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115A_practice_midterm1_solns

# 115A_practice_midterm1_solns - practice midterm solutions 1...

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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 α = 1 , β = 1 , v = (0 , 1). Then ( α + β ) · v = (1 + 1) · (0 , 1) = 2 · (0 , 1) = parenleftbigg 2 · 0 , 1 2 parenrightbigg = parenleftbigg 0 , 1 2 parenrightbigg , whereas α · v + β · v = 1 · (0 , 1) + 1 · (0 , 1) = parenleftbigg 1 · 0 , 1 1 parenrightbigg + parenleftbigg 1 · 0 , 1 1 parenrightbigg = (0 , 1) + (0 , 1) = (0 , 2) . In fact, you could have chosen any α, β such that α negationslash = 0 , β negationslash = 0 , α + β 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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115A_practice_midterm1_solns - practice midterm solutions 1...

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