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hwsol3 - Homework 3 Solutions Course Reader Exercises 10.16...

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Unformatted text preview: Homework 3 Solutions Course Reader Exercises 10.16 This is not a subspace, since it is not closed under scalar multiplication. For instance, v = 1 1 ∈ V but- v =- 1- 1 6∈ V 10.17 This is not a subspace since is it not closed under addition. For instance, v = 1 2 and w =- 2- 1 are in V , but v + w =- 1 1 is not in V . 10.21 Suppose V and W are subspaces of R n . Let v and w be vectors in V ∩ W . Since both v and w are in V , and V is a subspace, v + w is in V . Since both v and w are in W , and W is a subspace, v + w is in W . Thus v + w is in V ∩ W . Next let c ∈ R . Since v ∈ V and V is a subspace, c v is in V . Since v ∈ W and W is a subspace, c v is in W . Thus c v is in V ∩ W . This shows that V ∩ W is closed under addition and scalar multiplication, and is therefore a subspace. 11.1 The given matrix is A = 1 2 1 5 2 2 0 2 4 3 2 0 3 4 2 0- 2 2 5 2 0- 4 1 . Since rref( A ) = 1 0 0- 2- 1 0 1 0 3 3 0 0 1- 4 0 0 0 0 0 0 , the first three columns are the pivot columns, so 1 2 3 4 5 , 2 2 2 2 2 , 1 1 is a basis for C ( A ). From rref( A ), we see that N ( A ) is the solution set of the system x 1- 2 x 4- x 5 = 0 x 2 + 3 x 4 + 3 x 5 = 0 x 3- 4 x 4 = 0 . Thus, we can write the solution set for N ( A ) as x 1 x 2 x 3 x 4 x 5 = 2 x 4 + x 5- 3 x 4- 3 x 5 4 x 4 x 4 x 5 = x 4 2- 3 4 1 + x 5 1- 3 1 . We can take the two spanning vectors as a basis since they are independent, so 2- 3 4 1 , 1- 3 1...
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hwsol3 - Homework 3 Solutions Course Reader Exercises 10.16...

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