Prove that the rank of a nonzero $m \times n$ matrix $A$ is the largest number $r$ such that $A$ has an $r \times r$ minor with nonzero determinant.

Asked Nov 08 '16 at 21:25

Active Nov 08 '16 at 21:25

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I've been doing some googling and I believe that the largest $r$ such that $A$ has an $r \times r$ minor with nonzero determinant is called the determinantal rank. I want to prove that $detrank(A)=rank(A)$. I think a good way to do this would be to prove that $detrank(A) \le rank(A)$ and $detrank(A) \ge rank(A)$.

For $detrank(A) \le rank(A)$, we know that $r \le m$ and $r \le n$, so we must have that $detrank(A) \le r \le rank(A)$.

I'm not sure how to do the other way or that the first half is correct.

asked Nov 08 '16 at 21:25

AndroidFish

1,133

( $A$ has an $r \times r$ nonzero minor) if and only if ($A$ has $r$ linearly independent rows) if and only if (the vectors $A e_{1}, A e_{2}, \ldots, A e_{r}$--where $e_{k}$ is the $k$-th standard basis vector--are linearly independent). – avs Nov 08 '16 at 21:30
See this question/answer – Ben Grossmann Nov 08 '16 at 21:51

Prove that the rank of a nonzero $m \times n$ matrix $A$ is the largest number $r$ such that $A$ has an $r \times r$ minor with nonzero determinant.

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