Linear Algebra with Applications (3rd Edition)

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Math 417 homework 5 solutions Section 3.4 Problem 22 A = b - 3 4 4 3 B , S = b 1 - 2 2 1 B (the columns of S are the basis vectors). S 1 can be computed either by elementary row operations, or by formula which is faster here: S 1 = 1 1 · 1 - ( - 2) · 2 b 1 2 - 2 1 B = b 1 5 2 5 - 2 5 1 5 B . Thus, by Fact 3.4.4, B = S 1 AS = b 5 0 0 - 5 B . Section 5.1 Problem 17 A vector vx = x 1 x 2 x 3 x 4 is perpendicular to W if it is perpendicular to each vector in it: For all vy W , vx · v y = 0 . But obviously it is su±cient to check that for the two given vectors spanning W . (Reason: let W be spanned by y 1 , . . ., y k and assume that vx · v y j = 0 for all j = 1 , . . ., k . If v y W too, then vy = c 1 vy 1 + . . . + c k vy k for some real numbers c 1 , . . ., c k because the vy i span W (it does not matter whether they are actually a basis). Then vx · v y = vx · ( c 1 vy 1 + ··· + c k vy k ) = c 1 vx · vy 1 + ··· + c k vx · vy k = 0 1
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because vx · v y j = 0 for all j , by assumption. Thus vx vy .) A basis is a list of independent vectors spanning W . How many do we need? W is spanned by the two given vectors. They are clearly independent,
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This homework help was uploaded on 01/23/2008 for the course MATH 417 taught by Professor Elling during the Fall '07 term at University of Michigan.

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hw5sol - Math 417 homework 5 solutions Section 3.4 Problem...

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