# Linear algebra statment - rank of a matrix

How can I show that given an $m\times n$ matrix $A$ such that $\operatorname{rank} A = k \leq \min(m,n)$ , then there must exist a $k\times k$ minor of $A$ having $\det \neq 0$ .

I know that $rankA=k$ implies that there exist $k$ linearly independent rows and $k$ linearly independent columns , but how can I deduce from this the above statement ?

Thanks !

-

First take the $k$ linearly independent columns to get a $m \times k$ matrix of rank $k$. Since column rank equals row rank, there are $k$ linearly independent rows. Take those rows to get a $k \times k$ matrix. The rank is $k$ since there are $k$ linearly independent rows. (The latter also implies that the column rank $= k$.) By the invertible matrix theorem any $k \times k$ matrix of rank $k$ has determinant $\neq 0$.
But why these $k$ linearly independent rows must lie exactly in the $m\times k$ matrix ? Can't the restriction of the $k$ linearly independent rows to the $m\times k$ matrix be (for example) zero vectors ? Thanks! – homogenity Nov 10 '13 at 13:22
Only consider the $m \times k$ matrix. Its column rank is $k$, therefore its row rank is $k$, thus it has $k$ linearly independent rows. These are the rows you use. – Math Student 020 Nov 10 '13 at 14:43
What gurantees the column rank of the restricted matrix is also $k$ ? – homogenity Nov 10 '13 at 18:37
We know $A$ has $k$ linearly independent columns, because rank$(A) = k$. So we choose these columns to be the columns of the $m \times k$ matrix. Of course they remain linearly independent. – Math Student 020 Nov 10 '13 at 20:38