Vector Spaces 1.Vectors in Rn 2.Vector Spaces 3.Subspaces of Vector Spaces 4.Spanning Sets and Linear Independence 5.Basis and Dimension 4.1 1.Vectors in Rn a sequence of n real numbers) (, , 2 , 1 nx x x An ordered n-tuple : the set of all ordered n-tuples Rn-space : n = 4R4-space = set of all ordered quadruple of real numbers) , , , (4 3 2 1x x x xR1-space = set of all real numbersn = 1 n = 2R2-space = set of all ordered pair of real numbers ) , (2 1x xn = 3 R3-space = set of all ordered triple of real numbers ) , , (3 2 1x x x(R1-space can be represented geometrically by the x-axis) (R2-space can be represented geometrically by the xy-plane) (R3-space can be represented geometrically by the xyz-space) Notes: Ex: a point ( )2 1, x xa vector ( )2 1, x x( ) 0 , 0(1) An n-tuplecan be viewed as a point in Rn with the xis as its coordinates. (2) An n-tuplealso can be viewed as a vectorin Rn with the xis as its components. ) , , , (2 1 nx x x ) , , , (2 1 nx x x 1 2( , , , )nx x x = x A vector on the plane is represented geometrically by a directed line segment whose initial point is the origin and whose terminal point is the point (x1, x2) or ( ) ( )1 2 1 2, , , ,, , ,n nu u u v v v = = u v(two vectors in Rn) Equality: if and only if v u =n nv u v u v u = = = , , ,2 2 1 1 Vector addition (the sum of u and v): ( )n nv u v u v u + + + = + , , ,2 2 1 1 v u Scalar multiplication (the scalar multiple of u by c): ( )ncu cu cu c , , ,2 1 = u Notes: The sum of two vectors and the scalar multiple of a vector in Rn are called the standard operations in Rn Difference between u and v:1 1 2 2 3 3( 1) ( , , ,..., )n nu v u v u v u v + = u v u vZero vector: ) 0 ..., , 0 , 0 ( = 0Theorem 1 : Properties of vector addition and scalar multiplication Let u, v, and w be vectors in Rn, and let c and d be scalars (1)u+v is a vector in Rn (closure under vector addition) (2)u+v = v+u (commutative property of vector addition) (3)(u+v)+w = u+(v+w) (associative property of vector addition) (4)u+0 = u (additive identity property) (5)u+(u) = 0 (additive inverse property) (6)cu is a vector in Rn (closure under scalar multiplication) (7)c(u+v) = cu+cv (distributive property of scalar multiplication over vector addition) (8)(c+d)u = cu+du (distributive property of scalar multiplication over real-number addition) (9)c(du) = (cd)u (associative property of multiplication) (10) 1(u) = u (multiplicative identity property) Except Properties (1) and (6), these properties of vector addition and scalar multiplication actually inherit from the properties of matrix addition and scalar multiplication because we can treat vectors in Rn as special cases of matrices (Note that u is just the notation of the additive inverse of u, and u = (1)u will be proved in Thm. 4.4 ) Theorem 2: (Properties of additive identity and additive inverse) Let v be a vector in Rn and c be a scalar. Then the following properties are true (1) The additive identity is unique. That is, if v+u=v, u must be 0 (2) The additive inverse of v is unique. That is, if v+u=0, u must be v (3) 0v=0 (4) c0=0 (5) If cv=0, either c=0 or v=0 (6) (v) = v Notes: (1) The zero vector 0 in Rn is called the additive identity in Rn (see Property 4) (2) The vector u is called the additive inverse of u (The additive inverse of v is v, which implies that v and v are the additive inverses for each other) Notes: A vector incan be viewed as:) , , , (2 1 nu u u = unR1 2[ ]nuu u = u(((((

=nuuu21u Therefore, the matrix operations of addition and scalar multiplication generate the same results as the corresponding vector operations (see the next slide) or a n1 column matrix (column vector): a 1n row matrix (row vector): ) , , , () , , , ( ) , , , (2 2 1 12 1 2 1n nn nv u v u v uv v v u u u+ + + =+ = + v u1 2 1 21 1 2 2[ ] [ ][ ]n nn nu u u v v vu v u v u v+ = += + + +u v(((((

