Given a $n\times n$ matrix M. Prove that M has rank less than $k$ if and only if all of the determinants of its $k\times k$ minors are $0$.

My progress: I have thought about this problem for a while, but I only got that if the $k*k$ minors are $0$, then the $k$ columns of that submatrix are linearly dependent. I don't see how to relate this fact to the linear dependence of the original $k$ columns of matrix M. Can anyone please help me with a detailed explanation?


Hint: use the fact that row and column operations do not change the rank of a matrix. If you have a minor with nonzero determinant, you can use row and column operations to turn it into an identity submatrix.

  • $\begingroup$ Many thanks for your hint! But I think it only explains the fact that the submatrix will have $k$ linearly dependent columns/rows when all the determinants of $k\times k$ minors are 0 (since column rank=row rank). But how to prove that these columns/rows, when extended to the columns of matrix M, are still linearly dependent? Do we need to expand the det(M) along some rows above all those $k$ linearly dependent rows to show that det(M)=0? $\endgroup$ – user177196 Dec 12 '14 at 21:48
  • $\begingroup$ If there is a $k\times k$ minor of nonzero determinant, use row and column operations to turn it into a $k\times k$ identity matrix. Those $k$ columns will then be linearly independent. $\endgroup$ – Gyu Eun Lee Dec 12 '14 at 22:41
  • $\begingroup$ The thing is, after obtaining that $k\times k$ identity matrix, how to prove that the new $k$ columns of matrix M is L.I? Since all we have now is the $k\times k$ identity matrix with $k$ LI columns, but this matrix is INSIDE the big matrix M. So we are not sure if these $k$ LI columns correspond to the $k$ LI columns of the big matrix M. If we can show such correspondence, the problem is solved, but I can't prove it. $\endgroup$ – user177196 Dec 13 '14 at 0:54
  • $\begingroup$ Here's my attempt to prove that such correspondence exists: since these $k$ LI columns in the submatrix N will be part of $k$ big columns of M, if the $k$ big columns in M aren't LI, the $k$ columns in N are also not LI (contradiction). $\endgroup$ – user177196 Dec 13 '14 at 1:01
  • $\begingroup$ Just think: can $(1,0,a)$ and $(0,1,b)$ ever be linearly dependent? $\endgroup$ – Gyu Eun Lee Dec 13 '14 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.