If a linear map $A$ is injective, then there exists $c$ such that $|Ax|\geq c|x|\;\;\forall x$ If a linear map $A:\mathbb{R}^m\rightarrow \mathbb{R}^n$ is injective, then there exists $c>0$ such that $|Ax|\geq c|x|$ for all $x\in\mathbb{R}^m$
Could someone give any solution or hint?
Thanks.
 A: Hint: Consider
$$
\alpha:=\inf_{\|x\|=1}\|Ax\|.
$$
This must exist and be non-negative, since $\|Ax\|\geq 0$ for all $x\in\mathbb{R}^m$. If $\alpha>0$, then you can use linearity to prove the desired result.  What happens if $\alpha=0$?  
A: First note that if such an inequality holds on the unit sphere in the Euclidean norm then it is true for all $x\in\mathbb{R}^{n}$. This is because if we have
$\vert\vert Ax\vert\vert\ge C\vert\vert x\vert\vert=C$ for some constant $C>0$ on the unit sphere then we simply note that for any $x\in\mathbb{R}^{n}-\{0\}$ we have that $\frac{x}{\vert\vert x\vert\vert}$ has unit norm. So $\vert\vert A(\frac{x}{\vert\vert x\vert\vert})\vert\vert\ge \vert\vert \frac{x}{\vert\vert x\vert\vert}\vert$. Multiplying by $\vert\vert x\vert\vert$ which is simply a positive constant gives the inequality we desire.The inequallity holds trivially for $x=0$ so it holds for all of $\mathbb{R}^{n}$.
Note that $A$ is injective implies that $Ax=0$ only if $x=0$. Now we simply look at the function $f:S^{n-1}\to\mathbb{R}$ defined as $f(x)=\frac{\vert\vert Ax\vert\vert}{\vert\vert x\vert\vert}=\vert\vert Ax\vert\vert$. This is a continuous function on the compact surface $S^{n-1}$ and so attains a minimum on the unit sphere. Since A is injecive the lower bound is some $C>0$. This gives the bound we desire.
A: Since $\ker A = \ker (A^T A)$, we see that $A$ is injective iff $A A^T$ is invertible.
It is easy to see that $A^T A$ is symmetric and positive semi-definite. Since $A^T A$ is invertible, we see that $A^T A$ is positive definite, hence $A^T A \ge \lambda I$, where $\lambda$ is the smallest eigenvalue of $A^TA$ (and $\lambda >0$).
Consequently we have $\|Ax\|^2 = \langle A x, Ax \rangle = \langle x, A^T Ax \rangle \ge \lambda \langle x, x \rangle = \lambda \|x\|^2$. Setting $c = \sqrt{\lambda}$ gives the desired result.
