MATH 304Linear AlgebraLecture 13:Span (continued).Linear independence.Subspaces of vector spacesDefinition. A vector space V0is a subspace of avector space V if V0⊂ V and the linear operationson V0agree with the linear operations on V .Proposition A subset S of a vector space V is asubspace of V if and only if S is nonempty andclosed under linear operations, i.e.,x, y ∈ S =⇒ x + y ∈ S,x ∈ S =⇒ rx ∈ S for all r ∈ R.Remarks. The z ero vector in a subspace is thesame as the zero vector in V . Also, the subtractionin a subspace agrees with that in V .Examples of subspaces of M2,2(R): A =a bc d• diagonal matrices: b = c = 0• upper triangular matrices: c = 0• lower triangular m atrices: b = 0• symmetric matrices (AT= A): b = c• anti-symm etric matrices (AT= −A):a = d = 0 and c = −b• matrices with zero trace: a + d = 0(trace = the sum of diagonal entries)Span: implicit definitionLet S be a subset of a vector space V .Definition. The span of the set S, denotedSpan(S), is the smallest subspace of V thatcontains S. That is,• Span(S) is a subspace of V ;• for any subspace W ⊂ V one hasS ⊂ W =⇒ Span(S) ⊂ W .Remark. The span of any set S ⊂ V is well defined(it is the intersection of all subspaces of V thatcontain S).Span: effective descriptionLet S be a subset of a vector space V .• If S = {v1, v2, . . . , vn} then Span(S) is the setof all linear combinations r1v1+ r2v2+ · · · + rnvn,where r1, r2, . . . , rn∈ R.• If S is an infinite set then Span(S) is the set ofall linear combinations r1u1+ r2u2+ · · · + rkuk,where u1, u2, . . . , uk∈ S and r1, r2, . . . , rk∈ R(k ≥ 1).• If S is the empty set then Span(S) = {0}.Spanning setDefinition. A subset S of a vector space V iscalled a spanning set for V if Span(S) = V .Examples.• Vectors e1= (1, 0, 0) , e2= (0, 1, 0) , ande3= (0, 0, 1) form a spanning set for R3as(x, y , z) = xe1+ ye2+ ze3.• Matrices1 00 0,0 10 0,0 01 0,0 00 1form a spanning set for M2,2(R) asa bc d= a1 00 0+ b0 10 0+ c0 01 0+ d0 00 1.Examples of subspaces of M2,2(R):• The span of1 00 0and0 00 1consists of allmatrices of the forma1 00 0+ b0 00 1=a 00 b.This is the subspace of diagonal matrices.• The span of1 00 0,0 00 1, and0 11 0consists of all matrices of the forma1 00 0+ b0 00 1+ c0 11 0=a cc b.This is the subspace of symmetric matrices.Examples of subspaces of M2,2(R):• The span of0 −11 0is the subspace ofanti-symmetric matrices.• The span of1 00 0,0 00 1, and0 10 0is the subspace of upper triangular matrices.• The span of1 00 0,0 00 1,0 10 0,0 01 0is the entire space M2,2(R).Problem Let v1= (1, 2, 0) , v2= (3, 1, 1) , andw = (4, −7, 3). Determine whether w belongs toSpan(v1, v2).We have to check if there exist r1, r2∈ R such thatw = r1v1+ r2v2. This vector equation is equivalentto a system of linear equations:4 = r1+ 3r2−7 = 2r1+ r23 = 0r1+ r2⇐⇒r1= −5r2= 3Thus w = −5v1+ 3v2∈ Span(v1, v2).Problem Let v1= (2, 5) and v2= (1, 3). Showthat {v1, v2} is a