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Math416_HW6_sol

Math416_HW6_sol - Math 416 Abstract Linear Algebra Fall...

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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 . The transition matrix from the basis { 1 , 1 + t } to the basis { 1 - t, 2 t } is [1] { 1 - t, 2 t } [1 + t ] { 1 - t, 2 t } = 1 1 1 2 1 . Alternate method: The transition matrix from the basis { 1 , 1 + t } to the monomial basis { 1 , t } is 1 1 0 1 .

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Math416_HW6_sol - Math 416 Abstract Linear Algebra Fall...

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