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5:04 PM
I think that too much precision might obscure the general idea for people just trying to grasp this for the first time. I think that many-to-one gets the idea across as does many-to-one.
in my own early education there was far too much emphasis on relations that weren't functions. i blame underfunded public schools with a stockpile of textbooks from the 1960s.
Yeah, they wanted to differentiate the two and overdid it. functions are not so hard to grasp.
the beginning of the textbook would have one or two pages with the list of students names who had used it. sometimes you'd find your teacher on there.
@leslietownes well, you don't have to worry if they know the book...
i was randomly issued the same algebra 2 book that my sister had used. i had better luck with it.
my sister almost didn't graduate high school. she couldn't pass geometry. her teacher just made up a C so she could graduate. i think about that sometimes.
5:12 PM
@MoneySetsYouFree Are these called mapping diagrams or not?
that's a good name for them. i'd just worry that if a student were told a name they'd google it and find something different and maybe adopt the wrong guidance. better to just specify the form.
That is what we are told to, so I asked if it really was, but it seems that it shouldn't be
A: Show the ring is isomorphic to $k[x]$

Andreas CarantiHere's the part about the kernel. First recall that if $A$ is a domain, $A[y]$ is the polynomial ring in the indeterminate $y$, and $f \in A[y]$ is a monic polynomial, then you can do Euclidean division by $f$ in $A[y]$. Apply this to $A = k[x]$, and $f = y - x^{2}$. If $g \in \ker(\phi)$, divide...

