Homework SourseExercise 523 Prove or give a counterexample If v1 v2 vm
Exercise 5.23. Prove or give a counterexample: If v1, v2, . . . , vm is a linearly independent list of vectors in V , then 5v1 4v2, v2, v3, . . . , vm is linearly independent.
Exercise 5.24. Prove or give a counterexample: If v1, v2, . . . , vm and w1, w2, . . . , wm are linearly independent lists of vectors in V ,then v1 + w1, v2 + w2, . . . , vm + wm is linearly independent.
Exercise 5.25. Suppose v1, . . . , vm is linearly independent in V and w V . Show that v1, . . . , vm, w is linearly independent if and only if w / span(v1, . . . , vm).
Solution
5.23: Let {v1,v2,…,vm} be a linearly independent set of vectors in V. Also, let us assume that {5v14v2,v2,v3,.., vm } be a linearly dependent set. Then there exist scalars a1 , a2 , …am ( not all zero) such that a1(5v1 -4v2) + a2v2+… + am vm = 0 or, 5a1v1 + (a2- 4a1)v2 + …+am vm = 0. This means that a linear combination of v1,v2,…,vm equals zero so that the vectors v1,v2,…,vm are linearly dependent. This is a contradiction. Hence the assumption that {5v14v2,v2,v3,.., vm } is a linearly dependent set is incorrect. Thus {5v14v2,v2,v3,.., vm } is a linearly independent set.
5.24: Let {v1,v2,…,vm} and { w1, w2, . . . , wm} be linearly independent sets of vectors in V. Also, let us assume that the vectors v1+w1, v2+w2,,… vm+wm are linearly dependent. Then is there exist scalars a1 , a2, …, am ( not all zero) such that a1(v1+w1)+a2(v2+w2)+…+ am(vm+wm) = 0, then (a1v1+a2 v2+…+am vm)–(a1w1+a2w2+…+am wm) = 0. Hence a1v1+a2 v2+…+am vm = 0 and a1w1+a2w2+…+am wm = 0 (otherwise each vi = -wi which is not correct). This means that the vectors v1,v2,…,vm and the vectors w1, w2, . . . , wm are linearly dependent which is not correct. Hence the set { v1+w1, v2+w2,,… vm+wm} is linearly independent..
5.25: Let {v1,v2,…,vm} be a linearly independent set of vectors in V and let w be a vector which does not belong to span{ v1,v2,…,vm}. Then w cannot be expressed as a linear combination of the vectors v1,v2,…,vm This means that the vectors w, v1,v2,…,vm are linearly independent. Now let us assume that w,v1,v2,…,vm are linearly independent vectors. Then if there do not exist scalars a, a1 , a2, …, am ( not all zero) such that aw + a1v1+a2 v2+…+am vm = 0 or, we cannot have aw = -( a1v1+a2 v2+…+am vm) Then w - ( a1 /a)v1 – (a2 /a)v2 -…-(am /a)vm . This means that w cannot be expressed as a linear combination of v1,v2,…,vm. Thus w does not belong to span{ v1,v2,…,vm}.
