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May 23, 2014 at 20:50 comment added fedja I actually referred to "(or a small perturbation of)" :-) Sorry for the ambiguity.
May 23, 2014 at 19:09 history edited Joshuald1986 CC BY-SA 3.0
added 85 characters in body
May 23, 2014 at 13:35 comment added Joshuald1986 yes, that's what i want i mean. one can get some bounds in terms of matrix entries, but that makes sense only if you have large sizes and want to compute your bound way quicker than to just invert the matrix and look at the result
May 23, 2014 at 1:41 comment added fedja --- I m pretty sure also that the matrix A is a non-singular matrix.In your exemple, it's not the case. --- ??? Have you read the text in parentheses in my comment?.
May 22, 2014 at 18:37 comment added Joshuald1986 I m pretty sure also that the matrix $A$ is a non-singular matrix.In your exemple, it's not the case.
May 22, 2014 at 18:30 comment added Joshuald1986 Yes fedja, I want the bound in terms of $n$, the size of $A$ (something like $||A^{-1}||_{\infty} \leq 3n$ because i can have large matrix and calcute the inverse cost much.
May 22, 2014 at 18:15 comment added fedja So far your conditions do not prevent $A$ from being degenerate or as close to it as it wants: take (a small perturbation of) $\begin{bmatrix}1&-1&0\\-1&1&0\\0&0&1\end{bmatrix}$. So, can you state exactly what you are looking for? (I mean, one can get some bounds in terms of matrix entries, but that makes sense only if you have large sizes and want to compute your bound way quicker than to just invert the matrix and look at the result).
May 22, 2014 at 16:30 review First posts
May 22, 2014 at 16:30
S May 22, 2014 at 16:28 history suggested Jeremy Rouse
Removed the "number theory" tag, which is not relevant.
May 22, 2014 at 16:19 review Suggested edits
S May 22, 2014 at 16:28
May 22, 2014 at 16:11 history asked Joshuald1986 CC BY-SA 3.0