Timeline for Bound on the trace of inverse matrix
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2023 at 17:04 | comment | added | Iosif Pinelis | @Amin : Please finalize the current question (fully answered) before asking new ones. | |
| Feb 1, 2023 at 2:28 | comment | added | Amin | I generate samples and this behavior is observed | |
| Feb 1, 2023 at 2:27 | comment | added | Amin | Based on the search, I found that $tr(A^{-1})tr(A)\le n^2 \kappa$ where $\kappa$ is the condition number of matrix $A$. In the following post, I defined matrix $V$ and I wanna find an upper bound on trace of inverse of $A$ that depends on $t$. My guess is that this upper bound is a decreasing function of $t$ (I guess that $t$ should appear in the denominator, not in the nominator). I should mention that vectors $\mathbf{x}_s$ are not chosen by adversaries. math.stackexchange.com/questions/4626468/… | |
| Jan 31, 2023 at 21:57 | comment | added | Iosif Pinelis | @Amin : Do you have a further response to my answer and comment? | |
| Jan 29, 2023 at 1:53 | comment | added | Iosif Pinelis | Any lower bound on the eigenvalues $t_j$ of $A$ must be $\le L/n$ for $tr(A)$ to be $\le L$. So, the values $t_j=L/n$ for all $j$ will be allowed, and these values provide for the exact lower bound $n^2/L$, given in the answer. So, the answer to your comment is: No, of course not -- any admissible lower bound on the eigenvalues will not have any impact. | |
| Jan 27, 2023 at 23:40 | comment | added | Amin | Thanks for your response. I think this answer does not assume any lower bound on the eigenvalues of $A$. I am wondering if lower bound on eigenvalues has any impact? | |
| Jan 26, 2023 at 17:32 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 185 characters in body
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| Jan 26, 2023 at 17:26 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |