your original question with the $\lambda$ being given by multiplication by scalars has positive answer

if $\alpha$ is not a root of unity

It looks like I lied: the original question said $W$ was stable under $\lambda$ for all $\alpha\in k^\times$ ....

if $V=\bigoplus_{n\in\mathbb N}V_i$ and $\lambda:V\to V$ is multiplication by $\alpha^i$ on $V_i$, then each stable subspace is the direct sum of its intersections with the $V_i$.

(provided $\alpha$ is not a root of unity)

how can I see that?

let $W$ be a $\lambda$-invariant subspace

and let $w\in W$ be a fixed vector

then $w=v_{i_1}+\cdots+ v_{i_k}$ for some homogeneous $v_i$s and $\lambda^r w=\alpha^{i_1r}v_{i_1}+\cdots +\alpha^{i_1r}v_{i_k}$ for all $r\geq0$

and we do a linear combination of these to isolate a homogeneous vector...

each homog vector in the sum defining $w$, in fact... I see.

the set of the equalities $\lambda^r w=\alpha^{i_1r}v_{i_1}+\cdots +\alpha^{i_1r}v_{i_k}$, with $r=0,\dots,k-1$ can be solved for the $v_i$ in terms of the $\lambda^r w$.

this works if $\alpha$ is not a root of unity

thanks Mar

because then the powers of $\alpha$ are distinct

(Vandermonde comes into play here)

the general statement is: if $f:V\to V$ is a semisimple endomorphism, then for every $W\subseteq V$ which is $f$-invariant, we have $f:W\to W$ also semisimple

(your $\lambda$ is semisimple; in fact, diagonalizable)

It is diagonal, innit?

in one basis it is diagonal

in others, it is not

right

the only diagonal maps are the scalar maps :D

«this number, as defined by Cantor, is 0» ?

I am not a great fan of his questions...

Oh, I missed that :D (the Cantor definition)

the trisector

@tb Without even looking: GarouDan?

sigh

there was only one post today that annoyed be, though

@BrianMScott en plein dans le mille!

@tb I'm very impressed by the commenter who apparently understood the trisector's instructions well enough to carry them out and measure the result...

I like N.S.’s comment on the trisector.

@MarianoSuárezAlvarez I was surprised yesterday to find that in proving that if you have $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ exact and that $M' $ and $M''$ finitely generated means that $M$ is also, the proof was just using some basic knowledge of linear algebra!!

déjà vu

don't tell anyone, but everything is mostly linear algebra!

we talk about derived categories and such things only to confuse the analysts

(Some time ago, while discussing the complexity of the proof the Drozd's dichotomy theorem with a few people, a Russian guy in the group complained «but it is just linear algebra» with a dismissive tone only Russian mathematicians are able to get correctly :) )

@Brian: this MO thread might suit your taste. I found it quite surprising.

Anyone have a better original for the product of an odd number of transpositions being nontrivial?

Drozd: that's tame versus wild, right?

@HenningMakholm I intend to let it ride for now, in case someone comes up with a particularly nice argument or exposition. Failing that, one of the answers to the question that you found is adequate.

yeah

Googling led here. "Loosely, these two possibilities correspond to: hoping for a classification theorem, or throwing up our hands in dismay."

heh

well, the theorem shows that there is a lot of interesting in picking up nice subcategories of modules which we can describe

like BGG's category $\mathcal O$ in the Lie algebra case, say

we don't throw up our hands in dismay because of the halting problem :D

@tb Thanks! The only reason that I’m not too surprised is that I’ve been surprised too many times by what can happen in Čech-Stone compactifications.

people still work on the classification of pairs of matrices up to simultaneous conjugation, knowing that it is very very wild (in fact, wild means «at least as complicated as the classification of pairs of matrices up to simultaneous conjugation»)

«provide an example of a locally connected Hausdorff space not consisting of a single point.» hm?

The OP's comment is very strange

I was just about to point to it!

it always amazes me that people may understand what «locally connected Hausdorff topological space» means and not know an example

well, if they're confused about what can be interpreted as a point...

