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math110s-hw3sol

# math110s-hw3sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

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MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 3 SOLUTIONS 1. Let W 1 , . . . , W n be non-trivial subspaces of a vector space V over R . Show that there exists a vector v V such that v / W i for all i = 1 , . . . , n . Solution. If W 1 W 2 ∪ · · · ∪ W n , then any v / W i for all i = 2 , . . . , n will automatically satisfy v / W 1 . If W 2 ∪ · · · ∪ W n W 1 , then any v / W 1 will automatically satisfy v / W i for all i = 2 , . . . , n . Hence we may assume without loss of generality that W 1 * W 2 ∪ · · · ∪ W n and W 2 ∪ · · · ∪ W n * W 1 . Let v 1 W 1 and v 1 / W 2 ∪ · · · ∪ W n . Let v 2 W 2 ∪ · · · ∪ W n and v 2 / W 1 . Consider the set L = { λ v 1 + v 2 | λ R } . For any λ , λ v 1 + v 2 / W 1 since otherwise v 2 = ( λ v 1 + v 2 ) - λ v 1 W 1 — a contradiction. Now let i ∈ { 2 , . . . , n } . If λ i v 1 + v 2 W i for some λ i , then for any λ 6 = λ i , λ v 1 + v 2 / W i since otherwise v 1 = ( λ - λ i ) - 1 [( λ v 1 + v 2 ) - ( λ i v 1 + v 2 )] W i — a contradiction. Hence each W i contains at most one element of L . Since L has infinitely many elements, there exists v L such that v / W i for all i = 1 , . . . , n . 2. Let W 1 , . . . , W n be subspaces of a vector space V . Recall from Problem Set 2 that W 1 + · · · + W n is a direct sum (and hence may be denoted W 1 ⊕ · · · ⊕ W n ) if W i j 6 = i W j = { 0 } for all i = 1 , . . . , n. Show that the following statements are equivalent. (i) W 1 + · · · + W n is a direct sum. (ii) The function defined by f : W 1 × · · · × W n V, f ( w 1 , . . . , w n ) = w 1 + · · · + w n is injective. (iii) W 1 , . . . , W n satisfies ( W 1 + · · · + W i ) W i +1 = { 0 } for all i = 1 , . . . , n - 1 . Solution. We will make use of the three equivalent statements characterizing direct sum that we proved in Problem Set 2 . (i) (ii): Let f ( w 1 , . . . , w n ) = f ( w 0 1 , . . . , w 0 n ). Then w 1 + · · · + w n = w 0 1 + · · · + w 0 n and so w 1 = w 0 1 , . . . , w n = w 0 n since W 1 + · · · + W n is a direct sum. Hence ( w 1 , . . . , w n ) = ( w 0 1 , . . . , w 0 n ) .

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math110s-hw3sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

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