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The following is an exercise.

Suppose that $A$ is an $n\times n$ matrix and that $m$ rows of $A$ are selected to form an $m\times n$ submatrix $B$. By considering the number of zero rows in the normal form, prove that $\text{rank}B \ge m - n + \text{rank} A$.

My solution showed that $A$ could also be a non-square matrix and the result would still follow. Is my observation correct?

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  • $\begingroup$ Yes. In general, if $B$ is chosen in such a fashion from the $n \times k$ matrix $A$, then $\operatorname{rank}(B) \geq m-n+ \operatorname{rank}(A)$ $\endgroup$ Aug 5 at 21:26
  • $\begingroup$ @BenGrossmann thanks $\endgroup$
    – Philomath
    Aug 5 at 21:29

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