110 3131 A=                 0000 0000 1000 0110 3131 3 2 1 w w w B = .E.G bbbbaaaa 43214321 Finding a basis for a row space EXAMPLE A basis for RS(A) = {the nonzero row vectors of B} (Thm 3) = {w1, w2, w3} = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)}
34 THEOREM 4 If A and B are row equivalent matrices, then: (a) A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent. (b) A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B. If a matrix A is row equivalent to a matrix B in row-echelon form, then the nonzero row vectors of B form a basis for the row space of A. THEOREM 5
35 Finding a basis for the column space of a matrix                    2402 1243 1603 0110 3131 A Sol: 3 2 1 .. 00000 11100 65910 23301 21103 42611 04013 23301 w w w                                 BA EGT EXAMPLE
36 CS(A)=RS(AT) (a basis for the column space of A) A Basis For CS(A) = a basis for RS(AT) = {the nonzero vectors of B} = {w1, w2, w3}                                                                        1 1 1 0 0 , 6 5 9 1 0 , 2 3 3 0 1
37 Finding the solution space of a homogeneous system Ex: Find the null space of the matrix A. Sol: The null space of A is the solution space of Ax = 0.              3021 4563 1221 A                          0000 1100 3021 3021 4563 1221 .. EJG A x1 = –2s – 3t, x2 = s, x3 = –t, x4 = t 21 vvx tsts t t s ts x x x x                                                          1 1 0 3 0 0 1 232 4 3 2 1 },|{)( 21 RtstsANS  vv
38 Rank and Nullity If A is an mn matrix, then the row space and the column space of A have the same dimension. dim(RS(A)) = dim(CS(A)) THEOREM The dimension of the row (or column) space of a matrix A is called the rank of A and is denoted by rank(A). rank(A) = dim(RS(A)) = dim(CS(A)) Rank: Nullity: The dimension of the null space of A is called the nullity of A. nullity(A) = dim(NS(A))
39 THEOREM If A is an mn matrix of rank r, then the dimension of the solution space of Ax = 0 is n – r. That is n = rank(A) + nullity(A) • Notes: (1) rank(A): The number of leading variables in the solution of Ax=0. (The number of nonzero rows in the row-echelon form of A) (2) nullity (A): The number of free variables in the solution of Ax = 0.
40 Rank and nullity of a matrix Let the column vectors of the matrix A be denoted by a1, a2, a3, a4, and a5.                  120930 31112 31310 01201 A a1 a2 a3 a4 a5 EXAMPLE Find the Rank and nullity of the matrix. Sol: Let B be the reduced row-echelon form of A.                                  00000 11000 40310 10201 120930 31112 31310 01201 BA a1 a2 a3 a4 a5 b1 b2 b3 b4 b5 235)(rank)(nuillity  AnA G.E. rank(A) = 3 (the number of nonzero rows in B) 235)(rank)(nuillity  AnA
41 Coordinates and change of basis • Coordinate representation relative to a basis Let B = {v1, v2, …, vn} be an ordered basis for a vector space V and let x be a vector in V such that .2211 nnccc vvvx   The scalars c1, c2, …, cn are called the coordinates of x relative to the basis B. The coordinate matrix (or coordinate vector) of x relative to B is the column matrix in Rn whose components are the coordinates of x.                n B c c c  2 1 x
Find the coordinate matrix of x=(1, 2, –1) in R3 relative to the (nonstandard) basis B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)} Sol: 2100 8010 5001 1521 2310 1201 E.G.J.                         )5,3,2()2,1,0()1,0,1()1,2,1( 321332211  cccccc uuux                                       1 2 1 521 310 201 i.e. 152 23 12 3 2 1 321 32 31 c c c ccc cc cc              2 8 5 ][ B x Finding a coordinate matrix relative to a nonstandard basis
