Ii we are going to show that β is a basis for v

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(ii) We are going to show that β is a basis for V which contains β W . Since β is linearly independent, so it suffices to show that β spans V . 51 PNU-MATH
1.7. If v V and v is not contained in β (i.e. v is not in span( β ) ), then β ∪ { v } is linearly independent. This contradicts to the maximality of β . v V, v span ( β ) V span ( β ) = V . β is a basis for V which contains β W . 5. ( ) ( ) Let β = { v 1 , v 2 , · · · } is a basis for V . If v V , then v span V Thus v is a linear combination of the vectors of β (that is v can be expressed by some finite vectors of β and scalars in F ) Suppose that v = a 1 v 1 + a 2 v 2 + · · · + a n v n and v = b 1 v 1 + b 2 v 2 + · · · + b n v n are two such representation of v . Then 0 = ( a 1 - b 1 ) v 1 + ( a 2 - b 2 ) v 2 + · · · + ( a n - b n ) u n Since β is linearly independent, it follows that a i = b i , where 1 i n ( ) Suppose each v V can be uniquely expressed as a linear combination of vectors of β , then clearly β spans V . If 0 = a 1 u 1 + a 2 u 2 + · · · + a n u n And we also have 0 = 0 · u 1 + 0 · u 2 + · · · + 0 · u n By hypothesis, the representation of zero as a linear combination of the u i is unique. Hence each a i = 0, and the u i are linearly independent. 52 PNU-MATH
1.7. 6. (i) Since S 1 ∈ F , F 6 = Let C be a chain in F and U = ∪{ A | A C } Clearly S 1 U S 2 Let a 1 u 1 + · · · + a n u n = 0 s.t. u i U, a i F, i = 1 , · · · , n Since u i A, A i C s.t. u i A i Since C is a chain, A k a.t. A i A k Thus u i A k F a 1 = · · · = a n = 0 So U is n upper bound of C By the maximal principle, F has a maximal element β i.e. β is a maximal linearly independent subset of V (ii) v S 2 , if v β , then v span ( β ) If v / β, β ∪ { v } is linearly independent This contradicts to the maximality of β S 2 span ( β ) Then V = span ( S 2 ) span ( β ) V span ( β ) = V Therefore β is a basis for V s.t. S 1 β S 2 53 PNU-MATH
1.7. 7. Let S = { A β | A S = , A S is linearly independent } S 6 = ( A = ) Let C be a nonempty chain in S and B = ∪{ A | A C } If x B S , then x B and x S Since A B, x A for some A C thus x A S = It’s a contradiction, so B S = If a 1 u 1 + · · · + a m u m + b 1 v 1 + · · · + b n v n = 0 , a i , b j F, u i B, v j S Since B = ∪{ A | A C } , A 1 , · · · , A m C s.t. u i A i So we may assume that u 1 , · · · , u m A m then u 1 , · · · , u m , v 1 , · · · , u n A m S is linearly independent a i = 0 , b j = 0 B S is linearly independent i.e. B is an upper bound of C By the maximal principle, S has a maximal element, say S 1 Clearly S 1 β, S 1 S = , S 1 S 2 is linearly independent So we only need to show that either { S 1 S is a maximal linearly independent subset of V or V } or { β span ( S 1 S ) } v β, If v S 1 S , then v span ( S 1 S ) If v / S 1 S , then S 1 S ∪ { v } is linearly independent It’s a contradiction, β span ( S 1 S )

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