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I'm reading a proof that the map $A\mapsto A^{-1}$ is continuous in the operator norm. A part of the proof is that if $A,B$ are matrices such that $A$ is invertible and $\beta=\parallel B-A\parallel_{op},\alpha=1/\parallel A^{-1}\parallel_{op}, \beta<\alpha$ then $B$ is invertible and $\parallel B^{-1}\parallel_{op}\leq\frac{1}{\alpha-\beta}$. I'm having some hard time with the final conclusion of that part which is: $$|B^{-1}x|\leq\frac{1}{\alpha-\beta}|x|\Rightarrow\parallel B^{-1}\parallel_{op}\leq\frac{1}{\alpha-\beta}$$ The conclusion seems wrong to me since I can't prove that equality here in the general case: $\forall v\in\mathbb{R}^n, |Av|\leq\parallel{A}\parallel_{op}\cdot|v| $. Granted, I haven't given it much thought but intuitively it should hold for eigenvectors of the largest eigenvalue and there's no reason it should hold otherwise.

I guess my question would be, why is this result correct?

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What's your definition of $||\cdot||_{op}$? A common definition is that $||A||_{op} = \sup |Ax|/|x|$, the supremum taken over all nonzero vectors $x$. Or equivalently, $||A||_{op}$ is the smallest number $C$ such that $|Ax| \le C|x|$ for all $x$. From these definitions your conclusion is immediate. –  Nate Eldredge Oct 3 '11 at 12:34
Thanks! that was very helpful! –  Donjim Oct 3 '11 at 13:55
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