MATH 304Linear AlgebraLecture 13:Span. Spanning set.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 .Propositio n A s ubs et S of a vector space V is asubspace of V if and only if S is nonempty andclosed under linear opera tions, i.e.,x, y ∈ S =⇒ x + y ∈ S,x ∈ S =⇒ r x ∈ S for a ll r ∈ R.Remarks. The zero vector in a subspace is thesame as the zer o vector in V . Also, the subtractionin a subspace agrees with that in V .Examples of subspaces• F (R): all functions f : R → R• C (R): all continuous functions f : R → RC (R) is a subspace of F (R).• P: polynomials p(x) = a0+ a1x + · · · + an−1xn−1• Pn: polynomials of degree less than nPnis a subs pace of P.• Any vector space V• {0}, where 0 is the zero vector in VThe trivial space {0} is a subspace of V .Examples of subspaces of M2,2(R): A =a bc d• diago na l matrices: b = c = 0• upper triangular matrices: c = 0• lower triangular matrices: b = 0• symmetric matrices (AT= A): b = c• anti-sym metric matrices (AT= −A):a = d = 0 and c = −b• matrices with zero trace: a + d = 0(trace = the sum of diagonal entries)Let V be a vector space and v1, v2, . . . , vn∈ V .Consider the set L of all linear combinationsr1v1+ r2v2+ · · · + rnvn, where r1, r2, . . . , rn∈ R.Theorem L is a subspace of V .Proof: First of all, L is no t empty. For example,0 = 0v1+ 0v2+ · · · + 0vnbelongs to L.The set L is closed under addition since(r1v1+r2v2+ · · · +rnvn) + (s1v1+s2v2+ · · · +snvn) == (r1+s1)v1+ (r2+s2)v2+ · · · + (rn+sn)vn.The set L is closed under scalar multiplication sincet(r1v1+r2v2+ · · · +rnvn) = (tr1)v1+(tr2)v2+ · · · +(trn)vn.Thus L is a subspace of V .Span: implicit definitio nLet 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 welldefined (namely, it is the intersection of allsubspaces of V that contain 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 em pty set then Span(S) = {0}.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-symmetr ic 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).Spanning setDefinition. A s ubs et 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+ y e2+ 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.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 ther e exist r1, r2∈ R such thatw = r1v1+ r2v2. This vector equation is equival entto a system of linear equations:4 = r1+ 3r2−7 = 2r1+ r23 = 0r1+ r2⇐⇒r1= −5r2= 3Thus w = −5v1+ 3v2is in 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 sol ution: 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 solution: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 v ectors 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).More properties of spanLet S0and S be subsets of a vector space V .• S0⊂ S =⇒ Span(S0) ⊂ Span(S).• Span(S0) = V and S0⊂ S =⇒ Span(S) = V .• If v0, v1, . . . , vkis a spanning set for V and v0is a linear combination of vectors v1, . . . , vkthenv1, . . . , vkis also a spanning set for V .Indeed, if v0= r1v1+ · · · + rkvk, thent0v0+ t1v1+ · · · + tkvk= (t0r1+ t1)v1+ · · · + (t0rk+ tk)vk.• Span(S0∪ {v0}) = Span(S0) if and only ifv0∈ Span(S0).If v0∈ Span(S0), then S0∪ v0⊂ Span(S0), which impliesSpan(S0∪ {v0}) ⊂ Span(S0). On the other hand,Span(S0) ⊂ Span(S0∪
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