43 Change of basis problem You were given the coordinates of a vector relative to one basis B and were asked to find the coordinates relative to another basis B'. Transition matrix from B' to B: V BB nn spacevectorafor basestwobe}...,,{nda},...,,{etL 2121 uuuuuu  BB P  ][][hent vv         BBnBB v,...,,  1 uuu 2       BnBB P uuu 2  ...,,,1 where is called the transition matrix from B' to B If [v]B is the coordinate matrix of v relative to B [v]B‘ is the coordinate matrix of v relative to B'
44 If P is the transition matrix from a basis B' to a basis B in Rn, then (1) P is invertible (2) The transition matrix from B to B' is P –1 THEOREM 1 The inverse of a transition matrix THEOREM 2 Let B={v1, v2, … , vn} and B' ={u1, u2, … , un} be two bases for R n . Then the transition matrix P–1 from B to B' can be found by using Gauss-Jordan elimination on the n×2n matrix as follows. Transition matrix from B to B'  BB   1 PIn 
B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two bases for R 2 (a) Find the transition matrix from B' to B. (b) (c) Find the transition matrix from B to B' . BB ][find, 2 1 ][Let ' vv        Ex : (Finding a transition matrix) Sol:           22 21 22 43           12 23 10 01  G.J.E. B B' I P          12 23 P (the transition matrix from B' to B) (a) Sol:           22 21 22 43   G.J.E. B B' (a)
                            0 1 2 1 12 23 ][][ 2 1 ][ BBB P vvv (b)           22 43 22 21           32 21 10 01  G.J.E. B' B I P -1 (the transition matrix from B to B')          32 211 P  Check: 2 1 10 01 32 21 12 23 IPP                        )2,3()2,4)(0()2,3)(1( 0 1 ][ )2,3()2,2)(2()2,1)(1( 2 1 ][ :Check               vv vv B B (c)
References: Anton-linear algebra book & . .. . Dhrumil Patel(L.D. College of engineering)

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ppt on Vector spaces (VCLA) by dhrumil patel and harshid panchal

  • 1. Guided by: Prof. L.D. College Of Engineering Ahmedabad TOPIC : VECTOR SPACES BRANCH : CHEMICAL DIVISION : F SEM : 2ND ACADEMIC YEAR : 2014-15(EVEN) Active Learning assignment LINEAR ALIGEBRA AND VECTOR CALCULAS ACTIVE LEARNING ASSIGNMENT 1
  • 2. NAME ENROLLMENT NO. PATEL DHRUMIL P. 140280105035 SHAIKH AZIZUL AMINUL HAQUE 140280105048 SHAIKH MAHOMMED BILAL 140280105049 PANCHAL SAHIL C. 140280105031 SONI SWARUP J. 140280105056 RAMANI AJAY M. 140280105040 VAKIL JAYADEEP A. 140280105060 UMARALIYA KETAN B. 140280105057 MODI KATHAN J. 140280105027 SONI KEVIN R. 140280105054 PANCHAL SARTHAK G. 140280105032 2
  • 3. CONTENTS 1) Real Vector Spaces 2) Sub Spaces 3) Linear combination 4) Linear independence 5) Span Of Set Of Vectors 6) Basis 7) Dimension 8) Row Space, Column Space, Null Space 9) Rank And Nullity 10)Coordinate and change of basis 3
  • 4. 4 Vector Space Axioms Real Vector Spaces Let V be an arbitrary non empty set of objects on which two operations are defined, addition and multiplication by scalar (number). If the following axioms are satisfied by all objects u, v, w in V and all scalars k and l, then we call V a vector space and we call the objects in V vectors. 1) If u and v are objects in V, then u + v is in V 2) u + v = v + u 3) u + (v + w) = (u + v) + w 4) There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u for all u in V 5) For each u in V, there is an object –u in V, called a negative of u, such that u + (-u) = (-u) + u = 0 6) If k is any scalar and u is any object in V then ku is in V 7) k(u+v) = ku + kv 8) (k+l)(u) = ku + lu 9) k(lu) = (kl)u 10) 1u = u
  • 5. Example: Rn is a vector space Let, V=Rn u+v = (u1,u2,u3,. . . . . un)+(v1,v2,v3,. . . . vn) = (u1+v1,u2+v2,u3+v3, . . . .un+vn ) Ku =(ku1,ku2,ku3,. . . . . kun) Step :1- identify the set V of objects that will become vectors. Step :2- identify the edition and scalar multiplication operations on V. Step :3- verify axioms 1 & 6, adding two vectors in a V produces a vector in a V , and multiplying a vector in a V by scalar produces a vector in a V. Step :4- confirm that axioms 2,3,4,5,7,8,9 and 10 hold. 5
  • 6. Example: Set Is Not A Vector 6
