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practice-midterm-solns

# practice-midterm-solns - MATH 110 SOLUTIONS TO PRACTICE...

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Unformatted text preview: MATH 110 - SOLUTIONS TO PRACTICE MIDTERM LECTURE 1, SUMMER 2009 GSI: SANTIAGO CA ˜ NEZ 1. Given vector spaces V and W , V × W is the vector space given by V × W = { ( v,w ) | v ∈ V and w ∈ W } , with addition and scalar multiplication defined componentwise; i.e. ( v,w ) + ( v ,w ) := ( v + v ,w + w ) and a ( v,w ) := ( av,aw ) . Prove that a map T : V → W is linear if and only if its graph, defined by graph T := { ( v,Tv ) ∈ V × W | v ∈ V } , is a subspace of V × W . Proof. Suppose that T is linear. Since T (0) = 0, we know that (0 , 0) = (0 ,T (0)) ∈ graph T. Now, let ( u,Tu ) , ( v,Tv ) ∈ graph T . Then Tu + Tv = T ( u + v ), so ( u,Tu ) + ( v,Tv ) = ( u + v,Tu + Tv ) = ( u + v,T ( u + v )) ∈ graph T, and thus graph T is closed under addition. Finally, let a ∈ F and let ( u,Tu ) ∈ graph T . Since aTu = T ( au ), we have a ( u,Tu ) = ( au,aTu ) = ( au,T ( au )) ∈ graph T, so graph T is closed under scalar multiplication. We conclude that the graph of T is a subspace of V × W . Conversely, suppose that graph T is a subspace of V × W . We must show that T preserves addition and scalar multiplication. First, let u,v ∈ V . Then ( u,Tu ) and ( v,Tv ) are in graph T. Since the graph is closed under addition, we must have ( u,Tu ) + ( v,Tv ) = ( u + v,Tu + Tv ) ∈ graph T. But for this to be in the graph of T , we must have Tu + Tv = T ( u + v ) according to the definition of the graph. Thus T preserves addition. Similarly, let a ∈ F and u ∈ V . Then a ( u,Tu ) = ( au,aTu ) ∈ graph T since graph T is closed under scalar multiplication. Again, this implies that aTu = T ( au ), so we conclude that T preserves scalar multiplication and is hence linear. 2. Let T ∈ L ( V ) and suppose that U and W are T-invariant subspaces of V . Prove that U ∩ W and U + W are also T-invariant. Date : July 10, 2009. 2 GSI: SANTIAGO CA ˜ NEZ Proof. First, let x ∈ U ∩ W . Then x ∈ U and x ∈ W . Since U and W are T-invariant, Tx ∈ U and Tx ∈ W . Thus Tx ∈ U ∩ W so U ∩ W is T-invariant. Now, let x ∈ U + W . Then x = u + w for some u ∈ U and w ∈ W . Hence Tx = T ( u + w ) = Tu + Tw. Since U and W are T-invariant, Tu ∈ U and Tw ∈ W . Thus Tx can be written as something in U plus something in W , so Tx ∈ U + W and U + W is T-invariant. 3. Axler, 2.7 Proof. Suppose first that V is infinite dimensional and let v 1 be any nonzero element of V . Then v 1 does not span V since V is infinite dimensional, so there exists v 2 ∈ V such that v 2 / ∈ span( v 1 ). Again, since V is infinite dimensional, ( v 1 ,v 2 ) does not span V , so we can pick v 3 ∈ V such that v 3 / ∈ span( v 1 ,v 2 ). Continuing in this manner we construct a sequence v 1 ,v 2 ,... which satisfies the required condition. Indeed, for each n , since v n / ∈ span( v 1 ,...,v n- 1 ) , the list ( v 1 ,...,v n ) is linearly independent....
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practice-midterm-solns - MATH 110 SOLUTIONS TO PRACTICE...

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