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Linearly Independent Columns of a Matrix

  1. May 13, 2008 #1
    Hi wondering if can anyone help me... I've gotten so bogged down in all the rules and stuff for singular/non-singular matrices I've completely confused myself!

    Can anyone tell me is it true to say that if I have a matrix P, det(P) is NOT EQUAL to 0, then the vectors that would form the columns of P are linearly independent?

  2. jcsd
  3. May 13, 2008 #2
    Yes because then P is invertible, and thus have maximal rank.
  4. May 13, 2008 #3
    Brilliant - cheers! :)
  5. May 13, 2008 #4


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    do you know why this is true? i.e. do you know what it really means that the determinant is not zero?

    recall that any matrix can be rendered into upper diagonal form by repeatedly performing row operations, hence also by repeatedly multiplying by special invertible matrices.

    then the determinant is non zero iff the final matrix is actually diagonal and has non zero entries on the diagonal. these can then be made 1's.

    hence it has been inverted by matrix multiplication, and the columns are visibly independent, hence were also originally.
    this is just one of many ways to see it.
  6. May 13, 2008 #5
    Thank you - it does make sense now - I had a Linear Maths exam this afternoon and even though at some stage I had understood the reasoning behind what made a matrix singular, my mind seemed to be blanking on me in the hours leading up to the exam!

    Thanks again for the help :)
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