5: Week 2.2 Subspaces/Spanning (continued) 2.3 Independence/Bases
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Linear independence/Bases and Dimension Vectors v1 , v2 , . . . , vk in a vector space V are linearly dependent if there are scalars c1 , c2 , . . . , ck so 0 = c1 v1 + c2 v2 + + ck vk , with at least one cj = 0. Such a vector equation is said to be a relation of linear dependence among the v1 , v2 , . . . , vk . A collection of vectors for which the only solution of the vector equation is the trivial solution, c1 = 0, . . . , ck = 0 is said to be linearly independent. Examples: (1) i, j in R2 ; and (2) {i, j, k} in R3 ; are linearly independent. Verification: c1 i + c2 j + c3 k = c1 (1, 0, 0) + c2 (0, 1, 0) + c3 (0, 0, 1) = (c1 , c2 , c3 ) = (0, 0, 0) = 0, only when c1 = 0, c2 = 0, c3 = 0.
Our main objective is the definition of basis. Vectors v1 , v2 , . . . , vk in a vector space V that are are both (1) a spanning set for V and (2) linearly independent are called a basis for V. So (1) i, j are a basis of R2 ; and (2) {i, j, k} is a basis of R3 . A vector space has many bases, but the main fact is that if v1 , v2 , . . . , vk is a basis of V and w1 , w2 , . . . , wj is a second basis of V, then the number of vectors is the same: k = j. The number of vectors in a basis of V is called the dimension of V . We start with verifications of linear independence in subspaces of Rn
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Problem Determine whether the vectors (1, 2, 3), (1, -1, 2) and (1, -4, 1) are LI or LD in R3 . If they are LD find a relation of linear dependence. Solution [again; same method!] Write the vector equation as an equation in column vectors, then 1 1 1 0 0 = c1 2 + c2 -1 + c3 -4 1 2 3 0 1 1 1 c1 = 2 -1 -4 c2 . 3 2 1 c3 Observe that this is the homog linear system with coef matrix A = (v1 v2 v3 ). For n-by-n systems the system has a unique
solution exactly when the determinant is non-zero. 1 We have 2 3 1 -1 2 1 1 -4 = 0 1 0 1 -3 -1 1 -6 = 0 -2
so the system has nontrivial solutions and the vectors are LD. To find a relation we continue the reduction 1 0 -1 2 , A 0 1 0 0 0 so a spanning vector is (1, -2, 1), and 0 = v1 - 2v2 + v3 is a nontrivial relation of LD.
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Example: Show that the vectors v1 = (1, 1) v2 = (1, -1) are a basis of R2 . Solution. We know that i, j are a basis of R2 ; so that the dimension of R2 is . . . .
We check that v1 = (1, 1) and v2 = (1, -1) are linearly independent (LI) since 1 1 1 -1
= -2 = 0. Checking that v1 and v2 span R2 by expressing each v = (x, y) R2 as a linear combination (as we did for i, j) takes a little bit of effort, and involves fractions. Checking that each vector equation v = c1 v1 + c2 v2 has a solution (without finding c1 , c2 ) is easier. (why?) But the main fact has a practical version that saves any computation, or even very much further thought. If v1 and v2 did not span R2 there would be a 3rd vector v3 not in the span of v1 , v2 . But that would make v1 , v2 , v3 linearly independent. We could consider adding further vectors, each new one giving a
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larger collection of indep vectors; but we already have too many, as we've shown dim(R2 ) 3. So v1 , v2 have to span (for free, given the main fact), so v1 , v2 is a basis for R2 . This is the use of the text's Theorem 2.12.(1) If the dim(V ) = n for a vector space V, then any LI set of n vectors is a basis of V.