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Unformatted text preview: MATH 304, Fall 2011 Linear Algebra Homework assignment #5 (due Thursday, October 6) All problems are from Leon’s book (8th edition). Section 3.3: 2b, 2c, 2e, 8c, 9d, 14 Section 3.4: 8c, 10, 14c, 15a MATH 304 Linear Algebra Lecture 10: Linear independence. Wronskian. Basis of a vector space. Spanning set Let S be a subset of a vector space V . Definition. The span of the set S is the smallest subspace W ⊂ V that contains S . If S is not empty then W = Span ( S ) consists of all linear combinations r 1 v 1 + r 2 v 2 + ··· + r k v k such that v 1 , . . ., v k ∈ S and r 1 , . . . , r k ∈ R . We say that the set S spans the subspace W or that S is a spanning set for W . Remark. If S 1 is a spanning set for a vector space V and S 1 ⊂ S 2 ⊂ V , then S 2 is also a spanning set for V . Linear independence Definition. Let V be a vector space. Vectors v 1 , v 2 , . . ., v k ∈ V are called linearly dependent if they satisfy a relation r 1 v 1 + r 2 v 2 + ··· + r k v k = , where the coefficients r 1 , . . ., r k ∈ R are not all equal to zero. Otherwise vectors v 1 , v 2 , . . ., v k are called linearly independent . That is, if r 1 v 1 + r 2 v 2 + ··· + r k v k = = ⇒ r 1 = ··· = r k = 0. An infinite set S ⊂ V is linearly dependent if there are some linearly dependent vectors v 1 , . . . , v k ∈ S . Otherwise S is linearly independent . Examples of linear independence • Vectors e 1 = (1 , , 0), e 2 = (0 , 1 , 0), and e 3 = (0 , , 1) in R 3 . x e 1 + y e 2 + z e 3 = = ⇒ ( x , y , z ) = = ⇒ x = y = z = 0 • Matrices E 11 = parenleftbigg 1 0 0 0 parenrightbigg , E 12 = parenleftbigg 0 1 0 0 parenrightbigg , E 21 = parenleftbigg 0 0 1 0 parenrightbigg , and E 22 = parenleftbigg 0 0 0 1 parenrightbigg . aE 11 + bE 12 + cE 21 + dE 22 = O = ⇒ parenleftbigg a b c d parenrightbigg = O = ⇒ a = b = c = d = 0 Examples of linear independence • Polynomials 1 , x , x 2 , . . . , x n ....
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 Spring '08
 HOBBS
 Linear Algebra, Algebra, Vectors, Linear Independence, Vector Space, basis

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