# Show that $r$ is the rank of the $n$x$n$ matrix $A\iff A$ has a nonsingular $r$x$r$ submatrix [duplicate]

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

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.