2.a1 - x in T ( V 1 ), there is u in V 1 such that T ( u )...

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Linear Algebra -115 Solutions to Fifth Homework Problem 11 (Section 2 . 1) We observe that { (1 , 1) , (2 , 3) } is a linearly inde- pendent set, therefore, a basis. So, we can express (8 , 11) as a linear combination of these: (8 , 11) = 2(1 , 1) + 3(2 , 3) . Now, if we apply T to both sides and use linearity, we have that T (8 , 11) = 2 T (1 , 1) + 3 T (2 , 3) = 2(1 , 0 , 2) + 3(1 , - 1 , 4) = (5 , - 3 , 16) . ± Problem 13 (Section 2 . 1) We want to show that { s 1 , . . . , s k } is linearly independent. So assume that for some scalars a 1 s 1 + . . . + a n s n = 0. Applying T to both sides we have that T ( a 1 s 1 + . . . + a n s n ) = T (0) = 0 T ( a 1 s 1 ) + . . . + T ( a n s n ) = 0 a 1 T ( s 1 ) + . . . + a n T ( s n ) = 0 a 1 w 1 + . . . + a n w n = 0 Using the fact that { w 1 , . . . , w n } is linearly independent we have that all a 1 = . . . = a n = 0. Henceforth, { s 1 , . . . , s n } linearly independent. ± Problem 20 (Section 2 . 1) We have to verify the 3 conditions of being a subspace: 1. Since 0 V 1 and T (0) = 0, it must be that 0 T ( V 1 ). 2. Say that x, y T ( V 1 ). Then there have to be some u, v V 1 such that T ( u ) = x, T ( v ) = y . We need to show that x + y = T ( u )+ T ( v ) = T ( u + v ) is in T ( V 1 ). But this is immediate, since u + v V 1 . 3. Similarly for
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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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2.a1 - x in T ( V 1 ), there is u in V 1 such that T ( u )...

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