Section 8 Notes

# Section 8 Notes - HANDOUT MATH 294 Vadim Zipunnikov1...

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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 let’s 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 . Let’s 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 0 - 1 0 E 1 , 0 1 0 - 1 E 2 , 1 0 2 0 E 3 , 0 1 0 2 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 0 - 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 0 1 0 - 1 = 0 0 0 0 T ( E 4 ) = 1 1 2 2 0 1 0 2 = 0 3 0 6 Step 2: It’s 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: [email protected] This is a handout for Sections 12, 13, and 14. Any typos and errors are solely the responsibility of the author. 1

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Hence, B= 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 3 . Now we can find Im(T) and Ker(T) as we usually do. And don’t forget to return to initial space R 2 × 2 . Modif Pr 4.3.21 Let T: P 2 P 2 given by T ( f ( t )) = f ( t ) - 3 f ( t ) and a nonstandard basis B of P 2 B = { t E 1 , t 2 - 1 E 2 , t 2 - t E 3 } . Solution: Again we start with Step 1: Calculate T ( E 1 ) , T ( E 2 ) , T ( E 3 ) T ( E 1 ) = T ( t ) = ( t ) - 3 t = 1 - 3 t = - 3 t + 1 T ( E 2 ) = T ( t 2 - 1) = ( t 2 - 1) - 3( t 2 - 1) = 2 t - 3 t 2 + 3 = - 3 t 2 + 2 t + 3 T ( E 3 ) = T ( t 2 - t ) = ( t 2 - t ) - 3( t 2 - t ) = 2 t - 1 - 3 t 2 + 3 t = - 3 t 2 + 5 t - 1.
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• Fall '05
• HUI
• Math, Orthogonal matrix, Ker, Vadim Zipunnikov1

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