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Unformatted text preview: x in T ( V 1 ), there is u in V 1 such that T ( u ) = x . So u V 1 cu V 1 T ( cu ) = cT ( u ) T ( V 1 ) . Similarly for Z = { x V : T ( x ) W 1 } we have 3 conditions to verify: 1. 0 Z , since T (0) = 0 is in W 1 (a subspace). 2. If x, y Z , then T ( x ) , T ( y ) W 1 and W 1 being a subspace gives that T ( x ) + T ( y ) = T ( x + y ) W 1 . This last one say that x + y Z . 3. As above for x Z , we get that T ( x ) W 1 . This gives cT ( x ) = T ( cx ) is in W 1 , which implies that cx Z . 1 Problem 22 (Section 2 . 1) We need to know T on a basis. Let T (1 , , 0) = a, T (0 , 1 , 0) = b, T (0 , , 1) = c. Then T ( x, y, z ) = T ( x (1 , , 0) + y (0 , 1 , 0) + z (0 , , 1)) = xT (1 , , 0) + yT (0 , 1 , 0) + zT (0 , , 1) = ax + by + cz 2...
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 Winter '07
 Liu
 Linear Algebra, Algebra

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