Could someone explain this answer? He did Euclidean division with two variable but I've only done in one variable polynomial
5:29 PM
He's applying the division algorithm in only one variable, $y$, treating elements of $k[x]$ as constants. What in particular do you not understand?
@MoneySetsYouFree You could call this a "mapping diagram."
I just understood what he said I just confused
Well, I cannot read your mind.
@leslietownes Oy.
did i say something stupid again?
i mean, i know the answer is yes, but particularly stupid?
@barista division algorithm works in general if the divisor is a monic polynomial in any ring
y-x^2 is a monic polynomial in y with coefficients in k[x]
@AshishAhuja Perhaps someone else said this. I understand the transpose to be defined by the fundamental formula $$\langle Ax,y\rangle = \langle x,A^\top y\rangle,$$ where the inner products may be in different spaces. I.e., if $A$ is $m\times n$, the first one is in $\Bbb R^m$ and the second one is in $\Bbb R^n$. Using this, you can get that formula as follows:
$$\langle (AB)x,y\rangle = \langle A(Bx),y\rangle = \langle Bx,A^\top y\rangle = \langle x,B^\top(A^\top y)\rangle = \langle x,(B^\top A^\top)y\rangle.$$
Did you look to see where the arrow led you, @Leslie?
5:37 PM
no, as a rule i do not.
Well, let that be a lesson to you.
hi chat
Hi @Astyx!
@TedShifrin wait what is happening
my grandfather's last words, just before they sprung the trap, were "you can’t cheat an honest man. never give a sucker an even break or smarten up a chump"
5:39 PM
or maybe not. I probably should ignore that
@soupless Are you talking about anything?
Nope. I asked and now I am not
OK. Just as I suspected.
@TedShifrin no. It is not called as a mapping diagram because not every relation is a mapping.
OK, just a suggestion.
5:50 PM
Arror diagram maybe
i do not have strong feelings in this area, it all seems slightly arbitrary. consensus appears to be that there's no 'standard' name.
6:16 PM
Are we arguing? I want in on all the fun.
Oh those diagrams.
Most common name I have seen used are "bubble diagrams".
Which only makes sense if the context of "relations" is sufficiently clear.
Hi @TedShifrin.
I ain't doin' no arguing.
Hi, @feyn
Hello everyone. I would like to learn some applied category theory and I've found two books: An Invitation to Applied Category Theory by David Spivak and Brendan Fong, and Category Theory for Scientists by David Spivak
Anyone know either of these?
6:31 PM
I had a question about grad school admission.
The answer may depend hugely on in what country and what school, but go ahead.
@TedShifrin I was just thinking about this. To me, the transpose is defined by (A^T)_{ij} = A_{ji}, so your formula v * Aw = A^T v * w requires proof. The most straightforward proof I know starting from that definition is to prove the formula (AB)^T = B^T A^T --- which comes down to symmetry of the dot product, as you can compute the ij'th entry to be (i'th column of B) * (j'th row of A) on one side and (j'th row of A) * (i'th column of B) on the other. Then your formula follows.
Oh, hey, stranger.
Let me try to understand your approach.
I got an offer from a Canadian school in early March. They only gave me two weeks to decide. (Most Canadian school don't adhere to the 15th April agreement like the US schools). Since I didn't have any other offer at that time (except a bunch of waitlists), it felt like the best bet was to take their offer (considering that this year has been unusual with many schools cutting down their intake). But last week I received an offer from a US school, which I really like.
6:37 PM
Yes, of course, but for conceptual reasons, what I said is the right thing (it's the most useful property and generalizes, of course, to functional analysis).
How bad would it be to turn back on this Canadian school?
OK, I agree.
I presented it the way I outlined above but emphasized that the point is that formula, use that formula repeatedly, and asked them to prove that formula characterizes the transpose on a HW.
But your row/column argument is what I have in the linear algebra book, I'm pretty sure, @MikeM. I then deduce Ted's favorite formula (that formula) from the observation that $x\cdot y = x^\top y$.
Or rather asked them to show that Av * w = v * Aw implies A is symmetric, which amounts to the same observation.
That's what I did too yeah
@feynhat People do less-than-ethical things all the time. I have known young mathematicians who delayed a postdoc for a year, promising to go for 2-3 years a year later, and then reneged completely when something better came along. I personally think that showed a bad character defect. But people do stuff to make things better for themselves. I'm sure you would not be the first.
And, honestly, the Canadian school is giving you only two weeks to force you into a decision that may not be optimal for you.
6:41 PM
But let's take your approach seriously. First, one needs to justify that for fixed v, v * Aw is a linear function of w. This is clear enough from linearity of A and bilinearity of the dot product. Then you need to justify that every linear function of w is given by dot with a unique vector u. Straightforward. Then define A^T v by your formula above: it is the unique vector so that v * Aw = A^T v * w. OK. T
Then linearity follows by manipulating the LHS, and then the formula follows by considering v = e_i, w = e_j. Great.
This probably takes 10 minutes longer than I'd like it to and IMO at most three people would like a definition like this in this class, but I do like it as a mathematician.
I'm glad the proof you gave of (AB)^T = B^T A^T is not circular, since I like that proof.
My point is that throughout my linear algebra course the dot product formula gets used dozens of times. Of course, we want the students to be able to transpose a numerical matrix.
But it's in keeping with my emphasis on geometry to make that formula important :P
I did try to convince them to extend the deadline. But the person said that he really needs a grad student this fall and if he extended the deadline to say 15th April and I turn down their offer then he wouldn't be able to find a student that late.
@feynhat Oh, there's a particular person you're committing to work with? That's highly unusual in mathematics (quite usual in lab sciences).
ted is speaking my language.
Well, it affects your life more than it ultimately affects his, so I say you should go with what you prefer for your professional development. Be warned that this guy might badmouth you if he's an immature ass.
We should ask Leslie to pontificate on immature asses in academia.
6:46 PM
it's really weird, in the non-lab sciences, to have to deal with that. and i have no patience for it even when it is normal.
The Canadian program is 4 years long and yes I have committed to an advisor
In my experience (both with undergraduates and particularly with graduate students) such commitments are not a good idea. Often one's interests change, or one finds out the field is not a good fit. And even more often, the adviser turns out to be incompatible with you.
(which is why I am even more wary of turning back)
I lean 80% to changing to the American option if the school and program are in fact stronger.
Sure, I agree it is the point, just as if [x]_beta means the coordinate expression with respect to a basis beta, the point of the matrix A expressed w/r/t basis beta is A_beta [x]_beta = [Ax]_beta.
6:48 PM
think about yourself and take care of yourself.
Though this one is more definitional.
@feynhat does the advisor seem like someone you'd want as an advisor?
@MikeM: By calling it "Ted's favorite formula" in my multivariable math class, I insured that my students never forgot it. Of course the truth is that Stokes's Theorem is truly Ted's favorite "formula." :D
Sounds a little pushy and not human.
Not to mention self-serving.
Maybe my 80% leans to 90%.
6:51 PM
it's an exercise in rebranding. a generation of students will think of ted and not of the progenitors.