But it’s even more puzzling that someone would have trouble with space consisting of a single point.

In English "not consisting of a single X" could be interpreted as not containing any Xs.

And your guess was right.

So... Is the empty space locally connected? It certainly is Hausdorff.

Vacuously both, I’d say.

Probably. But the empty space is not connected, IIRC

@BrianMScott I'd say so. Still an odd question though. Could be homework.

If the OP didn't understand what was being asked for, maybe he/she just reproduced it verbatim.

this is an interesting edit

@tb It is by my usual definition: it’s not the union of two non-empty disjoint clopen sets.

that is simply a convention

and in practice, it is better that it be not connected than connected

just as the zero module is not considered to be simple, nor indecomposable, in general

seems that edit is by the original answerer, given the gravatar...

that's why I approved it.

@MarianoSuárezAlvarez It’s not a convention in general topology, which uses the definition that I gave (or equivalents).

that depends on who you talk to

as most conventions

okay, guys, I leave the nullology to you. Have a nice week-end!

I meant that it’s not a generally accepted convention. For instance, Engelking doesn’t use it, and neither does Willard. I don’t think that I’ve encountered it before,

they then pepper all their theorems with silly «non-empty» hypotheses

(or, for some situations, consider the empty space to have no connected components, and deal with that)

I’d rather have the occasional non-empty hypothesis than an unnecessarily complicated definition.

if one has a commutative ring $R$ of endomorphisms of a complex vector space, it is a fact that all of the operators of $R$ share a common eigenvector. one can show this tediously by constructing, inductively, the common eigenvector. i would love to know a simpler existence proof. does anyone have any direction that i might pursue?

in any case, everyone is free to make his or her life as complicated as he wants: that is a basic human right

of a finite dimensional vector space, @EricGregor

otherwise the fact is not true

yes @MarianoSuárezAlvarez, a finite space. sorry

you often catch my hidden assumptions!

I know what you actually meant, but I like the idea of basic human rights of finite-dimensional vector spaces!

finite dimensional vector space, that is

lol

5-dimensional complex vector spaces in every home!

i guess i can see how two commuting endomorphisms share an eigenvector

since if $S,T$ commute and $v$ is a $\lambda$-eigenvector of $T$, then $TSv=\lambda Sv$.

Suppose A is a comm. ring of endomorphisms of a finite dim. vector space V

so if you're over an algebraically closed space and $S$ stabilizes the eigenspace then it must have an eigenvector

in that space

in order to find a common eigenvector, we can suppose inductively that no subspace of V is invariant under A — otherwise we just replace V by an invariant subspace and find a common eigenvector there

let a in A be non-zero and let lambda be an eigenvalue of a

since A is commutative, the subspace of V of eigenvectors of a is invariant under A

so it must coincide with V

in other words, a acts like multiplication by lambda

we see that every element acts by multiplication by a scalar

so V is one-dimensional

"since A is commutative, the subspace of V of eigenvectors of a is invariant under A" this is true by my argument, right?

yes

that is The Key Property of commuting pairs of operators :D

i almost follow you

@MarianoSuárezAlvarez, so you are supposing that no subspace of $V$ is invariant under $A$, since if it were we could restrict ourselves to the invariant subspace and be done quickly

yes

since in this case if $W$ were invariant under $A$, it is stable under all of the endomorphisms and then my argument applies

or perhaps not my argument but the Key Property!

is that right?

no, the reduction is:

if V has an A-invariant proper subspace W, then by induction on the dimension we can find a common eigenvector in W

it is of course also a common eigenvector in V

so we need only consider the case in which V has no proper invariant subspaces

i don't understand the reduction then, if it doesn't use my argument. if $W$ is an $A$-invariant subspace, how do we find the common eigenvector in $W$?

