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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 !

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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 $\neq 0$.

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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
    
And don't forget column rank = row rank. –  Math Student 020 Nov 10 '13 at 20:39

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