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12:28 AM
Has @copper returned from his safari ride on a mongoose?

Of the fun Halo variety no less

Is the definition of $B$ correct? I'm wondering if the proof accidentally left out the we want $\vert\beta_i\vert\leq 1$. Also, the finiteness of $B$ doesn't make sense to me... we can always tweak the $\beta_i$ slightly, which yields infinitely many options?

12:54 AM
I guess the result is so well-known that I will just look at an alternative proof

sha: i don't understand the definition of B as written, but if the betas are intended to be chosen from {alpha_1, ..., alpha_m} then B is a subset of the finite set {alpha_1, ..., alpha_m}^n [moreover, i do not understand what {alpha_1, ..., alpha_m} has to do with the rest of the argument, if it does not somehow enter into the definition of B]

I was wondering the same
however, I'm not sure if in that case $B$ would be non-empty

1:29 AM
@ShaVuklia what is D

can't access the Cambridge university press site

@onepotatotwopotato Have you tried hacking the mainframe?

don't know how to hack lol
oh I can access the main site but I just can't access the book page I try to look at

This is as useless to post here as Shaun’s endless queries about downvotes.

maybe the book page doesn't even exist!

1:45 AM
@mick Why? Who cares? Why should anyone care about that problem? Where does it come from? What motivates it? From the description in the question itself, it looks an awful lot like "Whoo! Big numbers! Why? How?"
What is the application?
Assuming that there are other solutions, so what?
For a lot of problems on this site, the motivation is "I am taking a class, this problem was assigned" (i.e. the intention of the problem is to help a student master some skill). For many other questions, the motivation is "I am trying to solve problem [X], and have run into trouble with [Y]. [X] is interesting because...".
So why should anyone care about a particular diophantine equation of the form $A^a + B^b + C^c + D^d = E^e$? What makes it special or interesting or worthwhile?

Certain basic diophantine equations have impact I understand somewhat. But the random ones? Who cares?

Is the stabilizer theorem that I studied in group a weaker version of the structure theorem I studied in module? I asked my prof this question before and he was too busy to answer it. But what I want to know is just a superficial yer or no.
Forgive me if my question is not accurate

what is the stabilizer theorem? let alone the structure theorem?

@TedShifrin Exactly.

1:57 AM
@TedShifrin Isn't that what number theory is about?

I know you hate it when we agree, just as leslie does.

I mean the orbit stabilizer theorem of group that we study to deduce the class function of finte group

@Jakobian Not the general ones, no.

structure theorem I mean for the finitely generated module over pid.

Why does the orbit/stabilizer theorem have anything in common with the structure of f.g. modules over a PID?

1:59 AM
@TedShifrin Look at Peter or someone. They're just looking for big numbers or big primes or numbers of some kind. I don't see this as out of line of what number theory cares about.
I personally don't care, but someone would

They all seem to deal with decomposing a finite group. But I did realized that the way they did it are quite different.

I know my share of top-notch research number theorists. I don’t judge a field by what shows up here.
@oscar writing a f.g. abelian group in canonical form has what to do with group actions on sets?

Lol, yes. But top-notch research number theorists also have to put out some substance out there for it to be research. Who knows, maybe some of them care about finding such numbers in their spare time.

What middle school kids think is number theory is far removed from serious number theory. Enough on this.
Obviously, things like Fermat are very deep.

I don't see how this is so different from Fermat's last theorem, seems to be related

2:06 AM
not at all? as far as I can tell.

You can make up any homogeneous polynomial on earth and look for rational points on the projective variety. But in high degree and high dimension, is it necessarily interesting?
@oscarmetalbreak Me either. So why do you ask?

I mean the methods they use to deduce the conclusion share some similarities I guess. One is playing with orbit another is playing with cyclic module...

Number theorists seem to be obsessed with things like "does it work for three numbers? What about for four?" so yeah I think it might be interesting for them

Orbit/stabilizer is directly analogous to the fund. Homomorphism theorem.

Indeed

2:10 AM
I don’t see what you are talking about unless you explain it.

