MATH 311-504Topics in Applied MathematicsLecture 2-3:Subspaces of vector spaces.Span.Vector spaceA vector space is a set V equipped with two operations,additionV × V ∋ (x, y ) 7→ x + y ∈ Vand scalar multiplicationR × V ∋ (r, x) 7→ rx ∈ V ,that have the following properties:A1. a + b = b + aA2. (a + b) + c = a + (b + c)A3. a + 0 = 0 + a = aA4. a + (−a) = (−a) + a = 0A5. r(a + b) = r a + rbA6. (r + s)a = ra + saA7. (rs)a = r (sa)A8. 1a = aExamples of vector spaces• Rn: n-dimensional coordinate vectors• Mm,n(R): m×n matrices with real entries• R∞: infi nite sequences (x1, x2, . . . ), xi∈ R• {0}: the trivial vector space• F (R): the set of all functions f : R → R• C (R): all continuous functions f : R → R• C1(R): all continuously differentiable f uncti onsf : R → R• C∞(R): all smooth functions f : R → R• P: all polynomials p( x) = a0+ a1x + · · · + anxnSubspaces o f 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 .Examples.• F (R): all functions f : R → R• C (R): all continuous functions f : R → RC (R) is a subspace of F (R).• P: polynomi als p(x) = a0+ a1x + · · · + anxn• Pn: polynomi als of degree at most nPnis a subs pace o f P.If S is a subset of a vector space V then S inheritsfrom V addition and scalar multiplication. HoweverS need not be clo s ed under these operations.Proposition A subset S o f 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 =⇒ r x ∈ S for all r ∈ R.Proof: “only if” is obvious.“if”: properties like associative, commutative, or distributivelaw ho ld for S because they hold for V . We o nl y need toverify properties A3 and A4. Take any x ∈ S (note that S isnonempty). Then 0 = 0x ∈ S. Also, −x = (−1)x ∈ S.System of linear equations:a11x1+ a12x2+ · · · + a1nxn= b1· · · · · · · · ·am1x1+ am2x2+ · · · + amnxn= bmAny solution (x1, x2, . . . , xn) is an element of Rn.Theorem The so lution set of the sys tem is asubspace of Rnif and only if all bi= 0.Proof: “only if”: the zero vector 0 = (0, 0, . . . , 0) is asolution only if all equations are homogeneous.“if”: a system of homogeneous linear equations is equivalentto a matrix equation Ax = 0.A0 = 0 =⇒ 0 is a solution =⇒ solution set is not empty.If Ax = 0 and Ay = 0 then A(x + y) = Ax + Ay = 0.If Ax = 0 then A(rx) = r(Ax) = 0.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 not 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 multi plication sincet(r1v1+r2v2+ · · · +rnvn) = (tr1)v1+(tr2)v2+ · · · +(trn)vn.Example. V = R3.• The plane z = 0 is a subspace of R3.• The plane z = 1 is no t a subspace of R3.• The line t(1, 1, 0), t ∈ R is a subspace of R3and a subspace of the plane z = 0.• The line (1, 1, 1) + t(1, −1, 0), t ∈ R is not asubspace of R3as it l ies in the plane x + y + z = 3,which does not contain 0.• The plane t1(1, 0, 0) + t2(0, 1, 1), t1, t2∈ R isa subspace of R3.• In general, a line or a plane in R3is a subspaceif and only if it passes through the origin.Examples of subspaces of M2(R): A =a bc d• diagonal matrices: b = c = 0• upper triangular matrices: c = 0• lower triangul ar matrices: b = 0• symmetric matrices (AT= A): b = c• anti-symmetric matrices (AT= −A):a = d = 0, c = −b• matrices with zero trace: a + d = 0(trace = the sum of diagonal entries)• matrices with zero determinant, ad − bc = 0,do not form a subspace:1 00 0+0 00 1=1 00 1.Span: impli cit defini ti onLet 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 o ne hasS ⊂ W =⇒ Span(S) ⊂ W .Remar k . 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 o f 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.• Matrices1 00 0,0 10 0,0 01 0,0 00 1form a spanning set for M2,2(R) asa bc d= a1 00 0+ b0 10 0+ c0 01 0+ d0 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 there ex ist 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 s panning set for R2.Notice that R2is spanned by vectors e1= (1, 0)and e2= (0, 1) since (x, y ) = xe1+ ye2.Hence it is enough to check that vector s e1and e2belong to Span(v1, v2). ThenSpan(v1, v2) ⊃ Span( e1, e2) = R2.e1= r1v1+r2v2⇐⇒2r1+ …