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Unformatted text preview: 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 ? 1 1 (a) (b) What is the matrix of T in the ordered basis B = { 1 , 2 } where 1 = (1 , 2) and 2 = (1 , 1)? 2 1 1 1 (b) (c) Prove that for every real number c the operator ( T cI ) is invertible. Well 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 1 1 c 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 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 = { 1 , 2 } where 1 = ( a,b ) and 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 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 m = 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 were 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 { 1 , 2 , 3 } where 1 = (1 , , 1), 2 = ( 1 , 2 , 1), and 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 ndimensional vector space over the field F , and let B = { 1 ,..., n } be an ordered basis for V ....
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This note was uploaded on 11/14/2010 for the course PHYCS 498 taught by Professor Aa during the Spring '10 term at University of Illinois, Urbana Champaign.
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