No I can't explain it. It is just a random question.

@Jakobian Fermat's last theorem "looks like" the Pythagorean theorem, but larger exponents. Given that it works for $n=2$, it is reasonable to ask if it works for $n>2$. There is a whole family of Diophantine equations being considered there.
I don't think that anyone is particularly interested in $a^{47} + b^{47} = c^{47}$.

@XanderHenderson Before that people considered the case $n = 3, 4, 5$ with more or less success

The proofs I know for the f.g. blah blah are basically linear algebra. They connect via the fundamental homomorphism theorem. OK, done.

Just because whole Fermat's last theorem was proven for a family of curves, doesn't mean they didn't want to know even the partial results like that

2:25 AM
If there is any plausible reason it might be to look for some pattern
Another could be that the question was solved in more general setting of which this was a special case of
But I think its just about trying to find something, like miner with a pickaxe and a shovel

2:44 AM
@Jakobian there's no pdf file of that book

3:06 AM
@TedShifrin Mongoose are lovely little creatures. Ferocious too!

So just curious whether or not mathematics still thinks unique representations are distinct from the ideas common to each or not. If yes, then the idea of "factors" is still fundamentally broken in math, objectively speaking.
If everyone still thinks $1^n \neq 1$, then no wonder things don't progress much in math regarding the "interesting" stuff these days.
You can't deny that philosophy and mathematics are inseparably complementary to each other. One is the realm of the real, the other is the realm of the ideal. There's a reason Pythagoras literally thought "all is number".

Everyone still thinks what? ... And no, not everything "is number."