+++=(((((

+(((((

= +n n n nv uv uv uvvvuuu 2 21 12121v uVector addition Scalar multiplication (((((

=(((((

=n ncucucuuuuc c 2121u) , , , () , , , (2 12 1nn cu cu cu u u u c c== u1 21 2[ ][ ]nnc c uuucu cu cu==uTreated as 1n row matrix Treated as n1 column matrix Vector Spaces Vectorspaces: Let V be a set on which two operations (vector addition and scalar multiplication) are defined. If the following axioms are satisfied for everyu, v, and w in V and every scalar (real number) c and d, then V is called a vector space. Addition: (1)u+v is in V (2)u+v=v+u (3)u+(v+w)=(u+v)+w (4)V has a zero vector 0 such that for every u in V, u+0=u (5) For every u in V, there is a vector in V denoted by usuch that u+(u)=0 Scalar multiplication: (6)is in V. u c(7)v u v u c c c + = + ) ((8) u u u d c d c + = + ) ((9) u u ) ( ) ( cd d c =(10)u u = ) ( 1 Notes: (1) A vector space consists of four entities: (2) { } 0 = Vzero vector space Vnonempty set cscalar ) , ( ) , (u uv u v uc c = -+ = +vector addition scalar multiplication ( ) - +, , Vis called a vector space a set of vectors, a set of scalars, and two operations Examples of vector spaces: (1) n-tuple space: Rn ) , , ( ) , , ( ) , , (2 2 1 1 2 1 2 1 n n n nv u v u v u v v v u u u + + + = + ) , , ( ) , , (2 1 2 1 n nku ku ku u u u k =(2) Matrix space:(the set of all mn matrices with real values) n mM V=Ex: (m = n = 2) ((

+ ++ +=((

+((

22 22 21 2112 12 11 1122 2112 1122 2112 11v u v uv u v uv vv vu uu u((

=((

22 2112 1122 2112 11ku kuku kuu uu ukvector addition scalar multiplication vector addition scalar multiplication (3) n-th degree polynomial space: (the set of all real polynomials of degree n or less) ) (x P Vn=nn nx b a x b a b a x q x p ) ( ) ( ) ( ) ( ) (1 1 0 0+ + + + + + = + nnx ka x ka ka x kp + + + = 1 0) ((4) Function space:(the set of all real-valued continuous functions defined on the entire real line.) ) ( ) ( ) )( ( x g x f x g f + = +) , ( = c V) ( ) )( ( x kf x kf = Thm 3 : (Properties of scalar multiplication) Let v be any element of a vector space V, and let c be any scalar. Then the following properties are true. v v0 v 0 v0 00 v = = = ===) 1 ( (4)or 0then, If (3) (2)0 (1)c cc Notes:To show that a set is not a vector space, you need only find one axiom that is not satisfied. Ex : The set of all second-degree polynomials is not a vector space. Pf:Letand 2) ( x x p = 1 ) (2+ + = x x x qV x x q x p e + = + 1 ) ( ) ((it is not closed under vector addition) R , V e e211V e =2121) 1 )( ((it is not closed under scalar multiplication) | | |scalar Pf: Ex :The set of all integer is not a vector space. integer noninteger Ex : V=R2=the set of all ordered pairs of real numbers vector addition: ) , ( ) , ( ) , (2 2 1 1 2 1 2 1v u v u v v u u + + = +scalar multiplication: ) 0 , ( ) , (1 2 1cu u u c =) 1 , 1 ( ) 0 , 1 ( ) 1 , 1 ( 1 = = the set (together with the two given operations) isnot a vector space