Ted is turning students against the system
"No more Euler!", they will shout, "down with Gauss!"
@Leslie: It's not like that adjoint formula has a classical name attached to it. :P
if i ever teach again i'll call it ted's favorite formula.
LOL ... I don't think in a million years with all my students piled together I've earned the points of Gauss.
Ted, on a scale of -10 to -100, how much do you enjoy computing integrals on Riemann surfaces?
6:52 PM
LOL, @Leslie; make sure to give me royalties. Or I'll sue.
@anakhro I mean, its very hard tell based on an interview and a couple of other zoom calls whether or not I would like him. But I have spoken some of the students at his school and they all have very positive things to say about him. (But they do tell me that he has very high expectations from his students).
I'm right now floating at a solid -75
@anakhro One hardly ever truly "computes."
we'll pass a collection plate and send it to you. it may be in coins.
Other than in complex analysis courses (applying the residue theorem on the R.S.).
@feynhat I think having high expectations is not a "but" but a strong plus.
I lower my 90% back to 80%.
6:53 PM
@feynhat I wonder if I know him.
@TedShifrin Yeah, i thought so too. That 'but' was sort of carried off from the tone of those students.
That said, some of my colleagues told their advisees not to take graduate courses from me because I expected too much work (like actually doing 10-15 homework problems over the course of the semester).
10-15 sounds like so very few, though
imagine that.
that's downright dickensian.
My courses have a problem set every week of 6 problems.
6:56 PM
that's what they had david copperfield doing until he broke out of that system.
i forget the name of the guy. the monster who ran the bottle factory. i guess i don't forget his name, it was ted shifrin.
In advanced courses, I didn't really expect them to do 6 problems every other week, so I made compromises. At UGA the custom was to give automatic A's in advanced courses so that 5 students would enroll and the course would run. I refused to do that. I said I'd give an automatic B if they attend every class, but that an A had to be earned.
smacks @Leslie
@anakhro UWO?
@feynhat I won't know him in that case.
OK, disappearing, now that I got no commitment to royalties.
I can only think of one person at UWO. Dan Christensen
6:59 PM
Yeah, he interviewed me too. Cool guy.
Oh, and Kapulkin.
Kapulkin is the chair of Graduate Program.
See how many people I will be upsetting.
I had a friend who did his M.Sc. there.
It's okay, UWO doesn't deserve a golden fellow such as yourself. Stick it to them.
lol no. I actually like these people.
I don't even know if I want the other offer. Who knows maybe if I had both the offers at the same time, I probably would've went for UWO.
But I didn't, and now I feel a bit cheated.
Hey @MikeMiller.
You live in NYC right?
Most people don't actually live in NYC, but are in a constant state of perpetual death.
7:13 PM
I do
Mike traded smokey California for smoggy NYC.
My sister will start her PhD at NYU this fall.
(well she was supposed to start last year but 'rona happened)
Is she mathy, too?
No. Psychology.
that didn't come out right.
It's okay, I never saw.
7:21 PM
I hope she enjoys her time here. That's a very different part of the city than I am in. Busier.
7:32 PM
is it possible for $\int_a^b f(x) dx$ to have a closed form but $\int_a^{b/1.001} f(x)dx$ to NOT have a closed form?
that's an interesting question. frequently when the bounds are infinity, that's needed to get anything 'closed' (not speaking formally). with finite bounds it seems similarly possible but i don't know an example.
it's just, certain definite integrals are nice numbers and others aren't. but i don't have an example. i'm thinking of one.
@leslietownes Just transport one of your examples
well there's stuff with e^(-x^2) for example, but in my example b is infinity so b/1.001 is also infinity.
$$\int_{b = -\pi/2}^{a = \pi/2} e^{-\tan^2 u} \sec^2 u du = \sqrt{\pi},$$ now good luck computing this for $a \in (0, \pi/2)$
that's why i said transport
you have identified a very good example.
i'm quite dumb. don't let my repartee fool you, i can't think myself out of a paper bag.
7:38 PM
If you live in a paper bag, it's not all quite so bad.
i live in a house, although the mortgage payments on a paper bag would be preferable.
Water damage leaves it quite soggy, so you'd rather live somewhere dry.
very interesting...
@geocalc33 how have you been? What zany stuff have you been up to lately?
7:54 PM
Thanks for the advice @Ted. I don't know why I was expecting that you would tell me to stick with my decision.
@anakhro Trying to do a complexification of $\Bbb R_{>0}$
also asking robjohn to make a weird torus
@feynhat it's a long time and a lot of money to blow on something which could have been resolved (on their end) if they were a tad kinder.
@geocalc33 what are your vector space operations? x+y := xy and cx = x^c?
8:11 PM
@anakhro yeah
apparently the elements in the complexification could be: $e^1 \otimes (a + bi)$ but I don't get it
Well you can look at it purely formally just as the tensor product. So the vectors will all look like (sums of) $x\otimes\lambda$ for $x>0$ and $\lambda\in\mathbb C$.
What is troubling about the complexification for you, @geocalc33?
I guess I understand how to plot $(a+bi)$ but I don't know how to understand $e^1 \otimes (a+bi)$
@anakhro could you explain it differently?
8:32 PM
@TedShifrin I was going to mention that we have the complete set of Father Ted DVDs
@MikeMillerEismeier It is not too hard with $\mathrm{erf}$
@geocalc33 Are you familiar with the tensor product?
(e.g. this one is of vector spaces, so it's extra nice)
I prefer calmer products
@anakhro it's sort of like the cartesian product
you multiply vectors and then put them into a big list
8:54 PM
Can a Monte Carlo simulation act as a proof? (for non-trivial questions)
@SAJW it can for me
But not for most people
I guess it also depends on the problem itself
"We can run a Monte Carlo simulation to prove it:"
I think the hooman just means, "to make sure I didn't make a dumb mistake somewhere"
@robjohn $\int_0^1 \exp\big(\frac{1}{\log x}\big) dx$ has a closed form I know. Do you know if $\int_0^a \exp\big(\frac{1}{\log x}\big) dx$ has a closed form for any $0<a<1$?
Ah ok, was just wondering^^
8:58 PM
The stuff at the start is the actual proof
Yeah because that I found it confusing
@geocalc33 At first glance, I'd say no, but I'd have to give it more than a glance.
9:18 PM
@geocalc33 The closed form for $[0,1]$ is in terms of Bessel functions. I don't think the indefinite integral has a closed form.
9:37 PM
Hello everyone. I have not been introduced to algebra yet, and I'm considering Aluffi's "Algebra Chapter 0", but I'm not sure if the cat theory approach will help me when I actually attend algebra classes. Any thoughts? This way I learn cat theory alongside algebra so it's a win-win
Hi, I'm a little rusty in arguing. Why is it clear that $e$ is finite in this top answer math.stackexchange.com/questions/364038/… ? Because the probability to get heads is not infinitely small (or 1 for the other).
I think you can prove it by showing that the probability you don't get it in $k$ tries is less than $\alpha^k$ for some $\alpha<1$. So the expected value is bounded by a geometric series.
I think the expected value is $1 + \sum\limits_{i=1}^n p(i)$ where $p(i)$ is the chance you don't get it in the first $i$ tries.
To be completely honest I don't think proving it is finite is that much easier than actually solving it.
Love those "it's obvious that..." sentences :D
i'm always sure, but i'm not always right
I think saying that that part is clear is necessary so that the proof is dope
and who doesn't want that
9:48 PM
To be honest the only part I get is how to solve for $e$ in that one.
yeah there is a lot of proofs for calculating expected values of stopping times
where you assume it exists and solve an equation

