i don't know what you mean by efficient, but surely at least you could look at the set of lines determined by any two points and see if you can find any linear dependence
Hello guys! What's one way to go about proving that left and right eigenvectors need not be the same. I worked out the proof to show that left and right eigenvalues are the same, but I'm stuck on the eigenvectors.
Hey guys, quick question: if you have two sets A and B with some elements in common, what does the notation A\B mean? It appeared on an answer to one of my main page questions, but I'm slightly confused.
i have a stupid question: why in the definition of schwarz functions and other cases where you want to discuss some uniform bound of a function does one encouter $C(1+|x|)^a$ rather than just $C(|x|)^a$?
user19161
@Jordan Very good. Einstein did not bathe for days when working on relativity!
The smaller the userbase, the higher the quality of submissions for serious subreddits, it seems. I'm virtually unsubscribed from all front page subreddits.
askreddit use to be interesting when I first started going, but it turned into the same questions every day and the same stupid puns and memes at the top
I will be afk, but would appreciate if someone could answer my above question: essentially, why does one often encouter $C(1+x)^n$ rather than $C(x)^n$ when one talks about bounding functions. particularly with respect to defining Schwarz functions
I teld to appreciate more complaints about concrete situations rather than generic, close-to-empty declamations :/ If there is a problem, it should be address. But one can not address generic problems
@BillDubuque Please @Billl, make such complaints on a meta post, preferably with links to concrete problems. Such general statements as this, as you probably understand, will not help the situation which you perceive. Morevoer, someone with your long online experiece surely appreciates that phrasings like «pack» and «attack new users for sins they don't even know about», while colorful, rarely make communication or solution of problems any easier.
mariano, you are my favorite user on the site. you aren't a one-dimensional guy. i like your lack of nonsense and admire your attitude toward doing mathematics.
don't mean to make you blush or anything. just saying a lot of people wouldn't respond to Bill like that and would assume he had some deep point
apparently 6 people did
you were saying the other day that you think math is social. i found that comment interesting
but i'm not sure i entirely agree. at least i think it must be more than that. the trends and fashions should be informed by something deeper than "that's just what we're all working on at the moment"
i think a better reply to (i forgot who you were talking to, Alex?) would be to say that if point set topology were elevated to the highest place in mathematics the culture would be much more narrow and less capable of deep connections
i speak as an amateur. but this is how i understand it
again, you were kind of saying it. i just missed the conversation and wanted to add my two cents late
@MarianoSuárezAlvarez do you know what folland means when he says "let $\delta=\rho(K,U^c)$ (the distance from $K$ to $U^c$, which is positive since $K$ is compact"?
@AlexanderAmenta you might want to encourage them in the comments to supply this
@EricGregor ok, i answered a question just before (math.stackexchange.com/questions/141570/…) and did just that, but i'm concerned that i might have given too much detail
that is a complicated way of saying: the sequence of the $k_n$ is in $K$, which is compact, so by replacing the whole thing by subsequences, we can assume it converges
i don't know what you mean by "that is a complicated way of saying: the sequence of the kn is in K, which is compact, so by replacing the whole thing by subsequences, we can assume it converges"
you start with sequences $(k_n)$ and $(x_n)$ in $K$ and $U^c$, resp, such that $d(k_n,x_n)\to 0$. The first one is in $K$, which is compact, so it has a convergent subsequence $(k_{n_i})$. Call $k=\lim k_{n_i}$. Prove that $x_{n_i}\to k$ as $i\to\infty$. Since the $x$'s are in $U^c$, which is closed, this implies that $k$ is in $U^c$, which is impossible.
@MarianoSuárezAlvarez Can you help me with a question about Cyclotomic fields? I'm trying to determine the automorphisms that are in the quotient $$\operatorname{Gal}(\Bbb{Q}(\zeta_{13})/\Bbb{Q})/\operatorname{Gal}(\Bbb{Q}( \zeta_{5 } )/\Bbb{Q})$$
Here's my confusion: I form that quotient and I know I should be left with the order 3 automorphism. But $\operatorname{Gal}(\Bbb{Q}(\zeta_{13})/\Bbb{Q})$ is generated by $\sigma_{2}(\zeta)=\zeta^{2}$.
