is there a cheap way to estimate [math]\left \| A^{-1} \right \| = \max_{x \ne 0} \frac{ \left \| A^{-1} x \right \| }{ \left \| x \right \|}
\newline z= A^{-1} x \newline A z = x ?[/math]
>>8293936
[math]\left \| A^{-1} \right \| = \max_{x \ne 0} \frac{ \left \| A^{-1} x \right \| }{ \left \| x \right \|}[/math]
[math]z= A^{-1} x[/math]
A z = x [/math]
>>8293940
Dose this help?
https://en.wikipedia.org/wiki/Operator_norm#Equivalent_definitions
>>8293979
On [math]\mathbb{R}^{2}[/math] could be something like [math]||v||=1\iff v=e^{i\theta}[/math] so you have [math]||A^{-1}||=\sup_{\theta\in [0,\pi)}||A^{-1}(\theta)||[/math].
>>8293988
*[math]2\pi[/math].
>>8293936
Have you tried applying a Monte Carlo simulation?
>>8293936
useless /sci
http://www.cs.cornell.edu/~bindel/class/cs6210-f12/notes/lec11.pdf
Hager’s algorithm is the way
>>8293936
trial and error
>>8293940
It's just the largest eignvalue of A^-1 so the smallest eignvalue of A, no
>>8294463
\left \| . \right \|_1 is an operator norm
>>8294476
[math]\left \| . \right \|_1[/math]