Any body familiar with Neyhman-Pearson Lemma?
I have a question about one of the conditions of the Lemma
@SAJW Andre gives an argument in the comments, but in my answer to that question, there is no such assumption.
Let $ \pi _0,\pi _1 $ be two populations with probability distributions say $ f_0,f_1 $ (with measure $ \mu $ ). Then for testing $ H_0:f=f_0 $ against $ H_1:f=f_1 $ , we can define a test $ \phi $ with a constant $ k $ such that the expectation of the test under the null hypothesis $ H_0:f=f_0 $ denoted as $ E_0 $ is,
$ E_0\left[ \phi \left( x \right) \right] =\alpha $ (level of significance) (‎9.5)

$ \phi \left( x \right) =\begin{cases} 1& when\ f_1\left( x \right) >kf_0\left( x \right)\\ 0& when\ f_1\left( x \right) <kf_0\left( x \right)\\ \end{cases} $ (‎9.6)
Yes, that is the solution my olympiad students came up with last week @robjohn
10:01 PM
Casae#1: when $ 0<\alpha <1 $ . Let us define a quantity $ G\left( c \right) $ as,
$ G\left( c \right) =P_0\left\{ f_1\left( x \right) >cf_0\left( x \right) \right\} $
where $ P_0 $ is under $ H_0 $ . Since $ G\left( c \right) $ is computed when $ H_0 $ is true, so the inequality need to be considered only for the set when $ f_0\left( x \right) >0 $ ,
$ P_0\left( \frac{f_1\left( x \right)}{f_0\left( x \right)}>c \right) $
@SAJW But $E(X)$ is not the same as the expected duration until 5 in a row...
@HereToRelax wait, which solution?
The inequality here needs to be true when $f_0(x)$ >0, any hint why please?
the one you posted
@HereToRelax Ah, good. That one I like ;-)
the one where you get a hypergeometric series
10:05 PM
@Avra You know you can use $...$ and not take up so much vertical space?
using $$...$$ is so space intensive
@robjohn. I apologize, but my software automatically adds that!
which software?
change the $$s before you enter it here, please
Sure. Thanks for letting me notice
10:07 PM
it is also harder to read when it is short sentences so spread out
I was going to ask on the forum, but the question is not worthy as it's one point about the condition.
@robjohn does Mathematica tell you about a closed form for $\int_0^{1/e} \exp\big(\frac{1}{\log x}\big) dx$? Wolfram alpha doesn't say anything
So, I will try to break the question into smaller piece. If $
G\left( c \right) =P_0\left\{ f_1\left( x \right) >cf_0\left( x \right) \right\}
$, then could you please explain why $1-G(c)$ is a CDF?
@geocalc33 Yes, Mma gives the answer in terms of A Bessel function of the second kind
@Avra When $x\to-\infty$ it gives $0$ and when $x\to\infty$ it gives $1$
So if you are working through problems in a textbook aimed at an undergrad audience and wind up with some nasty to resolve algebra, that is usually a hint you made a mistake?
10:21 PM
@Avra: I have edited your previous comments to remove the $$s
@robjohn thanks could you tell me what that answer is if you have time?
otherwise I can ask on the site
@geocalc33 $2K_1(2)$
@robjohn that's for bounds 0,1
I was asking about 0,1/e
@geocalc33 No closed form. Numerically: $0.20753352343482877323$
ISC gives nothing (as per usual)
10:31 PM
I think I have to integrate in a different way
@geocalc33 Do you think it has a closed form?
@robjohn maybe. If you slice the function by y=x then it halves the total area, which we know to be $K_1(2).$ Then the integral of x from x=0 to x=1/e is equal to 1/(2e^2)
then it follows that $\int_{1/e}^1 \exp\big(\frac{1}{\log x}\big)dx=K_1(2)-1/(2e^2)$
So I know that it has a closed form for x=1/e to x=1
not sure about other intervals
So I suppose that by symmetry there is a closed form for the interval 0,1/e
and that closed form is $K_1(2)+1/(2e^2)$
which agrees with your numerical results
10:53 PM
@geocalc33 Since the integral over $[0,1]$ is $2K_1(2)$, that would indicate that the integral over $[0,1/e]$ would be $K_1(2)+\frac1{2e^2}$, which indeed does match the numerical result.
nice! we out computed Mathematica and Wolfram alpha
@robjohn TIL your mathjax bookmark renders unrendered Tex when printing in firefox
example incoming
before chatjax:
@AndrewMicallef Better for that is the "render MathJax" bookmark. It doesn't start an unnecessary loop.
after chatjax
noted, though I don't have that bookmarked :P
@AndrewMicallef well, the extra overhead is not that much
but it will check every second to see if that page needs to be re-rendered.
11:00 PM
anythoughts on why some renderers handle this expression but others dont (with jupyters mathjax renderer, and yours two I see latex, but when I export as html it gets garbled):

