HW3_solution - -1 1-1 1 where k is a constant This tells us...

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EE441 HW#3 Solution Spring 2007 Instructor : Dr. Jonckheere TA: Ben Raskob Due Date 2/1/2007 Problem 2.3.2 v 1 , v 2 , v 3 are independent, and v 4 = v 2 - v 1 , v 5 = v 3 - v 1 , v 6 = v 3 - v 2 . Other choices are possible, but only three vectors can be independent. The number of independent vectors is the dimension of the space spanned by the v ’s. Problem 2.3.16 Solving the equation c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = 0 is equivalent to solving the system of equations: 1 1 0 0 1 0 0 1 0 1 1 0 0 0 1 1 c 1 c 2 c 3 c 4 = 0 where the columns of the matrix are the vectors v i . Reducing to row echelon form gives: 1 1 0 0 0 1 0 - 1 0 0 1 1 0 0 0 0 1
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This gives a solution other than zero for the constants. Namely, c 1 c 2 c 3 c 4 = - c 4 c 4 - c 4 c 4
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Unformatted text preview: -1 1-1 1 where k is a constant. This tells us that-v 1 + v 2-v 3 + v 4 = 0. Therefore, they are not linearly independent. No solution exists for 1 1 0 0 1 0 0 1 0 1 1 0 0 0 1 1 c 1 c 2 c 3 c 4 = 1 so they do not span R 4 . Problem 2.3.26 a) True, any vector linearly independent of the basis vectors for S will extend the basis to R 6 . b) False, because the basis vectors may not be in S . Problem 2.3.42 Basis 1 , x, x 2 , x 3 ; basis x-1 , x 2-1 , x 3-1 when p (1) = 0. 2...
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