Bounded (finite-rank) Inverse Operators

Question

While studying for my functional analysis course I encountered bounded inverse theorem which states that given a bijective bounded linear operator $T:X\rightarrow Y$, then $T^{-1}:Y\rightarrow X$ is bounded too. (Actually at lesson it has been introduced as a corollary to open map theorem.) In particular, I was trying to relate this to finite rank operators like a matrix $A\in\mathbb{R}^{n\times m}$, which defines a linear map $T_A:\mathbb{R}^m\rightarrow \mathbb{R}^n$. Assuming such operator is bounded, I was thinking if it is reasonable asking that $T^{-1}_A x = A^{-1}x$ is bounded too in the case $m\neq n$, considering $A^{-1}$ in terms of its pseudoinverse.