spanning set for R2.Take any vector w = ( a, b) ∈ R2. We have tocheck that there exist r1, r2∈ R such thatw = r1v1+r2v2⇐⇒2r1+ r2= a5r1+ 3r2= bCoefficient matrix: C =2 15 3. det C = 1 6= 0.Since the matrix C is invertible, the system has aunique solution for any a and b.Thus Span(v1, v2) = R2.Problem Let v1= (2, 5) and v2= (1, 3). Showthat {v1, v2} is a spanning set for R2.Alternative solutio n: First let us show that vectorse1= (1, 0) and e2= (0, 1) belong to Span(v1, v2).e1= r1v1+r2v2⇐⇒2r1+ r2= 15r1+ 3r2= 0⇐⇒r1= 3r2= −5e2= r1v1+r2v2⇐⇒2r1+ r2= 05r1+ 3r2= 1⇐⇒r1= −1r2= 2Thus e1= 3v1− 5v2and e2= −v1+ 2v2.Then for any vector w = (a, b) ∈ R2we havew = ae1+ be2= a(3v1− 5v2) + b(−v1+ 2v2)= (3a − b) v1+ (−5a + 2b)v2.Problem Let v1= (2, 5) and v2= (1, 3). Showthat {v1, v2} is a spanning set for R2.Remarks on the alternative s olution:Notice that R2is spanned by vectors e1= (1, 0)and e2= (0, 1) since (a, b) = ae1+ be2.This is why we have checked that vectors e1and e2belong to Span(v1, v2). Thene1, e2∈ Span(v1, v2) =⇒ Span(e1, e2) ⊂ Span(v1, v2)=⇒ R2⊂ Span(v1, v2) =⇒ Span(v1, v2) = R2.In general, to show that Span(S1) = Span(S2),it is enough to check that S1⊂ Span(S2) andS2⊂ Span(S1).Linear independenceDefinition. Let V be a vector space. Vectorsv1, v2, . . . , vk∈ V are called linearly dependent ifthey satisfy a relationr1v1+ r2v2+ · · · + rkvk= 0,where the coefficients r1, . . . , rk∈ R are not allequal to zero. Otherwise vectors v1, v2, . . . , vkarecalled linearly independent. That is, ifr1v1+r2v2+ · · · +rkvk= 0 =⇒ r1= · · · = rk= 0.An infinite set S ⊂ V is linearly dependent ifthere ar e some linearly dependent vectors v1, . . . , vk∈ S.Otherwise S is linearly independent.Theor em The following conditions are equivalent:(i) vectors v1, . . . , vkare l inearly dependent;(ii) one of vectors v1, . . . , vkis a linear combinationof the other k − 1 vectors.Proof: (i) =⇒ (ii) Suppose thatr1v1+ r2v2+ · · · + rkvk= 0,where ri6= 0 for some 1 ≤ i ≤ k. Thenvi= −r1riv1− · · · −ri −1rivi−1−ri +1rivi+1− · · · −rkrivk.(ii) =⇒ (i) Suppose thatvi= s1v1+ · · · + si−1vi−1+ si+1vi+1+ · · · + skvkfor some scalars sj. Thens1v1+ · · · + si−1vi−1− vi+ si+1vi+1+ · · · + skvk= 0.Examples of linear independence• Vectors e1= (1, 0, 0) , e2= (0, 1, 0) , ande3= (0, 0, 1) in R3.xe1+ ye2+ ze3= 0 =⇒ (x, y, z) = 0=⇒ x = y = z = 0• Matrices E11=1 00 0, E12=0 10 0,E21=0 01 0, and E22=0 00 1.aE11+ bE12+ cE21+ dE22= O =⇒a bc d= O=⇒ a = b = c = d = 0Examples of linear independence• Polynomials 1, x, x2, . . . , xn.a0+ a1x + a2x2+ · · · + anxn= 0 identically=⇒ ai= 0 for 0 ≤ i ≤ n• The infinite set {1, x, x2, . . . , xn, . . . }.• Polynomials p1(x) = 1, p2(x) = x − 1, andp3(x) = (x − 1)2.a1p1(x) + a2p2(x) + a3p3(x) = a1+ a2(x − 1) + a3(x − 1)2== (a1− a2+ a3) + (a2− 2a3)x + a3x2.Hence a1p1(x) + a2p2(x) + a3p3(x) = 0 identically=⇒ a1− a2+ a3= a2− 2a3= a3= 0=⇒ a1= a2= a3=
View Full Document
Unlocking...