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Linear Independence Linear Independence. Definition. Let v ,v ,...,v ∈ Rn be a set of vectors. 1 2 k • The vectors are linearly dependent if there exist scalars λ1,λ2,...,λk ∈ R, not all zero, such that λ v +λ v +···+λ v =0. 1 1 2 2 k k • The vectors are linearly independent if λ v + ··· + λ v = 0 for scalars 1 1 k k λ1,...,λk implies that λ1 = λ2 = ··· = λk = 0. N Example1. 1. • Thevectors 1 0 0 0 1 0 e = ,e = ,...,e = 1 . 2 . n . . . . . . . 0 0 1 in Rn are linearly independent. • Toshowthis, consider any scalars λ ,...,λ such that λ e + λ e ··· + 1 n 1 1 2 2 λ e =0,i.e.suchthat n n 1 0 0 λ1 0 0 1 0 λ2 0 λ +λ +···+λ = = 1 . 2 . n . . . . . . . . . . . . . 0 0 1 λn 0 • Thelast equation implies that λ1 = λ2 = ··· = λn = 0. 2. • Thevectors 1 4 7 w = 2 ,w = 5 , andw = 8 1 2 n 3 6 9 in R3 are linearly dependent, since w1 − 2w2 + w3 = 0. • Howdidwecomeupwiththis? Startwiththeequation in the definition of linear dependence 1 4 7 0 λ 2 +λ 5 +λ 8 = 0 1 2 3 3 6 9 0 and solve this system for possible values of λ ,λ ,λ . 1 2 3 2. • Wecanwritethis system in matrix form as 1 4 7 λ1 0 2 5 8 λ2 = 0 . 3 6 9 λ3 0 • The coefficient matrix A is simply the matrix with columns w ,w ,w 1 2 3 which we can write as A = (w1w2w3). If we reduce A to row echelon form we get 1 4 7 0 −3 −6 . 0 0 0 • We have found that rankA is not maximal, and so the system is singular and has a nonzero solution (many in fact). • Thefollowing theorem generalizes the last example. Theorem1(S&B11.1). Vectorsv ,...,v ∈ Rn arelinearly dependent iff the linear 1 k system λ 1 λ A 2 =0, . . . λk where A = (v v ···v ), has a nonzero solution (λ ,...,λ ). 1 2 k 1 k • The following theorem gives us a way of checking that a set of n vectors in Rn are linearly independent. Theorem2(S&B11.2). Vectorsv1,...,vn ∈ Rn are linearly independent iff det(v v ···v ) 6= 0. 1 2 n Theorem3(S&B11.3). Ifk > n,anysetofk vectors in Rn is linearly dependent. Proof. Let v1,...,vk be k vectors in Rn with k > n. By theorem 1, this set of vectors is linearly dependent if and only if the system λ1 λ2 Aλ=A =0 . . . λk where A = (v v ···v ), has a nonzero solution (λ ,...,λ ). 1 2 k 1 k • But, since the vectors belong to Rn, the matrix A has n rows and k columns with k > n. 2 • Since A has more columns than rows, there must be some column, say column j, in the row echelon forms of A without a pivot. • But then λj is a free variable, and since Aλ = 0 has a solution, it must have infinitely many solutions all but one of which is nonzero. • Hencethevectors are linearly dependent. • Instead of the above, we could have used fact 7.11(a)(i) of S&B which says that if there are more unknowns than equations, the system Aλ = 0 must have infinitely many solutions. SpanningSets. • Asubset of V of the vector space Rn is a subspace of Rn if it is – closed under addition (the sum of any two elements in V is also in V ), and – closed under scalar multiplication (any scalar multiple of an element in V is also in V ). Example2. – Theset{(1,1,1)}isnotasubspaceofR3,becauseforexample(0,0,0) ∈/ {(1,1,1)}. – ThesetV = L[(1,1,1)] = {λ(1,1,1) | λ ∈ R} is a subspace of R3 as you can check. Definition. Let v ,...,v ∈ Rn be a set of vectors and let V ⊆ Rn be a subspace. 1 k • Theset of all linear combinations of these vectors L[v ,...,v ] = {λ v +···+λ v |λ ,...,λ ∈ R} 1 k 1 1 k k 1 k is called the set generated or spanned by v ,...,v . 1 k • Equivalently, the span of {v ,...,v } is the set L[v ,...,v ] (sometimes 1 k 1 k written span(v ,...,v )). 1 k • If V =L[v ,...,v ], 1 k wesayv ,...,v spanV. N 1 k • A set of vectors spans (or is a spanning set of) a subspace V of Rn if we can write any element in V as a linear combination of the vectors. Example3. 1. The n-dimensional Euclidean space is spanned by the vectors e1,...,en of our earlier example. Take any vector a = (a ,...,a ) ∈ Rn. Then a = a e + 1 n 1 1 · · · + a e . n n 2. There are many sets of vectors that span the same space. Each of the following sets spans the space R2. 3 (a) 1 0 1 0 56 0 , 1 . (c) 0 , 1 , −24 . −1 0 6 31 (b) 0 , 1 . (d) −6 , 31 . Theorem4(S&B11.4). Letv ,...v ∈ Rn beasetofvectors. LetA = (v v ···v ) 1 k 1 2 k andletb ∈ Rn beavector. ThenbliesinthespaceL[v ...,v ]spannedbyv ,...,v 1 k 1 k iff the system Aλ = b has a solution λ. • The first corollary to this theorem provides a way of checking whether or not a set of vectors spans all of Rn, while the second gives the minimum number of vectors needed in a set spanning Rn. Corollary1(S&B11.5). Letv ,...v ∈ Rnbeasetofvectors. LetA = (v v ···v ). 1 k 1 2 k Then v ,...,v span Rn iff the system of equations Ax = b has a solution x for every 1 k b ∈ Rn. Corollary2(S&B11.6). AsetofvectorsthatspansRn mustcontainatleastnvectors. Basis and Dimension. • Once we have a spanning set of vectors, we can always make the spanning set larger by including more vectors. • We want to find the smallest set of vectors that spans a subset V of the n- dimensional Euclidean space. Definition. Let v ,...v ∈ V be a set of vectors and let V ⊆ Rn. Then v ,...v 1 k 1 k forms a basis of V if 1. v ,...v span V, and 1 k 2. v ,...v are linearly independent. N 1 k Example4. 1. Fromourpreviousexamples,wecanseethattheunitvectorse1,...,en form a basis of Rn. This natural basis, is called the canonical basis of Rn. 2 2. In a previous example (2), we listed four spanning sets of R . The set (2c) is not linearly independent, since it contains more than two vectors. The collections (2a), (2b), and (2d) are linearly independent and so each forms a basis for R2. Theorem5(S&B11.7). EverybasisofRn containsexactly n vectors. • Using theorems 1, 2 and corollary 1, together with the fact that a square matrix is nonsingular iff its determinant is nonzero gives the following theorem. Theorem6(S&B11.8). Letv ,...,v ∈ RnbeasetofvectorsandletA = (v v ···v ). 1 n 1 2 n Thenthe following statements are equivalent. 4
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