  • 7. Examples of vector spaces (1) n-tuple space: Rn ),,,(),,,(),,,( 22112121 nnnn vuvuvuvvvuuu   ),,,(),,,( 2121 nn kukukuuuuk   (2) Matrix space: (the set of all m×n matrices with real values) nmMV  Ex: ：(m = n = 2)                     22222121 12121111 2221 1211 2221 1211 vuvu vuvu vv vv uu uu             2221 1211 2221 1211 kuku kuku uu uu k vector addition scalar multiplication vector addition scalar multiplication 7
  • 8. (3) n-th degree polynomial space: (the set of all real polynomials of degree n or less) n nn xbaxbabaxqxp )()()()()( 1100   n n xkaxkakaxkp  10)( (4) Function space:(the set of all real-valued continuous functions defined on the entire real line.) )()())(( xgxfxgf  ),(  cV )())(( xkfxkf  vector addition scalar multiplication vector addition scalar multiplication 8
  • 9. 9 PROPERTIES OF VECTORS Let v be any element of a vector space V, and let c be any scalar. Then the following properties are true. vv 0v0v 00 0v     )1((4) or0then,If(3) (2) 0(1) cc c
  • 10. 10 Definition: ),,( V : a vector space      VW W  : a non empty subset ),,( W ：a vector space (under the operations of addition and scalar multiplication defined in V)  W is a subspace of V Subspaces If W is a set of one or more vectors in a vector space V, then W is a sub space of V if and only if the following condition hold; a)If u,v are vectors in a W then u+v is in a W. b)If k is any scalar and u is any vector In a W then ku is in W.
  • 11. 11 Every vector space V has at least two subspaces (1)Zero vector space {0} is a subspace of V. (2) V is a subspace of V.  Ex: Subspace of R2    00,(1) 00 originhethrough tLines(2) 2 (3) R • Ex: Subspace of R3 originhethrough tPlanes(3) 3 (4) R    00,0,(1) 00 originhethrough tLines(2) If w1,w2,. . .. wr subspaces of vector space V then the intersection is this subspaces is also subspace of V.
  • 12. Let W be the set of all 2×2 symmetric matrices. Show that W is a subspace of the vector space M2×2, with the standard operations of matrix addition and scalar multiplication. sapcesvector:2222  MMW Sol: )(Let 221121 AA,AAWA,A TT  )( 21212121 AAAAAAWAW,A TTT  )( kAkAkAWA,Rk TT  22ofsubspaceais  MW )( 21 WAA  )( WkA Ex : (A subspace of M2×2) 12
  • 13. WBA        10 01 222 ofsubspaceanotis  MW Let W be the set of singular matrices of order 2 Show that W is not a subspace of M2×2 with the standard operations. WB,WA              10 00 00 01 Sol: Ex : (The set of singular matrices is not a subspace of M2×2) 13
  • 14. 14 kkccc uuuv  2211 formin thewrittenbecanifinvectorsthe ofncombinatiolinearacalledisspacevectorainA vector 21 vuuu v V,,, V k Definition scalars:21 k,c,,cc  Linear combination If S = (w1,w2,w3, . . . . , wr) is a nonempty set of vectors in a vector space V, then: (a) The set W of all possible linear combinations of the vectors in S is a subspace of V. (b) The set W in part (a) is the “smallest” subspace of V that contains all of the vectors in S in the sense that any other subspace that contains those vectors contains W.
  • 15. 15              2123 2012 1101    nEliminatioJordanGuass             7000 4210 1101 )70(solutionnohassystemthis   332211 vvvw ccc  V1=(1,2,3) v2=(0,0,2) v3= (-1,0,1) Prove W=(1,-2,2) is not a linear combination of v1,v2,v3 w = c1v1 + c2v2 + c3v3 Finding a linear combination Sol :
  • 16. dependent.linearlycalledisthen zeros),allnot(i.e.,solutionnontrivialahasequationtheIf(2) t.independenlinearlycalledisthen )0(solutiontrivialonly thehasequationtheIf(1) 21 S S ccc k     0vvv vvv   kk k ccc S   2211 21 ,,, : a set of vectors in a vector space V Linear Independent (L.I.) and Linear Dependent (L.D.): Definition: 16 Theorem A set S with two or more vectors is (a) Linearly dependent if and only if at least one of the vectors in S is expressible as a linear combination of the other vectors in S. (b) Linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S.