3:22 AM
0

Let $V$ be a vector space over a field $F.$ Let $f$ and $g$ be two non-zero functionals defined on $V$ such that $f(x)= 0$ implies that $g(x)=0$ for all $x\in V.$ Show that $g=kf$ for some $k\in F.$ I think by the word functionals they mean that $f,g$ are functions defined from $V$ to $V$ i.e $f,... I have no idea on how to solve this problem. No, you need to get a correct definition of functional. It is not about what you think. @TedShifrin If 1 is not prime because "bla bla$1^n$in front of every number" then by the same logic we should say no number has a "unique representation" because each is$x + 0^n | 0^0 = 0$. Also I'm just quoting Pythagorus. Numbers are logical beings. #Aristotelianism #Thomism @AMDG Hey, it's always 420 somewhere. The more reasonable thing is that each number is a unique idea, and each has infinitely many representations. In short, that means that both$1+1$and$2$are the same idea represented in two unique representations. If you are into Platonism, sure. 3:38 AM I'm into realism because I'm a Thomist. According to reality, this is a fact. It's not my opinion. If you want to call it an opinion, you can call it the opinion of Aristotle, refined over 2000 years. To abandon the tried and true path for the jungle and then call that progress is absurd nonsense to any reasonable person. @ThomasFinley Do you know about dihedral groups? @AMDG Okie dokie. @TedShifrin Can you give me a clue? @SoumikMukherjee Yes, I know the group representation of a dihedral group as the definition of a dihedral group and some examples of them. 3:53 AM Right, so if a prime is a thing that is better defined as a number consisting of one irreducible whole: itself whose logarithm is nominally 1 in its own base, then prime(1) (or prime(0) if you like) is 1. I rest my case. If you get the definition right, then it’s easy. @TedShifrin should I search it up on web? I am afraid because I can get cluttered up in a pool of informations which may be irrelevant. My final exam is in two parts: the first part is required. The second part is required if you want better than a C. About half the class finished the first part, then bailed. rant: profs adding redundant information to test problems makes me paranoid, because if there's any ambiguity then said piece of redundant info becomes an inconsistent statement You were told the correct definition in the comments. 3:57 AM Right. It is 9pm. I'm going home. @XanderHenderson That's sorta sad, but also sorta understandable in today's world. Oh, hey, @Semiclassic. Long time! o/ it's all well and good to want to give students extra information to make solving tedious equations faster I don't believe exam questions are the place for those games. to be fair, i don't think it was intentional @TedShifrin Indeed. Every question from the first part of the exam was taken directly from the book. 3:58 AM but the problem needed words to make the statement clear But, at a certain level, I also am a fan of "prove or give a counterexample" questions. Not in typical first-year levels, of course. And, moreover, directly from the homework problems that I actually assigned. Ah, I was about to ask that. Most books have hard questions that most students can't do. @ThomasFinley Can you do the matrix question now? And now I am really going. 4:00 AM G'night @Xander. for context: this was a circuit problem involving three resistors and two batteries. at the level of equations it amounted to:$x = 30a+30c, y=10b+30c, a+b=c$. other information given allowed you to deduce$x=6$and$y=2$. at that point, there is a unique solution$(a,b,c)=(0.12,-0.04,0.08)$. the second is negative and the problem was predicated on discovering that minus So where is the irrelevant info? however, solving three equations in three unknowns in a physics exam seemed too hard to the prof. so he told them that a=0.12 Oh shit. thing is, the wording doesn't make clear that it's positive 0.12 4:02 AM That kind of thing needs to be doubly vetted ... especially in this day and age with Mathematica, etc. all it says is that the current in that resistor is 0.12 now, -0.12 isn't actually consistent with the other equations...but that's not something you should have to check during a 50 minute quiz Oh, but most students are going to just do the positive case anyhow. they are. but since the whole point was finding that the other current was not in the obvious direction that really isn't excusable I don't like questions at college level that amount to tedious arithmetic. This is not a linear algebra class with matrices that work out nicely. well, tbf, circuit problems like that are really just linear algebra 4:04 AM No, but there can't be two solutions to a nonsingular system. Yes, but tedious arithmetic is not what should be tested. more or less agree, but again that's what i think the prof was trying to avoid Anyhow, if you get a solution with the positive$a$, then you're done, as there cannot be another solution. I assume the system is inhomogeneous. :) And maximal rank. to be clear, the profs intention was: if you know x=6 and y=2 from the first part, then