mat67-2011-hw3solutions

mat67-2011-hw3solutions - Solutions to HW #3 Math 67 UC...

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Solutions to HW #3 Math 67 UC Davis, Fall 2011 1. Solve the following problems in the textbook: (a) Calculational exercises 1, 2, 3 in Chapter 5. (b) Proof-writing exercise 4 in Chapter 5. Solution to calculational exercise 1. If a 1 ,a 2 ,a 3 are scalars such that a 1 v 1 + a 2 v 2 + a 3 v 3 = 0, this gives the homogeneous linear system of equations a 1 + a 2 + a 3 = 0 a 1 + 2 a 2 - a 3 = 0 a 1 + 3 a 2 + a 3 = 0 By simplifying the system using Gaussian elimination, one can check that a 1 = a 2 = a 3 = 0 is the unique solution, hence the vectors v 1 ,v 2 ,v 3 are linearly independent. Solution to calculational exercise 2. (a) Denote e 1 = (1 , 0 , 0), e 2 = (0 , 1 , 0), e 3 = (0 , 0 , 1), the standard basis vectors. Note that e 1 = ( - i ) v 1 , e 2 = v 2 - v 1 , e 3 = - v 3 + v 1 + ie 2 = - v 3 + v 1 + i ( v 2 - v 1 ) , i.e., e 1 ,e 2 ,e 3 span( v 1 ,v 2 ,v 3 ), and therefore also V = C 3 , which is the span of e 1 ,e 2 ,e 3 , is contained in span( v 1 ,v 2 ,v 3 ) (since the span is closed under linear com- binations). We also have the opposite containment relation span( v 1 ,v 2 ,v 3 ) V , so combining the two we get the claim that V = span( v 1 ,v 2 ,v 3 ). (b) We know from (a) above that ( v 1 ,v 2 ,v 3 ) is a spanning set of vectors, and since it has exactly 3 vectors (the dimension of V ), it must be linearly independent, since otherwise (by the “basis reduction theorem”, theorem 5.3.4 on page 55 of the textbook) we could discard some of the vectors and get a basis with less than 3 elements.

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This note was uploaded on 11/13/2011 for the course MATH 67 taught by Professor Schilling during the Fall '07 term at UC Davis.

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mat67-2011-hw3solutions - Solutions to HW #3 Math 67 UC...

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