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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 n-dimensional vector space over the field F , and let B = { 1 ,..., n } be an ordered basis for V ....
View Full Document

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.

Page1 / 8

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online