Math416_HW6_sol - Math 416 - Abstract Linear Algebra Fall...

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Unformatted text preview: Math 416 - Abstract Linear Algebra Fall 2011, section E1 Homework 6 solutions Section 2.8 8.2. Denoting the vectors respectively v 1 ,v 2 ,v 3 ,v 4 and choosing an order that will involve fewer computations, we have v 1 v 4 v 3 v 2 = 1 0 0 0 2 1 3 1 1 0 2 3 1 0 0 1 1 0 0 0 0 1 3 1 0 0 2 3 0 0 0 1 which has a pivot in every row and column. Therefore { v 1 ,v 2 ,v 3 ,v 4 } is a basis of R 4 . b. Because the vectors v i are already given in standard coordinates, i.e. [ v i ] standard = v i , the transition matrix from the basis { v i } to the standard basis is V = v 1 v 2 v 3 v 4 = 1 0 0 0 2 1 3 1 1 3 2 0 1 1 0 0 . 8.3. (2 pts) We must find the coordinates of 1 , 1 + t with respect to the basis { 1- t, 2 t } : 1 = 1- t + t = (1- t ) + 1 2 2 t [1] { 1- t, 2 t } = 1 1 2 1 + t = 1- t + 2 t = (1- t ) + (2 t ) [1 + t ] { 1- t, 2 t } = 1 1 ....
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This note was uploaded on 11/07/2011 for the course MATH 416 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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Math416_HW6_sol - Math 416 - Abstract Linear Algebra Fall...

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