This question already has an answer here:

Show that $r$ is the rank of the $n$x$n$ matrix $A\iff A$ has a nonsingular $r$x$r$ submatrix, but any larger square submatrix of $A$ is singular.

I know that to be nonsingular, det $ \neq 0$

I can see this to be true by writing out examples but I am unsure how to approach writing a proof for it.


marked as duplicate by Shahab, Daniel W. Farlow, hardmath, Em., user99914 Mar 31 '16 at 3:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What is your definition of rank? $\endgroup$ – Shahab Mar 31 '16 at 0:14
  • $\begingroup$ Where I grew up this is the definition of rank. Such equivalences usually mean "here are two different definitions, prove they imply each other." What is the other definition of rank that should be used here? $\endgroup$ – mathguy Mar 31 '16 at 0:16
  • $\begingroup$ Rank is the dimension of the row/column space, with dimension meaning the number of elements in the basis. $\endgroup$ – Lindsey G Mar 31 '16 at 0:20

Sketch (using that the column and row ranks are the same):

Since the rank of $A$ is $r$, there are $r$ independent columns in $A$. Consider the submatrix $B$ of $A$ formed by those $r$ columns. Then, the rank of $B$ is $r$ because the columns of $B$ are independent. Then, since the dimension of the row space of $B$ is $r$, there are $r$ independent rows. Form the submatrix $C$ by using those rows. This is an $r\times r$ submatrix which is nonsingular.

If there were a larger invertible submatrix $D$, then the columns of $A$ that include the columns of $D$ must be independent. This means that the rank of $A$ is larger than $r$, which is impossible.


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