I've got stuck on a result I am not able to prove. I will extrapolate it from its context:
Let$\space$ $f$ be a linear injective map between $\mathbb{R}^n $ and $\mathbb{R}^m $ with $n \leq m$ and $g$ be a linear isomorphism in $\mathbb{R}^n $.
Let $ || \space ||$ be some consistent $m \times n$ matrix norm. Then exists $\epsilon >0$ such that for every $A$, $||A|| < \epsilon$ we have that:
$f+Ag$ is still injective.
Can someone help?
Thanks :)
Because $f$ is injective, it is bounded below. Indeed, as $f:\mathbb R^n\to \text{Im}\,f$ is bijective, it is invertible (we are in finite dimension, so every linear map is continuous). Or, since $f$ achieves its max and min in the unit ball, there exist constants $c,d$ with $c\|x\|\leq\|fx\|\leq d\|x\|$ for all $x$.
Now, if $\|A\|<c/\|g\|$ and $(f+Ag)x=0$, we have $$ 0=\|fx+Agx\|\geq \|fx\|-\|Agx\|\geq c\|x\|-\|A\|\,\|g\|\,\|x\| =(c-\|A\|\,\|g\|)\,\|x\|. $$ As $c-\|A\|\,\|g\|>0$, we get $x=0$, and $f+Ag$ is injective.
In the argument above I'm using for $A$ the operator norm induced by the norms in $\mathbb R^n$ and $\mathbb R^m$. But since in a finite-dimensional space all norms are equivalent, we can replace the norm by any other, at the cost of changing the $\epsilon$ a bit.