If an $m\times n$ matrix with $m>n$ has full rank, does that imply that that uppermost $n\times n$ submatrix is invertible (also has full rank)?

If not, is it true at least that some combination of $n$ rows can be taken from the matrix to form an invertible matrix? How would one go about doing so?

  • $\begingroup$ BTW, the study of which subsets of vectors are independent are precisely what led to the development to matroid theory. $\endgroup$ – p.s. Jun 29 '14 at 3:06

One thing to note is that an $m \times n$ matrix with $n>m$ cannot have rank $n$, since otherwise we would have $n$ linearly independent vectors in $\mathbb{R}^m$. Having full rank (in this case rank $m$) implies that there is some $m \times m$ invertible submatrix, but this is not necessarily the uppermost submatrix.

  • $\begingroup$ How would you go about finding that submatrix though? $\endgroup$ – ruadath Jun 29 '14 at 2:53
  • $\begingroup$ You can use Gaussian elimination to put the matrix in row echelon form, and pick your rows so you get an invertible matrix. $\endgroup$ – Dylan Airey Jun 29 '14 at 2:55
  • $\begingroup$ Finding the submatrix depends on the values. You're going to have to be a bit more general than that. $\endgroup$ – Chris C Jun 29 '14 at 2:56

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