  • 17. 17 tindependenlinearlyis(1)  dependent.linearlyis(2) SS 0   tindependenlinearlyis(3) v0v  21(4) SS  dependentlinearlyisdependentlinearlyis 21 SS  tindependenlinearlyistindependenlinearlyis 12 SS  Notes
  • 18.       10,2,,21,0,,32,1, S 023 02 02 321 21 31    ccc cc cc  0vvv 332211 ccc Sol: Determine whether the following set of vectors in R 3 is L.I. or L.D.             0123 0012 0201   nEliminatioJordan-Gauss           0100 0010 0001  solutiontrivialonly the0321  ccc tindependenlinearlyisS v1 v2 v3 Ex : Testing for linearly independent 18
  • 19. 19 Span of set of vectors If S={v1, v2,…, vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S. )(Sspan   )invectorsofnscombinatiolinearallofset(the 2211 S Rcccc ikk  vvv  If every vector in a given vector space can be written as a linear combination of vectors in a given set S, then S is called a spanning set of the vector space. Definition: If S={v1, v2,…, vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S.
  • 20. 20 (a)span (S) is a subspace of V. (b)span (S) is the smallest subspace of V that contains S. (Every other subspace of V that contains S must contain span (S). If S={v1, v2,…, vk} is a set of vectors in a vector space V, then  0)((1) span )((2) SspanS  )()( ,(3) 2121 21 SspanSspanSS VSS    Notes: VS SV VS VS ofsetspanningais by)(generatedspannedis )(generatesspans )(span  
  • 21. 21 Ex: A spanning set for R3   sapns)1,0,2(),2,1,0(),3,2,1(setthat theShow 3 RS  .and,,ofncombinatiolinearaasbecanin ),,(vectorarbitraryanwhetherdeterminemustWe 321 3 321 vvv u R uuu Sol: 332211 3 vvvuu cccR  3321 221 131 23 2 2 uccc ucc ucc    .and,,ofvaluesallforconsistentis systemthiswhethergdeterminintoreducesthusproblemThe 321 uuu 0 123 012 201   A u.everyforsolutiononeexactlyhasbx  A 3 )( RSspan 
  • 22. 22 Basis • Definition: V：a vector space Generating Sets Bases Linearly Independent Sets  S is called a basis for V S ={v1, v2, …, vn}V • S spans V (i.e., span(S) = V ) • S is linearly independent (1) Ø is a basis for {0} (2) the standard basis for R3: {i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)  Notes:
  • 23. 23 (3) the standard basis for R n : {e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1) Ex: R4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} Ex: matrix space:                               10 00 , 01 00 , 00 10 , 00 01 22 (4) the standard basis for mn matrix space: { Eij | 1im , 1jn } (5) the standard basis for Pn(x): {1, x, x2, …, xn} Ex: P3(x) {1, x, x2, x3}
  • 24. 24 THEOREMS Uniqueness of basis representation If S= {v1,v2,…,vn} is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of vectors in S. If S= {v1,v2,…,vn} is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent. Bases and linear dependence
  • 25. 25 If a vector space V has one basis with n vectors, then every basis for V has n vectors. (All bases for a finite-dimensional vector space has the same number of vectors.) Number of vectors in a basis An INDEPENDENT set of vectors that SPANS a vector space V is called a BASIS for V.
  • 26. 26 Dimension  Definition: The dimension of a finite dimensional vector space V is defined to be the number of vectors in a basis for V. V: a vector space S: a basis for V Finite dimensional A vector space V is called finite dimensional, if it has a basis consisting of a finite number of elements Infinite dimensional If a vector space V is not finite dimensional,then it is called infinite dimensional. • Dimension of vector space V is denoted by dim(V).
