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# 235HW7 - Math 235 Linear Algebra HW 7 Problem 3.4.6 Problem...

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Math 235: Linear Algebra HW 7 Problem 3.4.6 Problem 3.4.6 Let T be the linear operator on R 2 defined by T ( x, y ) = ( - y, x ) (a) What is the matrix of T in the standard ordered basis for R 2 ? 0 - 1 1 0 (a) (b) What is the matrix of T in the ordered basis B = { α 1 , α 2 } where α 1 = (1 , 2) and α 2 = (1 , - 1)? - 2 1 1 1 (b) (c) Prove that for every real number c the operator ( T - cI ) is invertible. We’ll consider this with the standard basis on R 2 , and then note that since change of basis is an isomorphism if a linear transformation is invertible in one basis it must be invertible in every basis. In this case the matrix for T - cI is 0 - 1 1 0 - c 0 0 c = - c - 1 1 - c since c 2 +1 > 0 for every real number c the matrix for T - cI is invertible, therefore T - cI is invertible. (c) Problem 3.4.6 continued on next page. . . Page 1 of 8

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Math 235: Linear Algebra HW 7 Problem 3.4.6 [(d)] (d) Prove that if B is any ordered basis for R 2 and [ T ] B = A , then A 12 A 21 6 = 0. Let B = { α 1 , α 2 } where α 1 = ( a, b ) and α 2 = ( c, d ) be any basis for R 2 . Also let B s denote the standard basis. Then the matrix B = [ I ] B B s = a c b d is the change of basis matrix from the standard basis into B , and B - 1 = [ T ] B s B = 1 ad - bc d - c - b a is its inverse. (note that ad - bc 6 = 0 since change of basis is invertible). So the matrix C = [ T ] B B is given by BAB - 1 = 1 ad - bc a c b d · 0 1 - 1 0 · d - c - b a So C = m n r s where m, n, r and s are functions of a, b, c and d . (for example m = - ab - cd , can you figure out the rest?) From part (c) we know that T - cI is invertible. So letting c = m we see that C - mI = m n r s - m 0 0 m = 0 n r s - m is invertible. From previous work we know that this matrix is invertible if and only if 0 ( r - m ) - nr 6 = 0 nr 6 = 0. For our matrix n = A 2 , 1 and r = A 1 , 2 , so we’re done. (d) Problem 3.4.7 Let T be the linear operator on R 3 defined by T ( x, y, z ) = (3 x + z, - 2 x + y, - x + 2 y + 4 z ) (a) What is the matrix of T in the standard ordered basis on R 3 ? 3 0 1 - 2 1 0 - 1 2 4 (a) (b) What is the matrix of T in the ordered basis { α 1 , α 2 , α 3 } where α 1 = (1 , 0 , 1), α 2 = ( - 1 , 2 , 1), and α 3 = (2 , 1 , 1)? 4 - 2 7 - 2 4 - 3 3 9 4 (b) Problem 3.4.7 continued on next page. . . Page 2 of 8
Math 235: Linear Algebra HW 7 Problem 3.4.7 [(c)] (c) Prove that T is invertible and give a rule for T - 1 like the one which defines T . Row reducing the augmented matrix 3 0 1 1 0 0 - 2 1 0 0 1 0 - 1 2 4 0 0 1 we get 1 0 0 4 / 9 2 / 9 - 1 / 9 0 1 0 8 / 9 13 / 9 - 2 / 9 0 0 1 - 3 / 9 6 / 9 3 / 9 which tells us that T is invertible and that T - 1 ( x, y, z ) = 1 / 9 (4 x + 2 y - z, 8 x + 13 y - 2 z, - 3 x + 6 y + 3 z ) (c) Problem 3.4.12 Let V be an n -dimensional vector space over the field F , and let B = { α 1 , . . . , α n } be an ordered basis for V . (a) According to Theorem 1, there is a unique linear operator T on V such that T ( α j ) = α j +1 , j = 1 , . . . , n - 1 , T ( α n ) = 0 What is the matrix A of T in the ordered basis B The matrix for T is the identity matrix shifted down by one row, with a row of zeros added at the top.

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235HW7 - Math 235 Linear Algebra HW 7 Problem 3.4.6 Problem...

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