MATH 304Linear AlgebraLecture 14:Linear independence.Spanning setLet S be a subset of a vector space V .Definition. The span of the set S is the smallestsubspace W ⊂ V that contains S. If S is notempty then W = Span(S) consists of all linearcombinationsr1v1+ r2v2+ · · · + rkvksuch thatv1, . . . , vk∈ S and r1, . . . , rk∈ R.We say that the set S spans the subspace W orthat S is a spanning set for W .Remark. If S1is a spanning set for a vector spaceV and S1⊂ S2⊂ V , then S2is also a spanning setfor V .Linear independenceDefinition. Let V be a vector space. Vectorsv1, v2, . . . , vk∈ V are called linearly dependentif they 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 are some linearly dependent vectors v1, . . . , vk∈ S.Otherwise S is linearly independent.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= 0Problem Let v1= (1, 2, 0), v2= (3, 1, 1), andv3= (4, −7, 3). Determine whether vectorsv1, v2, v3are linearly independent.We have to check if there exist r1, r2, r3∈ R not allzero such that r1v1+ r2v2+ r3v3= 0.This vector equation is equivalent to a systemr1+ 3r2+ 4r3= 02r1+ r2− 7r3= 00r1+ r2+ 3r3= 01 3 402 1 −700 1 3 0The vectors v1, v2, v3are linearly dependent if andonly if the matrix A = (v1, v2, v3) is singular.We obtain that det A = 0.Theorem The following conditions are equivalent:(i) vectors v1, . . . , vkare linearly dependent;(ii) one of vectors v1, . . . , vkis a linearcombination of 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.Theorem Vectors v1, v2, . . . , vm∈ Rnare linearlydependent whenever m > n (i.e. , the number ofcoordinates is less than the number of vectors).Proof: Let vj= (a1j, a2j, . . . , anj) for j = 1, 2, . . . , m.Then the vector equality t1v1+ t2v2+ · · · + tmvm= 0is equivalent to the systema11t1+ a12t2+ · · · + a1mtm= 0,a21t1+ a22t2+ · · · + a2mtm= 0,· · · · · · · · ·an1t1+ an2t2+ · · · + anmtm= 0.Note that vectors v1, v2, . . . , vmare columns of the matrix(aij). The number of leading entries in the row echelon formis at most n. If m > n then there are free variables, thereforethe zero solution is not unique.Example. Consider vectors v1= (1, −1, 1),v2= (1, 0, 0), v3= (1, 1, 1), and v4= (1, 2, 4) in R3.Two vectors are linearly dependent if and only ifthey are parallel. Hence v1and v2are linearlyindependent.Vectors v1, v2, v3are linearly independent if andonly if the matrix A = (v1, v2, v3) is invertible.det A =1 1 1−1 0 11 0 1= −−1 11 1= 2 6= 0.Therefore v1, v2, v3are linearly independent.Four vectors in R3are always linearly dependent.Thus v1, v2, v3, v4are linearly dependent.Problem. Let A =−1 1−1 0. Determine whethermatrices A, A2, and A3are linearly independent.We ha ve A =−1 1−1 0, A2=0 −11 −1, A3=1 00 1.The task is to check if there exist r1, r2, r3∈ R not all zerosuch that r1A + r2A2+ r3A3= O.This matrix equation is equivalent to a system−r1+ 0r2+ r3= 0r1− r2+ 0r3= 0−r1+ r2+ 0r3= 00r1− r2+ r3= 0−1 0 101 −1 0 0−1 1 0 00 −1 1 0→1 −1 000 1 −1 00 0 0 00 0 0 0The row echelon form of the augmented matrix shows there isa free variable. Hence the system has a nonzero solution sothat the matrices are linearly dependent (one relation isA + A2+ A3=
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