# Help with a proof that the inversion mapping for matrices is continuous

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
@NateEldredge Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer May 6 '15 at 16:25

The usual definition of the operator norm is $\|A\|_{\mathrm{op}} = \sup_{x \ne 0} \frac{|Ax|}{|x|}$. If $|B^{-1} x| \le \frac{1}{\alpha - \beta} |x|$ then whenever $x \ne 0$, we can divide by $|x|$ to see $\frac{|B^{-1}x|}{|x|} \le \frac{1}{\alpha - \beta}$. Since this holds for all nonzero $x$, we must have $\|B^{-1}\|_{\mathrm{op}} \le \frac{1}{\alpha - \beta}$.