Verify V is not a vector space. Sol: Vector Subspaces Subspaces of Vector Spaces Subspace: ) , , ( - + V : a vector space )`_=V WW |: a nonempty subset ) , , ( - + W a vector space (under the operations of addition andscalar multiplication defined in V) W is a subspace of V Trivial subspace: Every vector space Vhas at least two subspaces. (1) Zero vector space {0} is a subspace ofV. (2)V is a subspace ofV. Thm 4 : (Test for a subspace) If W is a nonempty subset of a vector space V, then Wisa subspace of V if and only if the following conditions hold. (1) Ifuand vare in W, thenu+vis in W. (2) If u is in W and c is any scalar, then cu is in W. Ex:Subspace ofR3 Ex: Subspace ofR2 { } ( ) 0 0, (1) = 0 0origin he through t Lines(2)2(3) Rorigin he through t Planes(3)3(4) R{ } ( ) 0 0, 0, (1) = 0 0origin he through t Lines(2)Ex : (A subspace of M22) Let W be the set of all 22 symmetric matrices. Show that W is a subspace of the vector space M22, with the standard operations of matrix addition and scalar multiplication. sapces vector :2 2 2 2 _ M M W Sol: ) ( Let2 2 1 1 2 1A A , A A W A , AT T= = e) (2 1 2 1 2 1 2 1A A A A A A W A W, AT T T+ = + = + e e) ( kA kA kA W A , R kT T= = e e2 2of subspace a is M W) (2 1W A A e +) ( W kAe

W B A e((

= + 1 00 12 2 2 of subspace a not is M W Ex : (The set of singular matrices is not a subspace of M22) Let W be the set of singular matrices of order 2. Show that W is not a subspace ofM22 with the standard operations. W B , W A e((

= e((

=1 00 00 00 1Sol: Ex : (The set of first-quadrant vectors is not a subspace of R2) Show that, with the standard operations, is not a subspace of R2. Sol: W e = ) 1 , 1 ( Let uof subspace a not is2R W } 0 and 0 : ) , {(2 1 2 1> > = x x x x W( ) ( )( ) ( ) W e = = 1 , 1 1 , 1 1 1 u (not closed under scalar multiplication) Ex : (Determining subspaces of R2) Which of the following two subsets is a subspace of R2? (a) The set of points on the line given by x+2y=0. (b) The set of points on the line given by x+2y=1. Sol: { } { } R t t t y x y x W e = = + = ) , 2 ( 0 2 ) , ((a)( ) ( ) W t t Wv t t v e = e =2 2 2 1 1 1, 2 , 2 Let( ) ( ) W t t , t t v v e + + = +2 1 2 1 2 12 ( ) ( ) W kt , kt kv e =1 1 122of subspace a is R W (closed under addition) (closed under scalar multiplication) ( ) { }1 2 , = + = y x y x WW v e = ) 0 , 1 ( Let ( ) ( ) W , v e = 0 1 1 of subspace a not is2R W (b) (Note: the zero vector is not on the line) Ex : (Determining subspaces of R3) { }{ } R x x x x x x WR x x x x WRe + =e =3 1 3 3 1 12 1 2 13, ) , , ((b), ) 1 , , ((a)? of subspace a is subsets following the of Which Sol: W e = ) 1 , 0 , 0 ( Let (a) vW e = ) 1 , 0 , 0 ( ) 1 ( vof subspace a notis3R W W W e + = e + = ) u , u u , u (, ) v , v v , v ( Let (b)3 3 1 1 3 3 1 1u v( )( ) ( ) W u v , u v u v , u v3 3 3 3 1 1 1 1e + + + + + = +u v ( )( ) ( ) W v , v v , v3 3 1 1e + = k k k k kv3of subspace a is R W Thm 5 : (The intersection of two subspaces is a subspace) . of subspace a also is) by(denoted and of onintersecti then the, space vector a of subspaces bothare and IfUW V W VU W VSpanning Sets and Linear Independence 3.Spanning Sets and Linear Independence k kc c c u u u v + + + = 2 2 1 1form in the writtenbe canif invectors theof ncombinatio lineara called is spacevector a inA vector2 1v u u uvV , , ,VkLinear combination:scalars :2 1 k,c , ,c c Ex : (Finding a linear combination) 3 2 13 2 13 2 1, , of ncombinatio lineara notis 2,2) (1,(b), , of ncombinatio lineara is (1,1,1)(a) Prove1,0,1) ((0,1,2)(1,2,3)v v v wv v v wv v v == = = =Sol: 3 3 2 2 1 1 (a) v v v w c c c + + =( ) ( ) ( ) ( ) 1 0 1 2 1 0 3 2 1 1,1,13 2 1, , c , , c , , c + + =) 2 3 , 2 , (3 2 1 2 1 3 1c c c c c c c + + + = 1 2 31 21-