I can't seem to understand how I'm left with just $\sigma_{2}^{3}$.
OK. well I know what all the primitive generators are for all the subfields of $\Bbb{Q}(\zeta_{13})$ and I am trying to mod out so that I have a subfield that is isomorphic to $(\Bbb{Z}/3\Bbb{Z})^{\times}$. But the Galois groups of $\Bbb{Q}(\zeta_{13})$ is generated by $\sigma_{2}$ while the Galois group of $\Bbb{Q}(\zeta_{5})$ is cyclic with galios group $\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\}$.
Well $\Bbb{Z}/4\Bbb{Z}<\Bbb{Z}/12\Bbb{Z}$ and $\Bbb{Z}/4\Bbb{Z}\cong(\Bbb{Z}/5\Bbb{Z})^{\times}\cong\operatorname{Gal}(\Bbb{Q}( \zeta_{ 5 })/\Bbb{Q})$.
So $\Bbb{Z}/4\Bbb{Z}$ is a subgroup of $\Bbb{Z}/12\Bbb{Z}$ and $\Bbb{Z}/12\Bbb{Z}\cong(\Bbb{Z}/13\Bbb{Z})^{\times}\cong\operatorname{Gal}(\Bbb{Q}(\zeta_{13})/\Bbb{Q})$ and $\Bbb{Z}/4\Bbb{Z}\cong(\Bbb{Z}/5\Bbb{Z})^{\times}\cong\operatorname({Gal}\Bbb{Q}(\zeta_{5})/\Bbb{Q})$ But $$\operatorname({Gal}\Bbb{Q}(\zeta_{5})/\Bbb{Q})\not\subset\operatorname{Gal} (\Bbb{ Q }(\zeta_{13})/\Bbb{Q})$$?
Sorry to interrupt. I've tried multiple times to think of a particular matrix whose left and right eigenvectors are different, but I just keep failing. Anybody want to help me out?
It is. Q[x] is isomorphic to Q[x]^2--so you could think about Q[x] as being EQUAL to Q[x]^2--but it's also equal to Q[x] times {0} which lives inside Q[x]^2. In other words, there are a lot of ways in which Q[x] embeds into Q[x]^2. So, it's bad form to say Q[x] sits inside Q[x]^2 unless it's obvious in which way you mean. In your case it is obvious ONLY because there is a unique embedding Z_4 -->Z_12
@AlexYoucis Ok so, there is a subgroup $\Bbb{Z}/4\Bbb{Z}$ in $\Bbb{Z}/12\Bbb{Z}$ and $\Bbb{Z}/12\Bbb{Z}\cong(\Bbb{Z}/13\Bbb{Z})^{\times}\cong\text{Gal}(\Bbb{Q}(\zeta_{13})/\Bbb{Q})$. So I'm trying to find the subfield of $\Bbb{Q}(\zeta_{13})$ that corresponds to the quotient $\Bbb{Z}/3\Bbb{Z}$ in $\Bbb{Z}/12\Bbb{Z}$
@EricGregor how can I calculate the left eigenvectors using a calculator if the standard is using right. (I've chosen a non-zero determinant, non-symmetric 3x3 matrix)
@anilorap You clearly have that any group G of order 6 is a semi direct product Z_3\rtimes Z_2..there are clearly on two maps Z_2-->Aut(Z_3)=Z_2--thus only two distinct isomorphism classes (they are distinct since one is abelian and the other not)
Sure, which corresponds to the trivial map Z_2-->Aut(Z_3)
@tb What suggestion were you referring to when you said "I'd appreciate it if we established as a rule that only people should give book recommendations who actually answer questions on the topic. </rant> "
@tb He's taking the limit of the amount that remains after $n$ stages. He's right: you need to decrease the fraction taken at each stage. If you take a constant fraction, you end up with a null set.