  • 27. 27 Theorems for dimention THEOREM 1 All bases for a finite-dimensional vector space have the same number of vectors. THEOREM 2 Let V be a finite-dimensional vector space, and let be any basis. (a) If a set has more than n vectors, then it is linearly dependent. (b) If a set has fewer than n vectors, then it does not span V.
  • 28. 28 Dimensions of Some Familiar Vector Spaces (1) Vector space Rn  basis {e1 , e2 ,  , en} (2) Vector space Mm  basis {Eij | 1im , 1jn} (3) Vector space Pn(x)  basis {1, x, x2,  , xn} (4) Vector space P(x)  basis {1, x, x2, }  dim(Rn) = n  dim(Mmn)=mn  dim(Pn(x)) = n+1  dim(P(x)) = 
  • 29. 29 Dimension of a Solution Space EXAMPLE
  • 30. 30  Row space: The row space of A is the subspace of Rn spanned by the row vectors of A.  Column space: The column space of A is the subspace of Rm spanned by the column vectors of A.   },,{ 21 )((2) 2 (1) 1 RAAAACS n n n    }|{)( 0xx  ARANS n  Null space: The null space of A is the set of all solutions of Ax=0 and it is a subspace of Rn. Let A be an m×n matrix Raw space ,column space and null space The null space of A is also called the solution space of the homogeneous system Ax = 0.
  • 31.                                  mmnmm n n A A A aaa aaa aaa A      2 1 21 22221 11211 )( (2) (1) ],,,[ ],,,[ ],,,[ nmnm2m1 2n2221 1n1211 Aaaa Aaaa Aaaa        Row vectors of A Row Vectors:        n mnmm n n AAA aaa aaa aaa A      21 21 22221 11211                                                   mn n n mm a a a a a a a a a    2 1 2 22 12 1 21 11 Column vectors of A Column Vectors: || || || A (1) A (2) A (n)
  • 32. 32 THEOREM 1 Elementary row operations do not change the null space of a matrix. THEOREM 2 The row space of a matrix is not changed by elementary row operations. RS(r(A)) = RS(A) r: elementary row operations THEOREM 3 If a matrix R is in row echelon form, then the row vectors with the leading 1′s (the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1′s of the row vectors form a basis for the column space of R.
  • 33. 33 Find a basis of row space of A =                    2402 1243 1603 0110 3131 Sol:                    2402 1243 1603 0110 3131 A=                 0000 0000 1000 0110 3131 3 2 1 w w w B = .E.G bbbbaaaa 43214321 Finding a basis for a row space EXAMPLE A basis for RS(A) = {the nonzero row vectors of B} (Thm 3) = {w1, w2, w3} = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)}
  • 34. 34 THEOREM 4 If A and B are row equivalent matrices, then: (a) A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent. (b) A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B. If a matrix A is row equivalent to a matrix B in row-echelon form, then the nonzero row vectors of B form a basis for the row space of A. THEOREM 5
  • 35. 35 Finding a basis for the column space of a matrix                    2402 1243 1603 0110 3131 A Sol: 3 2 1 .. 00000 11100 65910 23301 21103 42611 04013 23301 w w w                                 BA EGT EXAMPLE
  • 36. 36 CS(A)=RS(AT) (a basis for the column space of A) A Basis For CS(A) = a basis for RS(AT) = {the nonzero vectors of B} = {w1, w2, w3}                                                                        1 1 1 0 0 , 6 5 9 1 0 , 2 3 3 0 1
  • 37. 37 Finding the solution space of a homogeneous system Ex: Find the null space of the matrix A. Sol: The null space of A is the solution space of Ax = 0.              3021 4563 1221 A                          0000 1100 3021 3021 4563 1221 .. EJG A x1 = –2s – 3t, x2 = s, x3 = –t, x4 = t 21 vvx tsts t t s ts x x x x                                                          1 1 0 3 0 0 1 232 4 3 2 1 },|{)( 21 RtstsANS  vv
  • 38. 38 Rank and Nullity If A is an mn matrix, then the row space and the column space of A have the same dimension. dim(RS(A)) = dim(CS(A)) THEOREM The dimension of the row (or column) space of a matrix A is called the rank of A and is denoted by rank(A). rank(A) = dim(RS(A)) = dim(CS(A)) Rank: Nullity: The dimension of the null space of A is called the nullity of A. nullity(A) = dim(NS(A))
  • 39. 39 THEOREM If A is an mn matrix of rank r, then the dimension of the solution space of Ax = 0 is n – r. That is n = rank(A) + nullity(A) • Notes: (1) rank(A): The number of leading variables in the solution of Ax=0. (The number of nonzero rows in the row-echelon form of A) (2) nullity (A): The number of free variables in the solution of Ax = 0.
