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Unformatted text preview: practice midterm solutions 1. V is not a vector space because it violates VS8, distributivity of multiplication over scalar addi- tion. We just need one counterexample. Choose = 1 , = 1 ,v = (0 , 1). Then ( + ) v = (1 + 1) (0 , 1) = 2 (0 , 1) = parenleftbigg 2 , 1 2 parenrightbigg = parenleftbigg , 1 2 parenrightbigg , whereas v + v = 1 (0 , 1) + 1 (0 , 1) = parenleftbigg 1 , 1 1 parenrightbigg + parenleftbigg 1 , 1 1 parenrightbigg = (0 , 1) + (0 , 1) = (0 , 2) . In fact, you could have chosen any , such that negationslash = 0 , negationslash = 0 , + negationslash = 0, and the second component of v negationslash = 0, and you would have demonstrated the same contradiction. 2. Because we know the set W is contained in the vector space R 3 , then we do not have to test VS1-8; instead we only have to test closure under addition and scalar multiplication, in which case W will be a subspace of R 3 . If we did not know W lived in some larger vector space, we would have to verify VS1-8. Right away we can tell this will be a subspace because we are given a linear relation between the components, and this relation has no constant terms (i.e. it is set equal to zero, not a non-zero constant)....
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