Review• Every solution to Ax = b is the sum of a fixed chosen particular solution and somesolution toAx = 0.For instance, letA =1 3 3 22 6 9 7−1 −3 3 4and b =155.Every solution toAx = b is of the form:x =x1x2x3x4=−2 − 3x2+ x4x21 − x4x4=−2010xp+ x2−3100+ x410−11elem ents of N u l(A)• Is span(111,123,−113)equal to R3?Linear independenceReview.•span{v1, v2,, vm} is the set of all linear combinationsc1v1+ c2v2++ cmvm.• span{v1, v2,, vm} is a vector space .Example 1. Is span(111,123,−113)equal to R3?Solution. Recall that the span i s equa l to1 1 −11 2 11 3 3x : x in R3.Hence, the span i s equa l to R3if and only if the system with a ugmented matrix1 1−1 b11 21 b21 33 b3is consistent for all b1, b2, b3.Armin [email protected] elimination :1 1 − 1 b11 2 1 b21 3 3 b3 1 1−1 b10 1 2 b2− b10 2 4 b3− b1 1 1−1 b10 1 2 b2− b10 0 0 b3− 2b2+ b1The system is only consistent if b3− 2b2+ b1= 0.Hence, the span d oes not equ al all of R3.• What went “w r o n g ”? span(111,123,−113)Well, the three v e ct ors in the span satisfy−113= −3111+ 2123.• Hence, span(111,123,−113)= span(111,123).• We are going to say th at the three vectors are linearly dependent because theysatisfy−3111+ 2123−−113= 0.Definition 2. Vectors v1,, vpare said to be linearly independent if the equationx1v1+ x2v2++ xpvp= 0has only the trivi al solu t i o n (n amely, x1= x2== xp= 0).Likewise,v1,, vpare said to be linearly dependent if there exist coefficients x1,, xp, not all zero,such thatx1v1+ x2v2++ xpvp= 0.Armin [email protected] 3.•Are the vectors111,123,−113independent?• If possible, find a linear dependenc e relation among th em.Solution. We need to ch e c k whether the equa t i o nx1111+ x2123+ x3−113=000has more than the tri vial solution.In other words, the three vectors are independent if and only if the system1 1 −11 2 11 3 3x = 0has no free variables.To find out, we reduce the matri x to echelon form:1 1 −11 2 11 3 3 1 1 −10 1 20 2 4 1 1 −10 1 20 0 0Since there is a colum n without pivot, we do have a free variable.Hence, the three vectors are not linearly independent.To find a linear dependence relation, we solve this system.Initial steps of Gaussian elimination are as before:1 1 −1 01 21 01 3 3 0 1 1−1 00 12 00 0 0 0 1 0−3 00 12 00 0 0 0x3is free. x2= −2x3, and x1= 3x3. Hence, for any x3,3x3111− 2x3123+ x3−113=000.Since we are only interested in one linear combi nation, we can set, say, x3= 1:3111− 2123+−113=000Armin [email protected] independence of matrix columns• Note that a linear dependence relation, such as3111− 2123+−113= 0,can be written in matrix form as1 1 −11 2 11 3 33−21= 0.• Hence, each linear dependence relation among the columns of a m atrix A corre-sponds to a nontrivial solu tion to Ax = 0.Theorem 4. Let A be an m × n matrix.The columns ofA are linearly independent.Ax = 0 has only the solution x = 0.Nul(A) = {0}A has n pivots. (one in each colum n)Example 5. Are the vectors111,123,−123independent?Solution. Put the vectors in a matrix, and produce an echelon form:1 1 −11 2 21 3 3 1 1 −10 1 30 2 4 1 1 −10 1 30 0 −2Since each column contains a pi vot, the three vectors are independent.Example 6. (once again, short version)Are the vectors111,123,−113independent?Solution. Put the vectors in a matrix, and produce an echelon form:1 1 −11 2 11 3 3 1 1 −10 1 20 2 4 1 1 −10 1 20 0 0Since the last column does not contain a pivot, the three vectors are linearly dependent.Armin [email protected] cases• A set of a single nonzero vect or {v1} is always linearly i ndependent.Why? Because x1v1= 0 only for x1= 0.• A set of two vectors {v1, v2} is linearly independent if and on ly if neither of thevectors is a multiple of the other.Why? Because if x1v1+ x2v2= 0 with, say, x20, then v2= −x1x2v1.• A set of vectors {v1,, vp} containing the zero vector is linearly de pendent.Why? Because if, say ,v1= 0, then v1+ 0v2++ 0vp= 0.• If a set contains m ore vectors than there are entries in each vector, then th e set islinearly dependent. In other words:Any set {v1,, vp} of vectors in Rnis linearly dependent if p > n.Why?Let A be the matrix with columns v1,, vp. This is a n × p matrix.The columns are linea rly independent if and only if each column contains a pivot.If p > n, then the ma trix can have at most n pivots.Thu s not allp columns can contain a pivot.In other words, the columns have to be linearly dependent.Example 7. With the least a mount of work possible, decide which of the following s e t sof vectors are linearly i ndependent.(a)(321,964)Linearl y indepen dent, because the two vectors are not multiples of each other.(b)(321)Linearl y indepen dent, because it is a single nonzero vector.(c) columns of1 2 3 45 6 7 89 8 7 6Linearl y dependent, because these are more than 3 (namely, 4) vectors in R3.(d)(321,964,000)Linearl y dependent, because the set includes the zero vector.Armin
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