I am proving the statement by induction on the dimension of V

The truth of P(1) is obvious

ah. so you are saying if $V$ were $n$ dimensional and there were an $m<n$ dimensional subspace $W$, then we are reduced to a simpler case

sorry, i missed that

yes

that's a very simple argument

thanks @MarianoSuárezAlvarez

i have only seen inductive arguments of this sort of thing. i understand that constructive proofs are nice sometimes, but i'd rather have existence arguments for this kind of thing

notice that this gives a construction

pick any element a of A

pick any eigenvalue lambda and construct the eigenspace of a for lambda

either the eigenspace is the whole thing, and we move onto another element of A, linearly independent with a

or restart the procedure, now in the smaller subspace

at some point we will reach either a subspace of dimension 1, or we will have exhausted A

in both cases, all the vectors which we have left are common eigenvalues

nice!

a simpler argument and the construction

thanks again @MarianoSuárezAlvarez, i appreciate it as always

np :)

Why are all my questions being down voted?

/me looks

the System does not see any strange ptterm of votes for you

neither for up nor down

well I just meant my last two questions had negative

well, the one about integrals is quite unclear

never say the book without saying what book

we do not read your mind!

@tb tb has ruined my sleeping habits

You seem to know that $\Delta x$ is $2/n$, be we do not know what $\Delta x$ is, because you did not tell us

So now you have $\frac{2}{n}\left(-2 +\frac{2k}{n}\right)$... why do you have that now?

what did you have before? What are you doing?

when writing a question, you have to keep in mind that mostpeople do not know what you are doing nor what you are thinking nor what you are reading

My guess is, the downvote is motivated by these things, or others

which question are you talking about?

You continue: you put that into the equation, but you have not written any equation!

I am talking abotu the integral at the top

I do not see any missing information except maybe I could have expanded on my incorrect work

well, I do see missing information

the one I mentioned, for example

I did not downvote, though

I cant get any of these problems on my own, not sure what to do, if I should just keep asking them here or what

the very best thing is to ask an actual human

nothing beats that

@Jordan i think you need someone who can give you a lot of one-on-one time

that costs too much

Hello.

to explain, what is going on in your question....shoot, i'd have to go over the entire basis of definition of the riemann integral, and it would take hours...and i can't draw pictures in this chat-room, at least not very well

I understand all that stuff

I tried tutors before and they cost a ton of money and most of them aren't good at math anyways

perhaps you do, perhaps not...i'm not judging you

I doubt (no offence)

it is a pretty simple concept, you add the area of rectangles together

Brian's answer is very clear

an alternative to a human is a book

there are many good books on calculus

(and many bad ones, sadly)

the approximations add rectangles together, but the integral itself is not one of those approximations

Is Stewarts Calculus book any good? I find it to be horrible

no the integral is the limit as x approaches inifnity of the approximation

it is pretty terrible

they broke google :/

@Mariano Who?

It seems like all the classes I am in use the worst books possible

google itself... if I start typing in the entrybox of the www.google.page it immediately jumps to the results page

amazingly annoying

man I haven't asked a question about integrals involving a coefficient yet

how do coefficients affect the (1+k/n) part?

would it be 3k/n for 3x?

proving a function is actually integrable can be a real chore....but IF a function is already KNOWN to be integrable (for example, if it is continuous on a closed interval [a,b]), we are free to use any "convenient" approximation...in this case, a "regular" partition of [-2,0] and using the value of f at the endpoints

if a function is differentiable isnt it integrable?

@Jordan one of the first results about integrals you should prove is that $\int_a^b c \cdot f(t) dt = c\int_a^b f(t) dt$

@Jordan yes, it is

I dont think I am that far yet

@Jordan one should always keep in mind "where" a function is differentiable. for example the function $f(x) = 1/x$ is differentiable....EXCEPT at 0. the "exceptional points" have to be watched out for.

in the case of $f(x) = x^2 + x$, on the interval [-2,0] we are indeed in good shape.

