Section 8 Notes - H ANDOUT M ATH 294 Vadim Zipunnikov 1...

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Unformatted text preview: H ANDOUT M ATH 294 Vadim Zipunnikov 1 Chapter 4.3 The matrix of a linear transformation Let T: V W is a linear transformation. If V = W , that is, T: V V (V can be P n , R n , R n n or another linear space) and V has a basis B = { E 1 ,...,E n } then we can construct B matrix by using fact 4.3.2 (this fact works only in the case V=W) which states that B = [ T ( E 1 )] B , [ T ( E 2 )] B ,..., [ T ( E n )] B . Most problems we deal with have n 4. So lets put n=4. Then B = { E 1 ,E 2 ,E 3 ,E 4 } and B = [ T ( E 1 )] B , [ T ( E 2 )] B , [ T ( E 3 )] B , [ T ( E 4 )] B . Lets split the algorithm into two steps. Step 1: Calculate T ( E 1 ) ,T ( E 2 ) ,T ( E 3 ), and T ( E 4 ). Step 2: Solve the following system (with respect to a 1 ,a 2 ,...d 3 ,d 4 ) T ( E 1 ) = a 1 E 1 + a 2 E 2 + a 3 E 3 + a 4 E 4 T ( E 2 ) = b 1 E 1 + b 2 E 2 + b 3 E 3 + b 4 E 4 T ( E 3 ) = c 1 E 1 + c 2 E 2 + c 3 E 3 + c 4 E 4 T ( E 4 ) = d 1 E 1 + d 2 E 2 + d 3 E 3 + d 4 E 4 After that, we can form matrix B= a 1 b 1 c 1 d 1 a 2 b 2 c 2 d 2 a 3 b 3 c 3 d 3 a 4 b 4 c 4 d 4 . Quiz 2.2 . We consider a linear transformation T(M): R 2 2 R 2 2 given by T ( M ) = " 1 1 2 2 # M . Find matrix of the transformation with respect to the basis B = { " 1- 1 0 # | {z } E 1 , " 1- 1 # | {z } E 2 , " 1 0 2 0 # | {z } E 3 , " 0 1 0 2 # | {z } E 4 } . Solution: Step 1: Calculate T ( E 1 ) ,T ( E 2 ) ,T ( E 3 ), and T ( E 4 ) T ( E 1 ) = " 1 1 2 2 #" 1- 1 0 # = " 0 0 0 0 # T ( E 3 ) = " 1 1 2 2 #" 1 0 2 0 # = " 3 0 6 0 # T ( E 2 ) = " 1 1 2 2 #" 1- 1 # = " 0 0 0 0 # T ( E 4 ) = " 1 1 2 2 #" 0 1 0 2 # = " 0 3 0 6 # Step 2: Its easy to see that T ( E 1 ) = " 0 0 0 0 # = 0 E 1 + 0 E 2 + 0 E 3 + 0 E 4 T ( E 2 ) = " 0 0 0 0 # = 0 E 1 + 0 E 2 + 0 E 3 + 0 E 4 T ( E 3 ) = " 3 0 6 0 # = 0 E 1 + 0 E 2 + 3 E 3 + 0 E 4 T ( E 4 ) = " 0 3 0 6 # = 0 E 1 + 0 E 2 + 0 E 3 + 3 E 4 . 1 Mailing address: Malott Hall 101, Cornell University, Ithaca, 14850, NY, USA. E-mail: vvz2@cornell.edu. This is a handout for Sections 12, 13, and 14. Any typos and errors are solely the responsibility of the author....
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Section 8 Notes - H ANDOUT M ATH 294 Vadim Zipunnikov 1...

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