Define Z as an m×n matrix. Show that Z has rank m if and only if the determinant of some m × m sub-matrix of Z, which is obtained by deleting the n − m columns of Z, is nonzero.

I have managed to show the forward direction for this proof. I'm stuck in showing the reverse however (Starting with the submatrix and the determinant of the submatrix being non-zero). I know that it has m linear independent columns, thus rank = m. How can I always know that adding the n-m columns back keeps the rank = m. Must I just assume they're linearly dependent from the other m columns? Thank you.

  • $\begingroup$ Well, can it decrease the rank? Can it increase the rank? $\endgroup$ – darij grinberg Mar 6 at 20:12
  • $\begingroup$ I'm sure it could increase the rank, but that wouldn't be supporting my proof. $\endgroup$ – opp333 Mar 6 at 20:16
  • 2
    $\begingroup$ Increase the rank beyond $m$ (the number of rows of the matrix)? That would be quite a surprise. $\endgroup$ – darij grinberg Mar 6 at 20:17
  • $\begingroup$ Sorry why is that not possible, isn't linear independency based upon columns and not rows. Sorry I'm very new to algebra like this. Thanks :) $\endgroup$ – opp333 Mar 6 at 20:52
  • $\begingroup$ Oh, you don't know "row rank = column rank" yet. Do you know that more than $m$ vectors in an $m$-dimensional space must always be linearly dependent? $\endgroup$ – darij grinberg Mar 6 at 20:53

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