@Jordan, well you asked how "coefficients affect the integral", that is the answer

oh

well I cant factor it out though

it isnt a factor of everything

ok, in the question you asked, you have a sum where k ranges from 0 to n-1, or from 1 to n (you don't say which).

and first we have to do some algebra to get the k's by themselves, so we can use these formulae:

I used the formula I have an entire page of reducing

$\sum_{k=1}^n k = n(n+1)/2$, and $\sum_{k=1}^n k^2 = n(n+1)(2n+1)/6$

yes, that's why Brian's answer is so good. it shows the algebra in a clean way.

my guess is that when algebra gets "involved" you get confused

@JM I like the animation of the zeros in your post that you linked :)

@AntonioVargas Heh, it's nice, no? I'm all the more impressed at Szegő that he conceived it without having a computer...

Oh hey it's JM. Just saw you on the Szego-y question.

Hi anon. I love that curve; can't help but comment...

@JM As am I. We're lucky to be able to make cool pictures like that.

I'm in the middle of a literature review... so far the trickiest limit curves seem to be those for L-functions.

oooh

The first example of such was studied in the monograph "Zeros of Sections of Power Series" by Edrei, Saff, and Varga

But those were the days before LaTeX and it's a little hard to get used to.

@AntonioVargas as in Dirichlet series?

@AntonioVargas Ah, Dick Varga. I like his papers...

I think he covered most of the truncations of elementary functions' power series...

This answer and its subsequent edit cracked me up.

Let me go grab my copy...

@anon Tsk, no edit history... no fun for me. :P

It was exactly the same, he just had f(k/n) instead of $f(k\delta)$ in the first line. Literally the only thing he did was put a delta in the first line, but in the wrong spot.

(On that note, is it really wise to hammer on the Riemann definition for somebody who isn't looking to be studying analysis deeply? I don't get these nonspecialist calculus textbooks sometimes...)

I wish Stewart of Stewart Calculus had contact information so I could complain about his book

Stewart is a celebrity

like Madonna

you just don't contact them

@JM They studied functions of the form $\prod_{k} \left(1+\frac{z}{x_k}\right)$

@Jordan I doubt that it would do much except make you feel better: it’s been a pretty popular book (in several different versions) for many years now. Believe it or not, it’s actually one of the more readable standard calculus texts.

@MarianoSuárezAlvarez That may very well be the first time I've seen a mathematician compared to Madonna...

@JM As far as I know he's done it for exp, sine, and cosine only

I figured it was so hard to contact him because he knew how bad his book was

@Jordan: Did you figure out where your calculations went off in the $\int_{-2}^0(x^2+x)dx$ problem? Your $\frac83$ is part of the right answer, so you were probably doing at least part of it right.

@AntonioVargas Sounds quite intriguing...

@JM I recently finished doing it for the Bessel functions, which surprisingly (or unsurprisingly, as they're asymptotic) had the same Szego curve as the cosine.

Stewart much be very very rich

and living in a private island in the pacific

no telephone

Does anyone actually get rich from textbooks? Even popular calculus books?

@AntonioVargas The first kind, I presume... (the modified ones are just "hyperbolic" versions after all...)

not rich, but some books do make money

his books surely do

What's the going rate for royalties these days?

@MarianoSuárezAlvarez I contacted Hofstadter once.

I know an author of a book on quantum groups who's made quite a bit

and that's quantum groups!

@JM Indeed. The analysis is simplified greatly when the function in question is entire.

I wonder how much my old dept. chairman Tom Hungerford has made from his graduate and undergraduate abstract algebra texts and his business math text.

@Brian No I keep getting it wrong

Hang on a bit, and I’ll add some calculations to my answer.

@AntonioVargas Reminds me... I notice that your specialty is studying polynomial roots; have you had occasion to look at Bessel polynomials (related to, but aren't, Bessel functions)?

@BrianMScott I imagine so. Even if the author just gets 50c per copy, imagine how many copies some of the big selling maths textbooks have sold.

@AntonioVargas (Also I suppose you consider orthogonal polynomials relatively boring... ;) )

@JM Actually there was a paper a few years ago which studied them. I believe they used Riemann-Hilbert methods, which is apparently a really reliable approach for this problem for orthogonal polynomials. I've seen a few other papers for different families which use it.

Let me find the reference...

@AntonioVargas That'd be cool (which I might look at again when I find more spare time); I had been working on this numerical method that kind of hinged on getting the roots accurately, but the zero-pattern is just rather unruly...

