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Linear independence of columns of a matrix

  1. May 4, 2007 #1
    1. The problem statement, all variables and given/known data
    Are the columns of this matrix linearly independent?
    1........3.........-2
    0.......-8.........11
    0........0.........1
    0....... 0......... 0
    (periods are just to make spacing clear)
    3. The attempt at a solution
    What is confusing me is the last row of zeros. If a set of vectors contains the zero vector, it is linearly dependent..but would this affect the linear independence of the columns of the matrix? If you augment the matrix with the zero vector, then the third row says that the only solution is the trivial one, which means that the columns of the matrix are linearly independent.
     
  2. jcsd
  3. May 4, 2007 #2

    cristo

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    If the columns are linearly dependent, then the third column can be made by adding multiples of the first and second columns together. Can this be done?
     
  4. May 4, 2007 #3
    No it can't, so the fact that there is a row of zeros doesn't matter for the columns of the matrix..
     
  5. May 4, 2007 #4

    cristo

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    Well no, because you're not looking at the rows here. To put the question another way, is the set of vectors {(1 0 0 0), (3 -8 0 0), (-2 11 1 0)} linearly independent?
     
  6. May 6, 2007 #5

    Fredrik

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    A row of zeros is significant because it immediately tells us that the dimension of the space spanned by the column vectors must be less than the number of columns (i.e. the number of components of the vectors). In this case however, that only means that the space spanned by the vectors is at most three-dimensional, but you knew that already since there are only three vectors in this problem.

    In other problems, a row of zeros may immediately give away the answer. For example: Are (a b 0), (c d 0) and (e f 0) linearly independent? No, they must be linearly dependent because there are three of them and they're all in the x-y plane, which is two-dimensional.
     
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