3 2 12 13 1= + += +=c c cc cc c((((

1 1 2 31 0 1 21 1 0 1 . . J G((((

0 0 0 01 2 1 01 1 0 13 2 113 2 v v v w + = = tt c t c t c = = + = 3 2 1 , 2 1 , 1(this system has infinitely many solutions) 3 3 2 2 1 1 ) (v v v w c c cb+ + =((((

2 1 2 32 0 1 21 1 0 1

n Eliminatio JordanGuass((((

7 0 0 04 2 1 01 1 0 1solution no has system this 3 3 2 2 1 1v v v w c c c + + = If S={v1, v2,, vk} is a set of vectors in a vector space V, then the span ofSis the set of all linear combinations of the vectors in S, the span of a set:span (S) = ) (S span { }) invectors of ns combinatio linearall of set(the2 2 1 1SR c c c ci k ke + + + v v v a spanning set of a vector space: 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. { } 0 = ) ((1) | span) ((2) S span S _) ( ) ( ,(3)2 1 2 12 1S span S span S SV S S_ _ _ Notes: V SS VV SV Sof setspanning a is by) (generated spanned is ) (generates spans) ( span = Notes: { } spans ) 1 , 0 , 2 ( ), 2 , 1 , 0 ( ), 3 , 2 , 1 ( set that the Show3R S = Ex : (A spanning set for R3) . and , , of ncombinatio lineara as be canin ) , , ( vectorarbitraryanwhetherdetermine mustWe3 2 133 2 1v v vuRu u u =Sol: 3 3 2 2 1 13v v v u u c c c R + + = e3 3 2 12 2 11 3 1 2 322u c c cu c cu c c= + += += . and , , of values all forconsistent issystem thiswhetherg determininto reducesthus problem The3 2 1u u u01 2 30 1 22 0 1== A u. everyforsolutionone exactlyhas b x = A3) ( R S span = { }1 2, , , : a set of vectors in a vector space kS V = v v v Definitions of Linear Independence (L.I.) and Linear Dependence (L.D.): 1 1 2 2For k kc c c + + + = v v v 0dependent linearlycalled is ) (orthenzeros) all not.e., solution(i trivial non ahas equation the (2)Ift. independen linearlycalled )is (orthen0) c ... c (c solution trivial only the has equation the If ) 1 (2 12 1k 2 1kk,..., v , v v S,..., v , v v S= = = ={ } ( ) ( ) ( ) { }1 2 3, , 1, 2, 3 , 0, 1, 2 , 2, 0, 1 S = = v v v Ex : Testing for linear independence 02 30 20 23 2 12 13 1= + += + += c c cc cc c = + + 0 v v v3 3 2 2 1 1c c cSol: Determine whether the following set of vectors in R3 is L.I. or L.D. ((((

