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Linear combination

  1. Sep 25, 2012 #1
    Given the vectors
    v1=(1, 1) ^t
    v2=(3, -1)^t

    setting up the matrix gives det≠0, thus any vector in R^n can be written as a linear combination of v1 and v2.

    This is where I'm getting confused.
    If the numbers in the matrix were changed so det=0, can you still right any vector in R^n as a linear combination of v1 and v2?
    If det=0, this would yield a free variable. In the examples in the book, they say you can write a vector as a linear combination of other vectors even if a free variable exists.
     
  2. jcsd
  3. Sep 25, 2012 #2
    I guess this leads to my next question:
    If v1 and v2 are linearly independent vectors
    in R^n and v3 cannot be written as a scalar
    multiple of v1, then v1, v2, and v3 are linearly
    independent.
    Why is this statement false?



    Never mind, I thought the statement said "and v3 cannot be written as a scalar multiple of v1&v2" instead of just v1. If it was v1 and v2, then it would be linearly independent.
     
  4. Sep 25, 2012 #3

    jbunniii

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    det=0 if and only if both columns are a scalar multiple of a particular vector, call it v. If that is the case, then any linear combination of the columns will also be a scalar multiple of v. Therefore any vector that is not a scalar multiple of v cannot be expressed as a linear combination of the columns.
     
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