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Unformatted text preview: 2.1. * Say dim V = n 2 { v 1 ,v 2 , ,v m } : a basis for W { v 1 ,v 2 , ,v m ,v m +1 , ,v n } : a basis for V Let W = span { v m +1 , ,v n } and W 00 = span { v m +1 v 1 ,v m +2 v 3 , ,v n v 5 } then V = W W = W W 00 Clearly W 6 = W 00 ( ) If W = W 00 , then v m +1 ,v m +1 v 1 W 00 v 1 W W = (0) Its a contradiction (b) Example (example 1) the projection on W along W 1 { ( a,b )) = (0 ,b 1 3 a ) + ( a, 1 3 a ) } (example 2) the projection on W along W 2 ( a,b ) = (0 ,b ) + ( a, 0) 28. (1) { } is Tinvariant x { } , T ( x ) = 0 { } ( T is linear) (2) V is Tinvariant (T(V) V) x V , T ( x ) V ( T : V V ) (3) R ( T ) is Tinvariant (T(T(V)) T(V)) 68 PNUMATH 2.1. , T ( x ) R ( T ), T ( T ( x )) T ( V ) R ( T ) (4) N ( T ) is Tinvariant x N ( T ), T( x ) = 0 N ( T ) ( N ( T ) V as subspace, so N ( T ) has zero) 29. If T ( W ) W , x,y W, and c F T W ( x + y ) = T ( x + y ) = T ( x ) + T ( y ) = T W ( x ) + T W ( y ) (W is Tinvariant and T is linear) T W ( cx ) = T ( cx ) = cT ( x ) = cT W ( x ) T W is linear 30. x V, T ( x ) = x 1 s.t x = x 1 + x 2 , x 1 W,x 2 W (1) W is Tinvariant x 1 W , T( x 1 ) = x 1 W T( W ) W (2) T W = I W T W : W W, x 1 W, T W ( x 1 ) = x 1 I W : W W, x 1 W, I W ( x 1 ) = x 1 x 1 W, T W ( x 1 ) = I W ( x 1 ) T W = I W 31. V = R ( T ) W , W is Tinvariant 69 PNUMATH 2.1. (a) T ( W ) T ( V ) W = (0) T ( W ) = 0 T ( W ) N ( T ) (b) Since V = R ( T ) W , so dim V = dim R ( T ) + dim W rank ( T ) + nullity ( T ) = dim V Since dim V < , dim W = nullity ( T ) (c) (Example 1) Exercise 21, left shift (Example 2) = { v 1 ,v 2 , } for V T : V...
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This note was uploaded on 02/14/2012 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.
 Spring '09
 RUDYAK

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