  • 40. 40 Rank and nullity of a matrix Let the column vectors of the matrix A be denoted by a1, a2, a3, a4, and a5.                  120930 31112 31310 01201 A a1 a2 a3 a4 a5 EXAMPLE Find the Rank and nullity of the matrix. Sol: Let B be the reduced row-echelon form of A.                                  00000 11000 40310 10201 120930 31112 31310 01201 BA a1 a2 a3 a4 a5 b1 b2 b3 b4 b5 235)(rank)(nuillity  AnA G.E. rank(A) = 3 (the number of nonzero rows in B) 235)(rank)(nuillity  AnA
  • 41. 41 Coordinates and change of basis • Coordinate representation relative to a basis Let B = {v1, v2, …, vn} be an ordered basis for a vector space V and let x be a vector in V such that .2211 nnccc vvvx   The scalars c1, c2, …, cn are called the coordinates of x relative to the basis B. The coordinate matrix (or coordinate vector) of x relative to B is the column matrix in Rn whose components are the coordinates of x.                n B c c c  2 1 x
  • 42. Find the coordinate matrix of x=(1, 2, –1) in R3 relative to the (nonstandard) basis B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)} Sol: 2100 8010 5001 1521 2310 1201 E.G.J.                         )5,3,2()2,1,0()1,0,1()1,2,1( 321332211  cccccc uuux                                       1 2 1 521 310 201 i.e. 152 23 12 3 2 1 321 32 31 c c c ccc cc cc              2 8 5 ][ B x Finding a coordinate matrix relative to a nonstandard basis
  • 43. 43 Change of basis problem You were given the coordinates of a vector relative to one basis B and were asked to find the coordinates relative to another basis B'. Transition matrix from B' to B: V BB nn spacevectorafor basestwobe}...,,{nda},...,,{etL 2121 uuuuuu  BB P  ][][hent vv         BBnBB v,...,,  1 uuu 2       BnBB P uuu 2  ...,,,1 where is called the transition matrix from B' to B If [v]B is the coordinate matrix of v relative to B [v]B‘ is the coordinate matrix of v relative to B'
  • 44. 44 If P is the transition matrix from a basis B' to a basis B in Rn, then (1) P is invertible (2) The transition matrix from B to B' is P –1 THEOREM 1 The inverse of a transition matrix THEOREM 2 Let B={v1, v2, … , vn} and B' ={u1, u2, … , un} be two bases for R n . Then the transition matrix P–1 from B to B' can be found by using Gauss-Jordan elimination on the n×2n matrix as follows. Transition matrix from B to B'  BB   1 PIn 
  • 45. B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two bases for R 2 (a) Find the transition matrix from B' to B. (b) (c) Find the transition matrix from B to B' . BB ][find, 2 1 ][Let ' vv        Ex : (Finding a transition matrix) Sol:           22 21 22 43           12 23 10 01  G.J.E. B B' I P          12 23 P (the transition matrix from B' to B) (a) Sol:           22 21 22 43   G.J.E. B B' (a)
  • 46.                             0 1 2 1 12 23 ][][ 2 1 ][ BBB P vvv (b)           22 43 22 21           32 21 10 01  G.J.E. B' B I P -1 (the transition matrix from B to B')          32 211 P  Check: 2 1 10 01 32 21 12 23 IPP                        )2,3()2,4)(0()2,3)(1( 0 1 ][ )2,3()2,2)(2()2,1)(1( 2 1 ][ :Check               vv vv B B (c)
  • 47. References: Anton-linear algebra book & . .. . Dhrumil Patel(L.D. College of engineering)