Characterization of injective linear transformations Seeing other questions here, I realized this is a known fact:

A linear transformation $T:\mathbb R^n\to \mathbb R^m$ is injective
  iff there is $\alpha\gt 0$ such that $\|T(x)\|\ge \alpha\|T(x)\|$ for
  all $x$.

I'm trying to prove it, I don't know if this can help but what I know is there is a $\beta\gt 0$ such that $\|T(x)\|\le \beta\|T(x)\|$ for all $x$, for every linear transformation $T$.
How can we prove this? why this is never taught in regular linear algebra classes?
There is a similar characterization to surjective linear transformations?
Thanks in advance
 A: Consider $T$ restriced to the unit sphere $S$ in $\Bbb R^n$.  Since $T$ is linear and acts on the finite dimensional space $\Bbb R^n$, $T$ is continuous, so $T$ restricted to $S$ is a continuous function on $S$.  $S$ being compact, $T$ attains its minimum value at some $z \in S$.  $Tz \ne 0$ since $T$ is injective, so $\Vert Tz \Vert = \alpha > 0$.  It follows that for any $y \in S$, $\Vert Ty \Vert \ge \alpha$.  Now use linearity:  if $0 \ne x \in \Bbb R^n$, then $x / \Vert x \Vert \in S$, so $\Vert T(x/ \Vert x \Vert) \Vert \ge \alpha$, whence $\Vert T(x) \Vert \ge \alpha \Vert x \Vert$.  Going the other way is easier, since $\Vert T(x) \Vert \ge \alpha \Vert x \Vert$ implies $T(x) \ne 0$ for $x \ne 0$.  QED.
Note:  We tacitly used $\Vert T(x) \Vert \le \beta \Vert x \Vert$ in the (equivalent) assertion that $T$ is continuous.  The title factoid is in fact taught sometimes, somewhere.  A similar result for surjective $T$ is problematic, since we might lose $\alpha > 0$, which depends on injectivity.   A careful analysis might reveal a useful formulation, however.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