0 1 2 30 0 1 20 2 0 1G.-J. E.1 0 0 00 1 0 00 0 1 0 ( ( ( ( ( ) solution trivial only the03 2 1= = = c c c1 2 3is (or, ,are) linearly independent S v v v(or det( ) 1 0,so there is only the trivial solution) A = = Ex : Testing for linear independence Determine whether the following set of vectors in P2 is L.I. or L.D. c1v1+c2v2+c3v3 = 0 i.e.,c1(1+x 2x2) + c2(2+5x x2) + c3(x+x2) = 0+0x+0x2 c1+2c2 = 0 c1+5c2+c3= 0 2c1 c2+c3= 0 Sol: This system has infinitely many solutions (i.e., this system has nontrivial solutions, e.g., c1=2, c2= 1, c3=3) S is (or v1, v2, v3 are) linearly dependent 1 2 0 01 5 1 02 1 1 0 ( ( ( ( 1 2 0 010 1 030 0 0 0 ( ( ( ( G. E.{ }{ }2 2 21 2 3, , 1 2 ,2 5 , S x x x x x x = = + + + v v v Ex : Testing for linear independence Determine whether the following set of vectors in the 22 matrix space is L.I. or L.D. { }1 2 32 1 3 0 1 0, , , ,0 1 2 1 2 0S (((= = ` ((( )v v vSol: ((

=((

+((

+((

0 00 00 20 11 20 31 01 23 2 1c c cc1v1+c2v2+c3v3 = 0 (This system has only the trivial solution)c1 = c2 = c3= 0 S is linearly independent 2c1+3c2+c3 = 0 c1 = 0 2c2+2c3 = 0 c1+ c2= 0 (((((

0 0 1 10 2 2 00 0 0 10 1 3 2(((((

0 0 0 00 1 0 00 0 1 00 0 0 1 G.-J. E.Basis and Dimension 4. Basis and Dimension Basis: V: a vector space (1)(2)S spans V (i.e., span(S) = V) S is linearly independent Spanning Sets Bases Linearly Independent Sets S is called a basis for V Notes: S ={v1, v2, , vn}_V A basis S must have enough vectors to span V, but not so many vectors that one of them could be written as a linear combination of the other vectors in S (For= , there is only the trivial solution (det( ) 0), see the definition on Slide 4.49)i ic A A = =v x 0V = = = e 0) ) (det( solutiona has , anyFor( A u Ax v c V ui i Notes: (1) the standard basis for R3: {i, j, k}i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) (2) the standard basis for Rn : {e1, e2, , en}e1=(1,0,,0), e2=(0,1,,0),, en=(0,0,,1) Ex:For R4, {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} Ex:matrix space: )`((

((

((

((

1 00 0,0 10 0,0 01 0,0 00 12 2(3) the standard basis for mn matrix space: { Eij | 1sism , 1sjsn }, and in Eij

(4) the standard basis for Pn(x):{1, x, x2, , xn} Ex: P3(x) {1, x, x2, x3} 1 other entries are zeroija = Ex: The nonstandard basis for R2 21 2Show that={ , }={(1,1),(1, 1)} is a basis forS R v v1 2 1 21 2 1 1 2 21 2 2=u(1) For any=(u ,u ) ,+ ==uc cR c cc c+e u v v uBecause the coefficient matrix of this system has a nonzero determinant, the system has a unique solution for each u. Thus you can conclude that S spans R2 1 21 1 2 21 2=0(2) For+ ==0c cc cc c+ v v 0Because the coefficient matrix of this system has a nonzero determinant, you know that the system has only the trivial solution. Thus you can conclude that S is linearly independent According to the above two arguments, we can conclude that S is a (nonstandard) basis for R2 Theorem 6: Uniqueness of basis representation for any vectors If 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 { }nS v v v , , ,2 1 =Pf: basis a is S (1) span(S) = V (2) S is linearly independent span(S) = VLetv = c1v1+c2v2++cnvn v = b1v1+b2v2++bnvn v+(1)v = 0 = (c1b1)v1+(c2 b2)v2++(cn bn)vn is linearly independent with only trivial solution S (i.e., unique basis representation) c1= b1 , c2= b2 ,, cn= bn coefficients for are all zeroi vIf is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent (In other words, every linearly independent set contains at most n vectors) Theorem 7 : Bases and linear dependence { }nS v v v , , ,2 1 = Finite dimensional: A vector space V is finite dimensional if it has a basis consisting of a finite number of elements Infinite dimensional: If a vector space Vis not finite dimensional, then it is called infinite dimensional Dimension:

