True or False: If {u, v} is linearly independent, but {u, v, w} is linearly depe

Question

True or False: If {u, v} is linearly independent, but {u, v, w} is linearly dependent, then w is in Span({u, v}). If true, explain why. If false, provide a counterexample. There are some of the generalizations we generated in class (from both classes.) If a set of vectors contains at least 2 vectors that are scalar multiples of each other, then the set is linearly dependent. If a set contains p vectors in R^n and p > n, then the set is linearly dependent. If a set of vectors contains the zero vector, then the set is linearly dependent. If a set contains exactly one vector and it is nonzero, then the set is linearly independent. A set of vectors in linearly dependent if and only if at least one vector in the set can be written as a linear combination of other vectors in the set. Choose two of these generalizations and for each, provide an explanation that justifies them as true. These explanations are like "mini-proofs, " so they should contain full sentences that help the reader following your mathematics. You can also use sketches, logic, computations, etc.-whatever helps! Complete the worksheet from class that asked you for examples of linearly independent and dependent sets. No explanation. For each example that you created, though, also describe the span of that set (e.g., you could say the span is the set of all vectors that are multiples of [3 5], which is a o line in R^2). The handout is replicated on the back, with the span part added in.

Explanation / Answer

2.The statement is true. If the set { u,v,w} is linearly dependent, then a linear combination of these vector equals zero. Let pu +qv +rw = 0 , where p,q,r are arbitraryscalars. Then rw = -(pu+qv) or w = -( p/r)u –(q/r)v. Thus w is a linear combination of qu and v. Therefore w is in span ({u,v}).

3. (a) Let S = { u,v} and let v = pu where p is an arbitrary scalar. Then v –pu = v –v= 0. This means that a linear combination of u and v equals 0. Then by the definition of linear dependence, the set S is linearly dependent.

   (e) LetS = {a1 , a2 , a3,…,an } be a set of vectors. Let us first presume that S is linearly dependent. Then there exist scalars p1 , p2, p3 , …,pn such that p1a1 +p2a2 + p3 a3 +…+pn an = 0. Then for some r< n, we have prar = - (p1a1 +p2a2 + p3 a3 +…+ pr-1 ar-1 + pr+1 ar+1 +…+pn an). Thus ar = (-p1/pr) a1+…+ (-pn/pr)an . This shows that a vector in the set S is a linear combination of the other vectors in the set. Now, let ar = p1a1 +p2a2 + p3 a3 +…+ pr-1 ar-1 + pr+1 ar+1 +…+pn an where r < n. Then p1a1 +p2a2 + p3 a3 +…+ pr-1 ar-1 - ar + pr+1 ar+1 +…+pn an = 0. Thus, by the definition of linear dependenmce, S is a linearly dependent set.

4. The work sheet is not available.

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