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Unformatted text preview: Section 4.6 Change of Basis Let B = { v 1 ,v 2 ,...,v n } be a basis for a finite dimensional vector space V. Let v ∈ V then we can express v as v = c 1 v 1 + c 2 v 2 + ... + c n v n . Recall the coordinate vector of v, which was denoted as ( v ) B = ( c 1 ,c 2 ,...,c n ) . In this section we will see how to coordinate vectors are changing with the change of basis. Example: B = { (1 , 0) , (0 , 1) } and B = { (1 , 1) , (2 , 1) } are bases for R 2 . Consider u = (2 , 3) ∈ R 2 . Then we have ( u ) B = (2 , 3) and ( u ) B = (4 , 1) What is the relation between the coordinates of a vector with respect to different bases? Solution: If we change the basis for a vector space V from an old basis B= { u 1 ,u 2 ,u 3 ,...,u n } to a new basis B = { u 1 ,u 2 ,...,u n } , then for each vector v in V, the old coordinate vector [ v ] B is related to the new coordinate vector [ v ] B by the equation [ v ] B = P [ v ] B where the columns of P are the coordinate vectors of the new basis vectors relative to the old basis; that is, the column vectors of P are [ u 1 ] B , [ u 2 ] B ,..., [ u n ] B P B → B = [[ u 1 ] B , [ u 2 ] B ,..., [ u n ] B ] , [ v ] B = P B → B [ v ] B P B → B = [[ u 1 ] B , [ u 2 ] B ,..., [ u n ] B ] , [ v ] B = P B → B [ v ] B Example: Consider the bases B = { u 1 ,u 2 ,u 3 } and B = { u 1 ,u 2 ,u 3 } for R 3 , where u 1 = ( 3 , , 3) ,u 2 = ( 3 , 2 , 1) ,u 3 = (1 , 6 , 1) and u 1 = ( 6 , 6 , 0) ,u 2 = ( 2 , 6 , 4) ,u 3 = ( 2 , 3 , 7) (a) Find the transition matrix from B to B (b) Compute the coordinate vector [ w ] B where w = ( 5 , 8 , 5) and use part (a) to compute [ w ] B . Solution: (a) We want P B → B = [[ u 1 ] B , [ u 2 ] B , [ u 3 ] B ] ; [ u 1 ] B : Consider u 1 = c 1 u 1 + c 2 u 2 + c 3 u 3 , from here we get a linear system of equations: 6 c 1 2 c 2 2 c 3 = 3 6 c 1 6 c 2 3 c 3 = 0 4 c 2 + 7 c 3 = 3 1 Solving this system gives c 1 = 3 4 ,c 2 = 3 4 ,c 3 = 0 , hence [ u 1 ] B = ( 3 4 , 3 4 , 0) . [ u 2 ] B : Consider u 2 = c 1 u 1 + c 2 u 2 + c 3 u 3 , from here we get a linear system of equations: 6 c 1 2 c 2 2 c 3 = 3 6 c 1 6 c 2 3 c 3 = 2 4 c 2 + 7 c 3 = 1 Solving this system gives c 1 = 3 4 ,c 2 = 17 12 ,c 3 = 2 3 , hence [ u 2 ] B = ( 3 4 , 17 12 , 2 3 ) . [ u 3 ] B : Consider u 3 = c 1 u 1 + c 2 u 2 + c 3 u 3 , from here we get a linear system of equations: 6 c 1 2 c 2 2 c 3 = 1 6 c 1 6 c 2 3 c 3 = 6 4 c 2 + 7 c 3 = 1 Solving this system gives c 1 = 1 12 ,c 2 = 17 12 ,c 3 = 2 3 , hence [ u 3 ] B = ( 1 12 , 17 12 , 2 3 ) ....
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This document was uploaded on 04/26/2011.
 Spring '11

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