# On the rank of a submatrix

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?

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