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Nov 4, 2020 at 17:18 vote accept FeedbackLooper
Nov 4, 2020 at 16:59 comment added RaphaelB4 For the $x^TA_0 x$ I also thing the answer is yes almost surely. I can share another idea but it is far from being complete : If you can also add a another random modification $P_i \rightarrow UP_iU^T$ with $U=I+\epsilon''$ an small perturbation, (the same $U$ for all $P_i$). Then this should be equivalent to keep the $P_i$ and change $A_0 \rightarrow U^T A_0 U$. With this perturbation hopefully $x^*$ is random with a continuous law on $E$ and then is regular almost surely.
Nov 4, 2020 at 15:05 comment added FeedbackLooper Thanks @RaphaelB4, this answer helped me a lot. I awarded the bounty to you. However, I would appreciate if you could also address the comments I made here. Thanks anyway!
Nov 4, 2020 at 15:04 history bounty ended FeedbackLooper
Nov 1, 2020 at 10:53 comment added FeedbackLooper Then, if $L_\tilde{\epsilon}$ contains regular points (with those vectors being linearly independent), thus the conclusion of $L_\tilde{\epsilon}$ being of Lebesgue measure 0, shouldn't correspond to the set of regular points on $E(\tilde{\epsilon})$ being of measure 0. Contrary to what is stated at the end of your answer?
Nov 1, 2020 at 9:04 comment added RaphaelB4 Indeed it should be $L_\tilde{\epsilon}$ and we use that for all $s$, $\{\epsilon:\{(P_i+\epsilon_i+s_i)x \}\text{ linearly independant}\}$ is lebesgue measure 0.
Oct 30, 2020 at 17:05 comment added FeedbackLooper Sorry for so much questions, but once we change the matrices to be $\tilde{\epsilon_i}=\epsilon_i+s_iI_n$, shouldn't we focus on the linear dependency of the vectors $\{(P_i+\tilde{\epsilon}_i)x\}_{i=1}^m$ instead of $\{(P_i+\epsilon_i)x\}_{i=1}^m$ so that the set $L_\epsilon(x)$ is not truly decoupled from $s_i$? Or am I missing something?
Oct 30, 2020 at 16:16 comment added FeedbackLooper Another comment, is it that $L_\epsilon(x)$ contain the indepedent vectors? did you mean dependent?
Oct 30, 2020 at 16:13 comment added FeedbackLooper This is awesome! This already gives me a lot of useful info. Just to make sure I understand: we can't rule out non regular points completely right? As a side question (your answer is already great): Do you think we can conclude something about the probably of this "optimal point" $x^*$ I mentioned in my question? Since it is optimal in the sense of a cost $x^TA_0x$, one can show similarly (I guess) that the set of "optimal" points lie in a set of measure zero with probably one. Can we show points $x^*$ are regular almost surely? (since we couldn't rule out nonregular ones completely)
Oct 30, 2020 at 15:48 history answered RaphaelB4 CC BY-SA 4.0