bio | website | |
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location | University of Sydney | |
age | ||
visits | member for | 4 years, 5 months |
seen | Mar 29 at 8:45 | |
stats | profile views | 1,038 |
I'm a theoretical physicist.
Apr 7 |
awarded | Popular Question |
Oct 26 |
awarded | Yearling |
Aug 20 |
answered | What are some deep theorems, and why are they considered deep? |
Jan 8 |
awarded | Popular Question |
Oct 26 |
awarded | Yearling |
Sep 17 |
awarded | Notable Question |
Jun 6 |
revised |
Eigenvalues of a sum of Hermitian positive definite circulant matrix and a positive diagonal matrix
clarification. |
Jun 6 |
answered | Eigenvalues of a sum of Hermitian positive definite circulant matrix and a positive diagonal matrix |
May 28 |
awarded | Notable Question |
May 27 |
accepted | Generating a group by randomly sampling generators |
May 26 |
answered | Generating a group by randomly sampling generators |
May 24 |
comment |
Generating a group by randomly sampling generators
A similar linear upper bound is $2^n k$, which is tighter for small $k$. These are all fantastic, but do you think it might be possible to get a strictly concave lower bound? The linear ones are not quite strong enough for my purposes, unfortunately. |
May 24 |
comment |
Generating a group by randomly sampling generators
Yes, of course. (I read "with" instead of "without".) In fact, $k \le 3^n$, the number of full elements in $G^n$. Then $N_{3^n} = 4^n \gg 3^n$. In that case, it seems the bound is rather loose, no? Plus one regardless! |
May 24 |
comment |
Generating a group by randomly sampling generators
I lost you after the first line... $N_k \le |G|^n = 4^n$ for all k, so the bound your wrote can't always hold. |
May 23 |
comment |
Generating a group by randomly sampling generators
@Gerhard, yes, you're right. I meant simple, sorry. Fixed. |
May 23 |
revised |
Generating a group by randomly sampling generators
fixed a typo |
May 23 |
asked | Generating a group by randomly sampling generators |
Oct 27 |
awarded | Yearling |
Aug 22 |
awarded | Self-Learner |
May 30 |
comment |
Generalization of scalar product to k-dimensional subspaces (as opposed to 1-dimensional subspaces, i.e. vectors)
Actually, this reduces to the squared inner product when $k=1$. So perhaps this isn't what you want. |