# Show that exists a constant $c>0$ such that: $\|f(x)\|\geq c\|x\| \forall x \in \mathbb{R^n}$

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!

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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,

$$\|f(x)\| = \|f(\|x\|y)\| = \|x\|\|f(y)\| \geq c\|x\|$$

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There are more ways to start.

1. Somebody mentioned in a comment that the unit sphere $B$ in $\mathbb R^n$ is compact, so is its $f$-image. It cannot contain $0$ because of linearity and injectivity, so $C:=\displaystyle\underset{x\in B}\min ||f(x)|| >0$ will be good.

2. As $f$ is linear, by fixing bases, it can be written in the form $x\mapsto A\cdot x$ for a matrix $A$. Since $f$ is also injective, it has a left inverse, a matrix $B$, its operator norm $||B||$ will be a good $C$.

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Note that $f$ is continuous. The unit sphere $S^{n-1}\subset\mathbb{R}^n$ is compact, hence its image $K\subset\mathbb{R}^m$. Since $f$ is injective, $0\not\in K$. Define $c$ to be the distance between $0$ and $K$. Compactness of $K$ implies $c>0$. The conclusion then follows from linearity.

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