being told a=0.12 from the outset lets you deduce c=0.08 from x=30a+30c and thus b=c-a = -0.04 but it's worded not as "the current in this resistor is 0.12 amps (up)" The trouble with these multi-part questions (and I remember having such in physics years ago) is that if you mess up part a), then the rest of the problem goes to hell quickly. just as "the current in this resistor is 0.12 amps" yeah granted, if you have time to check your work, then you can verify that 0.12 amps in the other direction does not work but that is not an ambiguity that should be present if you want that piece of information to help rather than harm 4:09 AM Thus, I carefully avoided such questions in my teaching. In linear algebra, if part a) was to reduce the matrix to reduced echelon form and later parts were based on that, I typically took points off for errors in a) and then graded the rest based on their wrong answer. A pain for me, but the only fair thing to do. Sometimes it ruined the question when they did that, but oh well. But I keep saying you don't need to check that. yeah, we try to do that too There is a unique solution, so why do you keep insisting you need to check$a=-.12$? mostly it's bad here b/c i don't know if i should say "taking it as 0.12 in the other direction isn't consistent with the rest of the problem, so wrong" because the prof's intention was effectively not for them to solve the system I don't care. but to solve the rest of said system using part of the solution 4:10 AM The system has a unique solution regardless. I don't get your point, unless you're unwilling to grant the linear algebra fact. also, a note of context: this is the intro physics course for pre-med and bio majors so not exactly sophisticatd Yeah, but they probably have some experience with at least 2x2 systems. And surely this came up implicitly in homework issues. i'm entirely willing to grant that -.12 doesn't work out. but my claim is that the fact that it doesn't work out is not immediately obvious This isn't a rigorous math course, so are you expecting a proof? I don't like the whole set-up, but I think you're wrong. Was there never a discussion in class/discussion sections about uniqueness of solutions to a system if there are not infinitely many solutions (or none)? why would there be? this is a physics course 4:13 AM Because one makes observations in working homework exercises. I'm not saying it should have been proved, but it may well have been commented. ... If you stumble on a solution, then you are done. Anyhow, if the prof's point was to facilitate things, let it be. Next time, criticize the problem before he publishes it. he didn't show us the problems before the quiz Yeah, probably because he didn't want you divulging anything accidentally to any student. probably, yes to be precise, here's the diagram that goes along with this Then voice your issue to him and see if he agrees with you or with me. Otherwise, let it go. @TedShifrin Oh! I didn't check the OP then. Ok, as I infer from the comments, that the phrase : "a functional f defined on a vector space V (, where V is a vector space over a field F)" means that f is a linear transformation from V to the vector space F (, this makes sense as F is a Vector Space over itself). Did I get it? Some comments seem to say otherwise, so I need some clarifications. 4:16 AM given that i'm the one grading it, "letting it go" is not entirely an option Linear functionals are always linear maps to the underlying field of scalars. but you're right that he's the one to make the call on this Why isn't it an option? If his point was to simplify things, you can make the judgment that you're going to go with the simplified approach. Period. @TedShifrin ok, thanks . It seems I have got it correctly. Let me try the problem again myself. I don't know where you came up with the idea that maps$V\to V$are ever called functionals. First step in mathematics — particularly in linear algebra — is to get definitions down 100%. 4:18 AM @TedShifrin when did I say that? I said V to F You most certainly did NOT. The title says$V\to V$and the text says$V\to V$. rolls$8\pi^5 + \sqrt e$eyes I was trying to do that @TedShifrin ok, now I get it. If you are talking about the title of to the OP, then please forgive me. It was just my misinterpretation to the problem. I did not get what you meant by " text says$V\to V$" though. I don't know what to do with$V\to V$You posted question said it You wrote your whole discussion based on that. Do you not even know what nonsense you wrote in the question? Seriously. 