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Rank of a matrix

  1. Aug 24, 2008 #1

    quasar987

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    If A is an nxk matrix of real numbers (n>=k) of rank k, is it true that we can eliminate n-k lines of A to obtain a matrix A' of nonvanishing determinant?

    I convinced myself of that one time while in the bus and now I can't find the proof.
     
  2. jcsd
  3. Aug 24, 2008 #2

    Defennder

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    Hmm, if A is of rank k, then that means that the row space of A is spanned by k vectors , and this means that we can eliminate (n-k) rows of A which are effectively linear combinations of the others. So when we do that we have A', which is a kxk matrix and of rank k, which implies that it is invertible which in turn implies its det is non-zero.
     
  4. Aug 24, 2008 #3

    quasar987

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    Ohhh.. yeah!

    Thanks!
     
  5. Aug 25, 2008 #4

    Defennder

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    Welcome. I'm trying to jog my linear algebra memory for a intermediate linear algebra class this semester.
     
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