TITLE: Show that exists a constant $c>0$ such that: $\|f(x)\|\geq c\|x\| \forall x \in \mathbb{R^n}$
QUESTION [5 upvotes]: So , this is the exercise: 
Let $f$ be a linear transformation and injective in $\mathbb {R^n}\rightarrow\mathbb{R^m}$. For abuse , let's denote by $\|.\|$ the norm in both sides.Show that exists a constant $c>0$ such that:
$$\|f(x)\|\geq c\|x\| \forall x \in \mathbb{R^n}$$
Someone explainded to me how to do this one, but honestly I didn't understand a thing...If anyone can help me how to get starded , Much appreciated!

REPLY [3 votes]: Let $D \subset \mathbb{R}^n$ be the unit sphere. Since $f$ is injective, $f(x) \neq 0$ for $x \neq 0$. In particular, $\|f(x)\| \neq 0$ for $x \in D$. But since $D$ is compact and $\|f(\cdot)\|$ is continuous and positive, $\|f(\cdot)\|$ attains a positive minimum $c$ on $D$.
Now, for every $x \in \mathbb{R}^n$, there exists a $y$ with $\|y\| = 1$ (i.e. $y \in D$) such that $x = \|x\|y$. In fact, $y = \frac{1}{\|x\|}x$. Hence,
$$\|f(x)\| = \|f(\|x\|y)\| = \|x\|\|f(y)\| \geq c\|x\|$$