4:22 AM @oscarmetalbreak The question actually said, "f, g are functionals defined on V" but I interpreted as$f,g:V\to V$@TedShifrin Ok, now I know what you were referring by it. Yes, now I can contemplate. I went on writing the things in the content to the OP based on my same misinterlretation, but the question in OP seems correct neverthless, as you helped me understand. Thanks! I am gonna delete that post now. @SoumikMukherjee are you talking about my previous question? Yes, I was able to do it. 4:37 AM Hello! I have a basic question if someone has the time to give me a simple answer. I am trying to grasp the concept of functions as vectors without going into much detail. @ThomasFinley yes, about the$T^3=0$one My question is, since the domain is a set and a vector has some order, how can a function from R2 be seen as a vector since for example [(1,1),(2,2)] -> [1,2] is different than [(2,2),(1,1)]->[1,2] What is the question giannisl9? This makes no sense. Precisely what is your course/text saying? Your context is not what I exoected. I can think of f(x) = x as an infinite vector because each coordinate is ordered. For example 1 comes before 1.2. 4:45 AM This still isn’t making sense. Is your course talking about continuous functions on some domain and saying they form a vector space? This is just because you can add two and get another and can multiply by a scalar. No, I am just trying to understand the concept of functions as vectors. I am studying the Fourier series. And you’re looking at the infinite vector of Fourier coefficients? I am looking about the generalization of the inner product to get the coefficients. So that is precisely what I talked about above. You’re looking at the space of continuous functions (or square-integrable functions) on$[-\pi,\pi]$and defining an inner product by the integral. Each function is a point (hence vector) in this space. Nothing about infinite vector. Nothing about ordering …. What are the dimensions of this space? 4:53 AM Maybe your teacher motivated thus with the integral as limit of Riemann sums. Each function is a point in this space you say. I get that I think. But doesn't this point have infinitely many coordinates? It’s infinite-dimensional, but you don’t write functions as literal vectors. You can look at the doubly infinite vector of Fourier coefficients and think of that. It doesn’t literally have coordinates. But you can label functions more or less by their Fourier coefficients, as I said, and think of those as coordinates in some sense. For example, there have to be conditions on those coordinates for the Fourier series to converge. Thanks for your time btw. Really appreciate it. My question is not so much about the Fourier series. Its about how to think of any N dimensional function as a vector. Well, without them, there is no vector of coordinates as you are envisioning. Functions are just vectors in an abstract way, not a geometric way with an arrow between two points. You can’t “draw”$\sin 3x$or some periodic version of$x$as arrows. They live abstractly… @SoumikMukherjee yes, I was able to solve it, using the linear transformation$L_A$where $$A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$. Show that the eigenvalues of a real symmetric matrix are all real. I think this is a very popular question. 5:00 AM @ThomasFinley yes It’s super important; that’s why. Can this question be solved without using conjugates and dual spaces. I haven't still learnt them? I am thinking of lets say f(x) = x as an infinite-dimensional vector <...,1,1.1,1.11,2,3,..> is that wrong? Yes. A nice proof using Lagrange Multipliers. @giannisl9 Yes, totally wrong. There is no such thing. @TedShifrin Do you have a source? That's becuase here in MSE, all the answers to this question seems to be using symbols like,$||x||,A^*,\langle Av,Av\rangle,\bar x,$etc (where$A$is a square matrix and$x$is a column vector) and honestly, I have no idea what any of the symbols mean. 5:04 AM For example, the space of continuous functions has uncountable dimension, more than integers many. 45 I am having a difficult time with the following question. Any help will be much appreciated. Let$A$be an$n×n$real matrix such that$A^T = A$. We call such matrices “symmetric.” Prove that the eigenvalues of a real symmetric matrix are real (i.e. if$\lambda$is an eigenvalue of$A$, show ... I am talking about answers in this post You can watch my lecture on it in YouTube. The calculus proof is certainly on this site. Oh god. But, my book said something along the lines of if we divide [a,b] into N parts then g = [g(t1), g(t2), .., g(tn)] is the vector representation of g as N goes to infinity. The linear algebra proof without complex inner oroducts is in my linear alg book with Adams. @giannisl9 They are motivating the inner product in terms of Riemann sums. But it is not anything correct or rigorous … @TedShifrin Is a pdf for the book you are referring to available on the internet? 5:08 AM There is no limiting infinite vector defined. @ThomasFinley Probably illegally so. I don’t want to know such things. So, what do you think I should do? Should I forget about this? for now? Probably. What field can I study to learn about this connection though? I suspect as I’ve said they’re trying to motivate why the integral should generalize usual dot product. Yes, and I thought I got that and then tried to apply it in 2 demensions. Thinking of each function of n dimensions as a infinite vector whose columns have dimensions n. 5:11 AM More advanced linear algrbra and analysis courses, eventually Lebesgue integrals … @TedShifrin ha ha...it's understandable. But I think, it is not. So, I guess there's no way to get hold of a "normal" proof unless someone directs me to a handout or a link where it's given. As of now, a superficial search on it on MSE seems to be much futile. Okay, thank you very much for your time. I am a bit disappointed but I guess it will have to wait. I personally am fond of the calculus proof. But the real linear alg proof is in books, just not many, as it’s sort of artificial. You still need complex numbers. @Thomas For the proof, if$a+bi$is an eigenvalue, consider$S=(A-(a+bi)I)(A-(a-bi)I)$.