+ A^2 \left[-\frac{ab^2}{a^2 + b^2} +\frac{4b^3}{3(a^2 + b^2)} +\frac{a^3}{3(a^2 + b^2)}\right],
and try to simplify further
$$A^2 \left[ \frac{a}{3} -\frac{ab^2}{a^2 + b^2}
+\frac{4b^3}{3(a^2 + b^2)} +\frac{a^3}{3(a^2 + b^2)}
chatjax didn't mind that I see
exporting html: i.imgur.com/tkW4aFD.png
The backslashes might confuse something
Ah... the $s are getting muxed somewhere
yeah, I guess I should leave it anyhow, the expression looks too complicated to be a correct answer anyway :P
The first and last $$ are not being used
yeah the text in between is getting rendered'
may try reformatting and see if that helps
yes, it is getting surrounded by the $$ in between the two that are not being used
11:05 PM
progress of a kind
The bullet might be breaking things up too much
Or the + is being changed to a bullet
what program are you using?
hmm seams the md renderer takes precedence
I had it formatted as
+ equation
+ equation
+ equation
fixed it by putting it all on one line
and + at the beginning of a line becomes a bullet?
I'm using jupyter and exporting to html
so fixed?
11:08 PM
the + at the begiining becomes a bullet and it just ignores the opening $$
still looks like a horrid eqiuation though, but a nicely formatted one at least
Let me edit the one above to show you something...
@AndrewMicallef Look at your equations above. Look better?
Hello! Small question: I saw this answer today on MSE. I was wondering if we can apply the same proof but $f(mx)$ as well as $g(mx)$ are $f(x^{1/m})$ and $g(x^{1/m})$.
11:11 PM
11 mins ago, by Andrew Micallef
anythoughts on why some renderers handle this expression but others dont (with jupyters mathjax renderer, and yours two I see latex, but when I export as html it gets garbled):

