# Prove a matrix norm inequality

Given an induced matrix norm $$||\cdot||$$ such that $$\exists_{\varepsilon >0} \forall_{x\in\mathbb{R} ^n} ||Ax||\ge\varepsilon ||x||$$ prove that $$||A^{-1}||\leq\frac1\varepsilon$$.

I figured out that $$||A||\ge \varepsilon$$ and that $$\varepsilon \varepsilon^{-1}=1 =||AA^{-1}||\le||A||||A^{-1}||$$, but however I manipulate those terms, I never can get the inequality right.

Abridged solution. Set $$y = A(x)$$ so that $$x = A^{-1}(y)$$ and so the hypothesis $$\|A(x)\|\geq \varepsilon \|x\|$$ is the desired conclusion. QED
• Right, I managed to figure it out just before checking if someone already responded. You were faster though, so the points are definitely yours! I would also add that one needs to take advantage of the fact that $||Ax||\le||A||||x||$ for appropriate matrix and vector to arrive at a fully formal proof. – Joald Nov 8 '18 at 23:26