$S$is real and singular. Work out$Sx\cdot x$. @TedShifrin so there is no dot product as integral? "This notion can be generalized to continuous functions: just as the inner product on vectors uses a sum over corresponding components, the inner product on functions is defined as an integral over some interval [a, b]" 5:30 AM @giannisl9 Huh? Go reread what I’ve said for the last half hour. although I can not interpret the function as an infinite dimensional-vector I can see the generalization but this way I can not understand how it relates to orthogonality I have reread everything rest assured I really value your time I pasted this from the wikipedia You said "defining an inner product by the integral" I am okay with that But I can't see the connection if my interpretation of infinite-dimensional vector is wrong You also said "each function is a point (hence vector) in this space. Nothing about infinite vector" but then you said "It’s infinite-dimensional, but you don’t write functions as literal vectors" So I am quite confused. If each function is a point in space with infinite dimensions, isn't it an infinite-dimensional vector? Maybe there is some confusion about the dimensions? @nickbros123 is this for me? 5:50 AM @giannisl9 Oh crap, I wanted to ask a question xD How to come to the "block matrix inversion" part in the above link Just wondering, can anyone do better with the Sylow's theorem? I have read an improvement that if$p^k | |G|$then the number of subgroups of order$p^k$is$\cong 1\mod{p}$. It is more general than the case of Sylow 2 where it states the number of sylow-p subgroup etc... 6:24 AM @nickbros123 what is your question? @copper How’s the mongoose? been a while since i saw one :-) i just rode my bike to the middle school track & back i like the sounds of kids running around & playing at 10:30 pm? what are they vandalizing at this hour :-), before dinner still dark, though it's dark so early now, i love this time of year i think our sunset starts moving later after this week :( 6:32 AM i timed my hip operation accordingly solstice in the day after its far ahead of you now, but get your kids to get their driving licenses as soon as practicable will save everybody $$How’re the leeks, @leslie? i love leek & vegetable soup it's been a minute since i've been to a grocery store that sells leeks, the nearest grocery doesn't it's on my to-do list 6:48 AM Even Ralph’s has them. I think you’ve procrastinated for 3 years. yes, but ralph's isn't the nearest grocery :) there is this web thing called amazing or something like that? 7:11 AM off for a quick cup of soylent green before bed. 7:59 AM @copper.hat that particular inverse given in the wiki link for the partitioned matrix, how to arrive at that 2 hours later… 10:17 AM @onepotatotwopotato sometimes they don't type in all the authors. Maybe that's it? Or did you just type it in google? 1 hour later… 11:20 AM @Jakobian and (f\circ g)' exists a.e. Okay I didn't think this assumption was important but it seems it is 11:40 AM @XanderHenderson really ? who big numbers, who cares ?? Is that your opinion about diophantine equations ? you could say the same about fermats last theorem since it also has high powes and low sums. YOU ask for applications of a diophantine equation ? what were the applications of fermats last theorem before it was proved ? WHAT are the applications of the taxicab number ? What is the application of beal conjecture apart from proving it ? seems you do not like diophantine equations. @XanderHenderson Well I am not a student and that is not an obligation for this site. Im not going to pretend to be a student. I do not lie 1 hour later… 12:53 PM @mick I am asking for motivation. In the case of FMT, (1) it is very reasonable to look at Pythagorean triples, and to ask if the result can be generalized, (2) there is a larger open problem in number theory: "Can an n-th power be represented as the sum of fewer than n n-th powers?", and (3) FMT is not a single Diophantine equation with a specific set of exponents arbitrarily fixed from the start, but a family of equations, one for each n\in\mathbb{N}---it is much more general. It is also worth mentioning that FMT was studied by one of the most famous mathematicians of the last half century. People are often attracted to problems which fascinated famous people, hence people pursued FMT simply because Fermat thought it was worthwhile, and it (seemingly) stumped him. As for the Beal conjecture, I am not sure that the problem is all that intrinsically interesting, but, again, it