+ A^2 \left[-\frac{ab^2}{a^2 + b^2} +\frac{4b^3}{3(a^2 + b^2)} +\frac{a^3}{3(a^2 + b^2)}\right],
and try to simplify further
$$A^2 \left[ \frac{a}{3} -\frac{ab^2}{a^2 + b^2}
+\frac{4b^3}{3(a^2 + b^2)} +\frac{a^3}{3(a^2 + b^2)}
right and left braces?
huh nice
I've just been throwing around \big on everything from time to time
ie : A^2 \Big[ \frac{3 b^2x + x^3}{3(b^2 + a^2)}\Big]_a^b
$$A^2 \Big[ \frac{3 b^2x + x^3}{3(b^2 + a^2)}\Big]_a^b$$
$$A^2 \left[ \frac{3 b^2x + x^3}{3(b^2 + a^2)}\right]_a^b$$
@robjohn thanks \left \right look way nicer
more text based procrastination ensues
looks like color coding everything wasn't a total waste of time,
I found my mistake...
11:35 PM
@robjohn one more thing. So before I got distracted by rendering mathjax I was trying to solve this integral:
(which in turn is part of an expression $C + I = 1$, where $C$ is real constant a value I have already computed and $I$ is the above integral)

So I made a mistake expanding out the integrand, which I know realise is:
A^2 \frac{b^2 -2bx + x^2}{b^2 -2ab + a^2}

As far as next steps go, my instinct is to split this up in order to make it easier to integrate, and I think I just answered my own question before asking it
What is the correct method to post a psq when you are pretty sure all your ideas are trash?
what's a psq and why are your ideas trash?
Well, not trash
just not useful for the problem
what is the problem?
I don't have one right now
but I can give you an example where it most likely happened
11:38 PM
Q: Is the set of integers so that $n!+1$ divides $(2012n)!$ finite or infinite?

