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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.

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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

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  • $\begingroup$ 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. $\endgroup$ – Joald Nov 8 '18 at 23:26

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