Suppose a linear transformation $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$ has rank $k$.Show that there exists a $\delta>0$ such that if for a linear transformation $S:\mathbb{R}^n\rightarrow \mathbb{R}^m$ we have that $\left \|S-T \right \|<\delta$, then $rank(S)\geqslant k$.
My hunch tells me that the proof should be simple, but a cannot figure this out. I'd appreciate a small hint.
The rank of a linear transformation can be expressed by Minors. The rank is $\geq k$ if and only if there is a non zero $k\times k$-minor. Because the determinant is a continous function, we get that all transformation in a neighbourhood have also a non zero $k\times k$-minor and therefore rank $\geq k$.