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Math Help - Linear Algebra True/False Questions

  1. #1
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    Linear Algebra True/False Questions

    So I was working on some true/false questions as practice for an upcoming quiz and I disagree with the textbook in a few cases. Please let me know if there is something I am missing. The questions and my reasoning are as follows:

    1. If u and v are vectors in R-m, then -u is in Span{u,v}.

    I say true, since Span{u,v} is the set of all vectors that can be written as c*u + d*v. If I choose c = -1 and d = 0, then I get that -u is in the span.

    2. If w is a linear combination of u and v in R-n, then u is a linear combination of v and w.

    I say false. Suppose that u is some nonzero vector and that v and w are the zero vector in R-n. Then w = 0*u + k*v, but u cannot be expressed as a linear combination of v and w since Span{u,v} is the zero vector.

    3. Suppose that v1, v2, and v3 are in R-5, v2 is not a multiple of v1, and v3 is not a linear combination of v1 and v2. Then {v1,v2,v3} is linearly independent.

    I say false. Simply let v1 be the zero vector in R-5 and let v2 and v3 be some nonzero vectors that are not multiples of one another. Then v2 is not a multiple of v1 since Span{v1} is the zero vector and v3 is not a linear combination of v1 and v2 since span {v1,v2} is the line in the direction of v2 through the origin. Also, the set {v1,v2,v3} is linearly dependent because it contains the zero vector. As an aside, I think that if v1 were not the zero vector and all of the conditions in the statement held, then {v1,v2,v3} would be linearly independent. In this case, we can apply the theorem that says if a set of vectors is linearly dependent, then some vector is a linear combination of the preceding vectors in the set. Obviously this is not the case, so the set would be linearly independent.

    4. A linear transformation is a function.

    I say true. Every transformation is a function and a linear transformation is just a transformation with certain properties, so it follows that a linear transformation is just a special type of function.

    The above seems well-reasoned to me. If there is something wrong with any of my arguments, please let me know. Also, if they are all sound, please let me know so that I can rest easy tonight. Thanks!
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  2. #2
    Super Member Gamma's Avatar
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    Looks good to me, i'm convinced.
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