Given an invertible, square matrix $A$, the induced 2-norm $$||A|| := \sup_{x \neq 0} \frac{||Ax||}{||x||}$$ can be calculated easily by finding the largest singular value of $A$, assuming we're using the $L^2$ norm on vectors above.
But what about $$||A||_\delta := \sup_{x : ||x|| > \delta} \frac{||Ax||}{||x||}$$ for some positive $\delta$? How can we calculate this? Can we say something about where the supremum is achieved?
If it makes the problem easier, I'm only interested in computing this quantity for 3x3 matrices $A$ (but would appreciate a solution for general square matrices).
Since the fraction is invariant to scaling $x$, perhaps these two "norms" are equal?
It is clear that $\left\|A\right\|_{\delta} \le \left\|A\right\|$, now let $x$ such that $$\frac{\left\|Ax\right\|}{\left\|x\right\|} \ge \left\|A\right\| - \epsilon$$ and $u = 2\delta \frac{x}{\left\|x\right\|}$, $\left\|u\right\| = 2\delta > \delta$ and $$\left\|A\right\|_{\delta} \ge \frac{\left\|Au\right\|}{\left\|u\right\|} = \frac{\left\|Ax\right\|}{\left\|x\right\|} \ge \left\|A\right\| -\epsilon$$