HereToRelaxI am having trouble with this problem. We have to determine whether the set of integers such that $n!+1$ divides $(2012n)!$ is finite or infinite. Basically we have to determine if the prime factors of $n!+1$ are small enough for a finite or infinite number of integers, but I have not been able t...

for example here
I had no idea what to do
Or more recently here
A: Sum of all elements in finite subgroup of matrices is $0$ if it has trace $0$.

HereToRelaxLet $S=\sum\limits_{i=1}^n A_i$. We have $A_iS=S$ for all $i$. To see this notice $A_iS = (A_iA_1+ \dots + A_iA_n)$, and because the set of $A_i$ is a group this just gives us a sum for the elements of the group in a different order. In particular $S^2 = A_1S + \dots + A_nS = nS$. It follows $S^2...

I once crafted a math contest mock test that contained it and I graded it for like 20 students
Half of them left the problem blank
Because most of the stuff people come up with is useless
that's just the way the problem is.
Well, at least that dude proved it had to be a subgroup of GL_n
I got nothing, both of those questions and your explaination contain too much that I don't understand
My main point is that a lot of people want to post a question but don't know how to aproach it
So the current requirement is that you post your life story or something
Hello, I was recently in my classroom and my professor gave us this assignment, I currently own this book and it has polynomial in it's name, so I guess that may be useful, kind regards.
And if you don't do that you get dunked on
Isn't the point to determine what level someone is coming from though?
Like if you don't put that stuff you get answers written by people who use way too much specific jargon that aren't really super usefull because you have no idea how to do integration by parts
But that's something that helps the person who posts the question
If you add context you can avoid that happening and get better solutions
Like the current situation is that if you don't do that your question gets downvoted and closed
Like for example, the question that I just answered on $GL_n$ already has two delete votes
11:48 PM
your qesution was closed because of duplication...
is it not a duplication?
(I can't tell both questions use symbols I don't know the meaning of)
Which one?
A: Sum of all elements in finite subgroup of matrices is $0$ if it has trace $0$.

HereToRelaxLet $S=\sum\limits_{i=1}^n A_i$. We have $A_iS=S$ for all $i$. To see this notice $A_iS = (A_iA_1+ \dots + A_iA_n)$, and because the set of $A_i$ is a group this just gives us a sum for the elements of the group in a different order. In particular $S^2 = A_1S + \dots + A_nS = nS$. It follows $S^2...

Q: Finite multiplicative group of matrices with sum of traces equal to zero.

M.HLet $G=\{M_1,M_2,...,M_k\}$ be a finite set such that $ M_i\in M_n(\mathbb R)$ and $(G,\;\cdot\:)$ is a group with matrix multiplication. If $\sum _{i=1}^k \operatorname{tr} (M_i)=0$ then how to prove $\sum _{i=1}^k M_i=0$ ?

Oh, it got closed for lack of context
But I guess someone found a duplicate after that
By the way the initial post is even more psq than the recent one
yeah I can't parse either
Because in the original one the group condition is given explicitly and in the new one it is deduced
11:50 PM
which is why I need you to tell me if they are actually duplicates
Yeah they are pretty much the same thing
But the question didn't get closed for that reason
The new one is definitely a dupe of the old one.
good point
maybe the guidlines were not imposed so stringently 7 years ago
7 years ago this site was awesome
anyway I should get back to my homwework
I was 15 and I posted a bunch of garbage and everyone treated me super nice
I learned a lot
11:52 PM
I feel like both questions could be improved by more context though. I have no idea what they are talking about :P
I don't really know what I could have added to the one I posted in 2015
I don't know what just happened
fixed font happened
I don't know either, maybe it just wasn't interesting enough to pique anyones interest at the time
Oh I was just commenting about it because I decided to add a solution to it as I'm trying to answer all of my old stuff
I mean you got some feedback on the problem in comments
and it got a close vote and downvote
11:56 PM
the internet can be harsh sometimes
yeah, it's just sad because it didn't use to be like this here
things change, for better or worse the universe is in constant flux
doesn't pay to get caught up in how things always used to be better
that's very true
Make SE great again
I am selling hats if you want one
Don't joke :)
11:59 PM
They come in red and each is emblazoned with a @leslietownes NFT
sorry, I'm off to do my homework
haha homework
(self imposed, but learning all the same)

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