is a much more general problem---it is not about having a fixed set of exponents, but rather about proving a very general result about an entire family of Diophantine equations. And if you asked a question about a specific set of exponents with respect to the Beal conjecture, that might actually give some sense of motivation to a problem. (Probably not to a level where it would be a good fit for Math SE, since (1) Math SE is not meant for questions on the cutting edge, and (2) it is not clear how a specific example would give insight into the conjecture, unless that example were a counterexample, in which case one probably doesn't want to publish it on Math SE, anyway). @mick Being a student is not required, no. But providing context is. Motivate the problem. Explain to the reader why they should care about the solution. Put a problem in a larger context of mathematical work. @mick No one asked you to. In any event, you asked for opinions. When you ask for opinions, you should be prepared to hear them. 1:11 PM chat is on 🔥 1:25 PM @XanderHenderson I don't think it stumped him if he wrote that he proved it. @Jakobian Hence the "seemingly". I think that he thought he solved it, but I am quite certain that he was wrong. That's for the cranks to decide New easy proof of Fermat's last theorem every day Seems unlikely. I just think its funny how it gave all the cranks more of a reason to write about it Oh, yeah, that too. But cranks gonna crank. Who else is gonna square those circles and trisect those angles? Perhaps that is a good reason to have a couple of big "prize problems"---attract the cranks and keep them distracted so that they don't bother people trying to get real work done. :D Like one of those mosquito zappers. 1:41 PM I wonder which is the best mosquito zapper, Collatz or RH? @Jakobian seems that this question may be of your intrest @SineoftheTime not interesting. Just remove a vector from an orthonormal basis Heck. Just choose a non-zero vector of norm 1 deleted lol Integration by substitution theorem (ultimate version?): Let F be indefinite integral of f\in \mathcal{R}^*(I) and \Phi be differentiable a.e. Then (f\circ \Phi)\cdot \Phi' \in \mathcal{R}^* and$$\int_{\Phi(\alpha)}^{\Phi(\beta)} f = \int_\alpha^\beta (f\circ \Phi)\cdot \Phi'$$for all$\alpha, \beta \in I$iff$F\circ \Phi$is an indefinite integral of a function in$\mathcal{R}^*(I)$. If$f\in \mathcal{L}^1(I)$then in above setting$F$is an absolutely continuous function such that$F' = f$a.e. and the equality and$(f\circ \Phi)\cdot \Phi'\in\mathcal{L}^1$holds iff$F\circ \Phi$is absolutely continuous. (indefinite integral of a Lebesgue integrable function is precisely an absolutely continuous function) Sorry,$\alpha, \beta\in J$lets say This follows from the a.e. version of chain rule @Jakobian$(f\circ \Phi)\cdot \Phi'$and the equality of integrals for all$\alpha, \beta$This theorem is general but pretty awkward to apply, there are special cases however. 2:32 PM Ugh... I was on campus until 9 last night. I don't wanna go back now. :( I DON'T WANNA! *stomps feet* 3:01 PM Somebody doesn't like me... One of your students? I see I might have idea who that might be, but I might be wrong anyway I just find it funny. @XanderHenderson ❤️ hi guys I'm not your guy, pal. 3:12 PM I am trying to follow the Birkhoff ergodic theorem from the point of view of sampling @Monty It's been several years since I looked at that stuff. I am no help. no worries 4:08 PM @nickbros123 have you tried multiplying the matrices? 4:49 PM @XanderHenderson At least what is showing is +2 I am confused, given a flow$\phi_t$which for any$t$leaves a measure$\mu$invariant, if$\mathcal{L}$is the generator of the flow why is it the case that$\int \mathcal{L} f d\mu =0 $for test functions$f$? @copper.hat I can show that such and such formula is true, but I would like to understand how one can arrive at it / motivate it actually got it. 5:13 PM @nickbros123 Here's the answer. If the matrix has the form$\begin{bmatrix} A & B \\ O & D\end{bmatrix}$then you can easily find the inverse (assuming$A$and$D$are invertible). But by row operations you can get the matrix with$C$in that form; indeed, write the row operations as premultiplying by an appropriate block matrix. 5:42 PM @TedShifrin Thank you. I'll look into this. I am reeading Hoffman Kunze for linear algebra, and they didn't cover this block method, which seems very nifty I was breaking my head over the inverse of the Hilbert matrix but with this block method it comes quite easily 5:54 PM Yes, block thinking is powerful. I have exercises on determinants with block matrices in both my books. The point is that by thinking in blocks, you can manipulate this just like with a regular$2\times 2$matrix, as long as you remember that the blocks do not commute. 2 It's well know that the function$y=\begin{cases} \text{exp}(-1/x^2) &x\neq0 \\ 0 &x=0\end{cases}\$ is smooth but not analytic. But this can't happen for complex functions. The reason is that analytic functions are defined as those function that can be expressed as a power series? Is my understanding correct?