HW3solutions_3

HW3solutions_3 - MATH 115 SOLUTION SET 3 1.6: 10a, b. Im...

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MATH 115 SOLUTION SET 3 § 1.6: 10a, b. I’m sure you can do these. .. 11. Suppose { u,v } is a basis for V . We claim that the set { u + v,au } is also a basis. We must show it is linearly independent and generates V . Suppose that c 1 ( u + v ) + c 2 ( au ) = 0, so that ( c 1 + c 2 a ) u + c 2 v = 0. Since { u,v } is linearly independent, we must have c 1 + c 2 a = 0 and c 2 = 0. This implies that c 1 = c 2 = 0, so that { u + v,au } is linearly independent. By must it generate V ? Here’s a shortcut: Let W = Span { u + v,au } . Then W is two-dimensional, and it’s a subspace of V , which is also two-dimensional, hence W = V . The case of { au,bv } is similar. 12. First we show that { u + v + w,v + w,w } is linearly independent. Suppose a ( u + v + w ) + b ( v + w ) + cw = 0. Then au + ( a + b ) v + ( a + b + c ) w = 0. We get a = a + b = a + b + c = 0 because { u,v,w } is linearly independent. Solving, we get a = b = c = 0.
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HW3solutions_3 - MATH 115 SOLUTION SET 3 1.6: 10a, b. Im...

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