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If I made no mistake, one can calculate the operator norm of the inverse of any given (invertible) operator $A: V\rightarrow V$ via:

\begin{align}\|A^{-1}\| & = \sup\left\{\frac{\|A^{-1}b\|}{\|b\|} : b\neq 0\right\} \\ & \qquad \left\downarrow A\text{ is a bijection }V\setminus\{0\}\rightarrow V\setminus\{0\}\right. \\ & = \sup\left\{\frac{\|A^{-1}Ab\|}{\|Ab\|} : b\neq 0\right\} \\ & = \sup\left\{\frac{\|b\|}{\|Ab\|} : b\neq 0\right\} \\ & = \frac{1}{\inf\left\{\frac{\|Ab\|}{\|b\|} : b\neq 0\right\}}\end{align}

Is there a name for the expression $\inf\left\{\frac{\|Ab\|}{\|b\|} : b\neq 0\right\}$ which is similar to the operator norm but with replacing $\sup$ with $\inf$?

Note: Above I assumed $\inf\left\{\frac{\|Ab\|}{\|b\|} : b\neq 0\right\} > 0$ which is always the case if $V$ is finite dimensional.

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    $\begingroup$ I'm not sure, but maybe you could start by playing with vector $p$-norms with negative $p$ values? So $||x||_{-p}=((\sum_i |1/x_i|^p)^{1/p})^{-1}$... $\endgroup$ – Tyler Streeter Oct 19 '17 at 14:48
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The short answer is yes. This is the smallest singular value, as a direct consequence of the min-max theorem for singular values.

For extra information you may want to consult: https://en.wikipedia.org/wiki/Singular_value (section basic properties), or for example Numerical Linear Algebra by Trefethen and Bau, see pages 94-95, if you prefer a real book.

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