V: a vector spaceS: a basis for V dim(V) = #(S) (the number of vectors in a basis S) The dimension of a vector space Vis defined to be the number of vectors in a basis for V Notes: (1) dim({0}) = 0 (2) dim(V) = n, S_V S: a spanning set #(S) > n S: a L.I. set#(S) s n S: a basis #(S) = n (3) dim(V) = n, if Wis a subspace of V dim(W) s nSpanning Sets Bases Linearly Independent Sets #(S) > n#(S) = n#(S) s n dim(V) = n (If V consists of the zero vector alone, the dimension of V is defined as zero) For example, if V = R3, you can infer the dim(V) is 3, which is the number of vectors in the standard basis Considering W = R2, which is a subspace of R3, due to the number of vectors in the standard basis, we know that the dim(W) is 2, that is smaller than dim(V)=3 (Since a basis is defined to be a set of L.I. vectors that can spans V, #(S) = dim(V) = n) (see the above figure) Ex: Find the dimension of a vector space according to the standard basis (1) Vector spaceRn basis {e1 , e2 , . , en} (2) Vector space Mmn basis {Eij | 1sism , 1sjsn} (3) Vector space Pn(x) basis {1, x, x2, . , xn} (4) Vector space P(x) basis {1, x, x2, .} dim(Rn) = n dim(Mmn)=mn dim(Pn(x)) = n+1 dim(P(x)) = Ex : Finding the dimension of a subspace of R3 (a) W={(d, cd, c):c and d are real numbers} (b) W={(2b, b, 0):b is a real number} Sol: (Hint: find a set of L.I. vectors that spans the subspace, i.e., find a basis for the subspace.) (a)(d, c d, c) = c(0, 1, 1) + d(1, 1, 0) S = {(0, 1, 1) , (1, 1, 0)} (S is L.I. and S spans W) S is a basis for W dim(W) = #(S) = 2 S = {(2, 1, 0)} spans W and S is L.I. S is a basis for W dim(W) = #(S) = 1 (2 , , 0) (2,1, 0) b b b =(b) Let Vbe a vector space of dimension n (1) If is a linearly independent set of vectors in V, then S is a basis for V (2) Ifspans V, then S is a basis for V (Both results are due to the fact that #(S) = n) Spanning Sets Bases Linearly Independent Sets dim(V) = n #(S) > n #(S) = n #(S) s n { }1 2, , ,nS = v v v{ }1 2, , ,nS = v v v Theorem8:Methodstoidentifyabasisinann-dimensional space Row Space and Column Space ( )( )( ) ((((((

=(((((

=mmn m mnnAAAa a aa a aa a aA 212 12 22 211 12 1111 12 1 (1)21 22 2 (2)1 2 ( )( , , , ) ( , , , )( , , , )nnm m mn ma a a Aa a a Aa a a A===row vectors of A row vectors: (with size 1n) ( ) ( ) ( )11 12 11 2 21 22 21 2nn nm m mna a aa a aA A A Aa a a ( ( ( (= = ( ( (((((

(((((

(((((

mnnnm maaa aaa aaa 212221212111column vectors of A column vectors: (with size m1) || |||| A(1) A(2)A(n) So, the row vectors are vectors in Rn So, the column vectors are vectors in Rm Let A be an mn matrix: Row space: } ,..., , | ... { ) (2 1 ) ( ) 2 ( 2 ) 1 ( 1RA A AA RSm m me + + + =o o o o o oColumn space: ( ) } , , {2 1) ( (2)2(1)1R A A A A CSnnne + + + = | | | | | | Note: dim(RS(A)) (or dim(CS(A)) equals the number of linearly independent row (or column) vectors of A (If A(1), A(2), , A(m) are linearly independent, A(1), A(2), , A(m) can form a basis for RS(A)) (If A(1), A(2), , A(n) are linearly independent, A(1), A(2), , A(n) can form a basis for CS(A)) The row space of A is a subspace of Rn spanned by the row vectors of A The column space of A is a subspace of Rm spanned bythe column vectors of A