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Unformatted text preview: MAT1341A: LECTURE NOTES FOR FALL 2008 MONICA NEVINS (HEAVILY BORROWING FROM HANDWRITTEN NOTES OF BARRY JESSUP) 7. Linear Dependence and Independence Today’s lecture is based on [N, 5.2]; see also [N, 4.2.1, 4.2.3]. Last time, we introduced the concept of the span of some vectors. Given finitely many vectors ~v 1 , ~v 2 , . . . , ~v m , we define their span to be the set of all linear combinations of these vectors. We write span { ~v 1 , ~v 2 , . . . , ~v m } = { a 1 ~v 1 + a 2 ~v 2 + ··· + a m ~v m  a i ∈ R } . Thus far, we should agree that: • span { ~ } = { ~ } • If ~v 6 = 0, then span { ~v } is the line through the origin in direction ~v . • If ~u, ~v 6 = 0 and ~u and ~v are not parallel, then span { ~u, ~v } is a plane (at least: in R 2 and R 3 ). (We also argued that three noncoplanar vectors should span all of R 3 , but this was only done verbally at the end of class, and I don’t expect you to feel convinced of that one, yet.) Notice the (geometric!) conditions we had to place on our statements. Today, let’s agree that they are necessary, and then work out what the correct algebraic version of these conditions should be, so that we can apply it to any vector space. (We’ll call the condition linear independence and one thing it will guarantee, as we’ll see, is that if { ~v 1 , ~v 2 , . . . , ~v m } are linearly independent, then span { ~v 1 , ~v 2 , . . . , ~v m } is strictly bigger than span { ~v 1 , ~v 2 , . . . , ~v m 1 } (or the span of any proper subset).) 7.1. Difficulties with span: Nonuniqueness of spanning sets. The first issue to men tion here is not one we’re going to be able to solve; but we need to discuss it to be aware of the issue. Example: span { (1 , 2) } = span { (2 , 4) } since these vectors are parallel and so span the same line. Example: If W = span { (1 , , 1) , (0 , 1 , 0) } and U = span { (1 , 1 , 1) , (1 , 1 , 1) } then in fact W = U . Why? Date : September 26, 2008. 1 2 MONICA NEVINS (HEAVILY BORROWING FROM HANDWRITTEN NOTES OF BARRY JESSUP) First note that (1 , 1 , 1) = (1 , , 1) , (0 , 1 , 0) and (1 , 1 , 1) = (1 , , 1) (0 , 1 , 0). Now apply the theorem from last time: The set W = span { (1 , , 1) , (0 , 1 , 0) } is a subspace of R 3 by theorem (1). We’ve just shown that (1 , 1 , 1) and (1 , 1 , 1) are in W (because they are linear combinations of a spanning set for W ). Thus by part (2) of the theorem, span { (1 , 1 , 1) , (1 , 1 , 1) } ⊆ W . So U ⊆ W . Next, write (1 , , 1) = 1 2 (1 , 1 , 1) + 1 2 (1 , 1 , 1) and (0 , 1 , 0) = 1 2 (1 , 1 , 1) 1 2 (1 , 1 , 1), and apply the same argument as above, with the roles of U and W reversed, to deduce that W ⊆ U . Hence W = U . Moral: You can’t usually tell if two subspaces are equal just by looking at their spanning sets! Every vector space (except the zero vector space) has infinitely many spanning sets....
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 Winter '10
 DAIGLE
 Linear Algebra, Vectors, Vector Space, Barry Jessup, Monica Nevins

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