Show linear Transformation $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is one-to-one So I’ve been stuck on this question for some time now and I can’t seem to get the ball rolling. It is clear to me that to prove either $T$ is onto or $S$ one-to-one, I must find whether if $n\geq m$. Nevertheless, I’m stumped on the next step to take to demonstrate this. Perhaps you might be able to help me out. 
Let $A$ be an $m\times n$ matrix and $B$ be an $n\times m$ matrix. 
Suppose $AB = I_{m}$, that is the $m\times m$ identity matrix. 
Consider the linear transformation $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ and $S: \mathbb{R}^m\rightarrow \mathbb{R}^n$ respectively defined by $[T] = A$ and $[S]= B$.
(a) Show that $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is onto.
(b) Show that $S:\mathbb{R}^m\rightarrow\mathbb{R}^n$ is one-to-one.
 A: The fact that you are dealing with matrices can be ignored. This is one of those cases where it may be easier to just use what you know about linear transformations.
Hints: 
$1).\ $ Show that in general, if $S:X\to Y$ is a linear transformation of vector spaces, then $S$ is injective if and only if $\text{ker}\ S=\{0\}.$ Now use this in the context of your problem to show that $S$ is injective. 
$2).\ $ In the context of your problem, suppose $T$ is not onto. Then, there is a $y\in \mathbb R^m$ such that no $x\in \mathbb R^n$ satisfies $Tx=y.$ On the other hand, by hypothesis, $TSy=y.$ 
A: Welcome to MSE. Your problem is all related to solve linear systems. Note that, if $S$ is one-to-one, then $B X = 0$ has a unique solution $X = 0$, to this happens you may solve this system, for which values of n, m this system has a unique solution?  And if $T$ is onto, given $Y \in \mathbb{R}^{m}$, you have to find solutions for a linear system like $AX = Y$ with $X \in \mathbb{R}^{n}$. Again, to which values of $n, m$ you can solve a system like this? Do you know how to do this calculation?
