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I have the following problem. If $$\|Ax\| \geq \theta \|x\|$$ for a square matrix A, $\theta$ a positive real number, $x$ a vector and a natural norm $\| \cdot \|$. Prove that $A$ is invertible and $$\|A^{-1}\| \leq \frac{1}{\theta}$$. My attempt, let $x=A^{-1}y$ so $$\|y\| \geq \theta \|A^{-1} y \|$$ and $$\|A^{-1} \| = \sup_{\|y\|=1} \| A^{-1} y \| 1/ \theta \leq \sup_{\|y \|=1} 1/ \theta \|y\|=1/ \theta$$. Is this proof OK? Also I can't seem to figure out how to prove that $A$ is invertible in the first place, how should I approach this?

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  • $\begingroup$ You need to assume that $\theta>0$. Is the matrix square. Can you show that $A$ is injective? $\endgroup$ – copper.hat Dec 2 '17 at 22:03
  • $\begingroup$ The matrix is square $\endgroup$ – Dimtsol Dec 2 '17 at 22:05
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    $\begingroup$ Your attempt is wrong since you write $x=A^{-1}y$ before proving that $A$ is invertible. $\endgroup$ – José Carlos Santos Dec 2 '17 at 22:05
  • $\begingroup$ So am I, but am I injective? $\endgroup$ – copper.hat Dec 2 '17 at 22:05
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Your result is true if $ A \in M(n\times n) $. This way, $ |A(x)| \ge \theta |x| $ implies that $ A $ in injective. By the rank-nullity theorem the dimension of image is $ n $ so the $ A $ is also surjective. Also, $ |A(x)| \ge \theta |x| $ implies $ \|A\| \ge \theta $. Finally, as $ A\cdot A^- = I $ we have $$ \|A\|\|A^-\| =1. $$ Then, $ \|A^{-1}\| \le 1/\theta. $

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  • $\begingroup$ Thanks for the answer,can you show why $|Ax| \geq \theta |x|$ implies $A$ being injective? $\endgroup$ – Dimtsol Dec 2 '17 at 22:38
  • $\begingroup$ Suppose $Ax=Ay$. Then, $0=|Ax-Ay|=|A(x-y)|\ge \theta |x-y|\ge 0$. So we must have $|x-y|=0$. This means $x=y$. $\endgroup$ – shall.i.am Dec 3 '17 at 7:45

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