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Unformatted text preview: Homework assignment 7 pp. 105 Exercise 2. Let B = { 1 , 2 , 3 } be the basis for C 3 defined by 1 = (1 , , 1) 2 = (1 , 1 , 1) 3 = (2 , 2 , 0) . Find the dual basis of B . Solution: The first element of the dual basis is the linear function * 1 such that * 1 ( 1 ) = 1 , * 1 ( 2 ) = 0 and * 1 ( 3 ) = 0 . To describe such a function more explicitly we need to find its values on the standard basis vectors e 1 , e 2 and e 3 . To do this express e 1 ,e 2 ,e 3 through 1 , 2 , 3 (refer to the solution of Exercise 1 pp. 5455 from Homework 6). For each i = 1 , 2 , 3 you will find the numbers a i ,b i ,c i such that e 1 = a i 1 + b i 2 + c i 3 (i.e. the coordinates of e i relative to the basis 1 , 2 , 3 ). Then by linearity of * 1 we get that * 1 ( e i ) = a i . Then * 2 ( e i ) = b i , and * 3 ( e i ) = c i . This is the answer. It can also be reformulated as follows. If P is the transition matrix from the standard basis e 1 ,e 2 ,e 3 to 1 , 2 , 3 , i.e. ( 1 , 2 , 3 ) = ( e 1 ,e 2 ,e 3 ) P , then ( P 1 ) t is the transition matrix from the dual basis e * 1 ,e * 2 ,e * 3 to the dual basis * 1 , * 2 ,a * 3 , i.e. ( * 1 , * 2 ,a * 3 ) = ( e * 1 ,e * 2 ,e * 3 )( P 1 ) t . Note that this problem is basically the change of coordinates problem: e.g....
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This note was uploaded on 10/27/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Differential Equations, Linear Algebra, Algebra, Equations

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