To determine Rank of Linear Transformation Question is to find the rank of $T_1 $and $T_2$ 
Since the composition is bijective so rank of $T_1T_2 = m$. But how do I get the ranks of$ T_1 $and$ T_2 $from here? Thanks.
 A: First of all, we do not know which is greater. If $m>n$, the composition cannot be bijective, since $T_2$ "loses information" by going from a greater space to a lesser one. So to answer the question, we need $m\leq n$.
In that case, to have a bijection as the composition, we will surely need the image of $T_2$ to be isomorphic to $\mathbb{R}^m$ whence we started, otherwise we will end up in a space of smaller dimension which cannot, under a linear map, be mapped bijectively onto $\mathbb{R}^m$, but at most injectively. Thus, $\mathrm{rank}(T_2)=m$.
For similar reasons, $T_1$ must be surjective, otherwise the composition cannot be surjective. This means $\mathrm{rank}(T_1)=m$ as well.
A: Hint: Use, along with the fact that $T_1T_2$ is bijective (rank is $m$ as you have pointed out already), the following theorems to find the rank of $T_2$:


*

*For a transformation $A:\mathbb{R}^n\rightarrow\mathbb{R}^m$, $\dim \mathcal{N}(A)+\dim \mathcal{R}(A) = n$.

*For any multiplication-compatible $A,B$ we have $\mathcal{N}(AB)\supseteq \mathcal{N}(B)$.


To get the rank of $T_1$ apply the same above to $(T_1T_2)^T$ along with the fact that the row rank is equal to the column rank.
NOTE: This is of course assuming you know how to prove the above theorems rigorously along with the knowledge of the four fundamental subspaces. 