...and to segue slightly: the current level of activity on the front page scares me...

@JM Ah, it was Varga's student, Carpenter. Here is the link.

I haven't had a chance to look at it yet; I've been focused on power series.

Ack, I've a copy of that already... :( Thanks for taking the trouble, though.

Heh.

(Clustering is nice for pictures, but terrible for numerics... :( )

I see. What kind of estimate are you looking for?

Well, to be not too restrictive, just about any nice estimate that won't require me to do Newton-Raphson to extended precision...

With clustering, you (almost) always have convergence to roots previously found, and though deflation works as a cure, it's terribly slow.

I've seen linear-algebraic routes, but they have their own different set of caveats.

@Jordan Finished. Take a look at the update, and see if you can work out where you went astray.

I just wish I could do it on my own

I just did the basic math wrong

I think that you’re getting a bit better. You’re setting these problems up properly now; you’re just having trouble keeping all of the calculations straight.

Yeah but I do not really understand what I am doing, I will forget it by tomorrow

@JM Are the Bessel polynomials orthogonal on some contour in the complex plane? Or just a real interval?

@AntonioVargas Yes, they satisfy complex orthogonality; let me check my copy of Chihara on this computer...

@Jordan It’s too bad that your school doesn’t have a free walk-in tutoring centre. Mine started one a few years before I retired, and it’s been a great success, both with students who just need it on rare occasions and with those who are in getting help almost every day.

My school does but it isn't very good, majority of the tutors are immigrants which isn't a problem but their accents are hard to understand especially for math

and then in general none of them really are that good at math anyways, they can do the problems but dont really understand the concepts

That can be a problem, yes. We were fortunate: we did have some foreign students tutoring, but they all had light accents, and most of our tutors were really quite good as tutors. Some were excellent.

(I know from first-hand experience: I used to hang out there and pitch in when it got busy.)

@AntonioVargas Ah, here: they are orthogonal with respect to the weight ${}_1 F_1\left({{1}\atop{\alpha+1}} \mid -\frac2{z}\right)$ with support over the unit circle.

yeah I just go to a community college, over half the math classes are pre college algebra

It's in the original paper by Krall/Frink.

the majority of the students are not white, lots of ethiopians and other african countries. Mostly people that go to this school don't know math well

If I may ask, where are you?

Minneapolis

Brrr!! :-) A brother and a sister of mine did graduate work there.

U of M?

Yes.

yeah that is a really hard school to get into, I gave up on trying

my grades are pretty bad and I will have to go to a local party school

I forget what other schools are in the area.

@JM Hey there!

Hey rob!

@JM Nice egg :-)

cooler than the last one :-)

There aren't really any other good schools around Minneapolis. Mankato and St Cloud is about it I think

at least cooler colors.

Well, it's nice to do pastels from time to time... :)

@Jordan Well, at least I recognize the names! Don’t really know anything about either of them, though.

Arturo is like the God Of Patience

amazing...

@MarianoSuárezAlvarez That's news? ;)

I am going to apply to like 6 good schools around the country anyways

maybe one of them will accidently accept me

I always said that something is terribly wrong with you if you manage to tick him off...

@JM It doesn't look like the zeros of Bessel polynomials have been accurately located (aside from their limit curve). I was definitely mistaking Carpenter's paper for another one.

@Jordan Try to cover a range of schools. Would you prefer a small one or a large one?

we have a few users who have something terribly wrong, by that standard :D

At least, that's what I gather. I have some papers which may be helpful but they're pretty impenetrable and I'm not sure if I want to inflict them on you.

I have no idea really, I just want to get into a school that will accept a < 3 gpa :P

@AntonioVargas Ah, I think I'll just keep looking when I have a lot of free time again. Thanks for the legwork, even if you didn't have to... :)

@JM I think that it may be even harder to tick off joriki.

@BrianMScott Yeah, you may be right...

@JM Not at all. I enjoy this kind of thing. Maybe sometime next year if it still hasn't been done I'll get in touch with you again :)