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We know that if a matrix $\operatorname{rank}(A) \leq r$ where $A \in \mathbb{R}^{N \times M}$ then each submatrix $B_i \in \mathbb{R}^{n \times m}$ has $\operatorname{rank}(B_i)\leq r$

The opposite is true also if we take all $B_i \in \mathbb{R}^{n \times m}$ where $N \geq n \geq r$ , $M \geq m \geq r$ and make sure all $\operatorname{rank}(B_i)\leq r$ then $\operatorname{rank}(A) \leq r$

My question is if we take a particular $m,n$ and fix them, say $m=r+1$, $n=r+1$, can we claim if all $\operatorname{rank}(B_i) \leq r$ then $\operatorname{rank}(A) \leq r$.

Thanks a lot.

Thomas Russell
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cdicle
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  • Yes, it is true as long as you fix $m=n$; that is, when you check only for square submatrices. Otherwise it will not be true. – a12345 Apr 17 '14 at 15:27
  • Can you give more details please? – cdicle Apr 17 '14 at 15:31
  • Assume that $rank(A)> r$. Then just pick $r+1$ rows that are linearly independent. We get an $r+1 \times M$ submatrix. Now this submatrix has rank $r+1$ so we can pick $r+1$ columns that are also linearly independent. We are done. – a12345 Apr 17 '14 at 15:51
  • I see. I got it. Thanks a lot. I think we do not need square case $m=n$, we just need strictly $m \geq r+1$ and $n \geq r+1$. Am I right? – cdicle Apr 17 '14 at 16:17
  • Yes, that is true – a12345 Apr 17 '14 at 17:45

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