Mathematics - 2017-01-17 (page 1 of 5)

PVAL-inactive

00:00

@Ted I'm pretty sure its trivial. It should correspond to $(2,2)$ in $\pi_1(SO(2)) \times \pi_1(SO(2))$.

Which is 0 inside $\pi_1(SO(4))$.

(through the correspondence coming from clutching functions).

@Mike TS^2 \oplus TS^2 is trivial right?

I'm forgetting the argument.

Mike Miller

Yeah your argument is correct. You can check SW numbers if you want.

PVAL-inactive

There all 0 that doesn't help

0 wait nvm I'm being silly

the nontrivial one has w_2\ne 0.

Mike Miller

yeah

robjohn

@amWhy: the first thing you need to do is to open a gravatar.com account.

Maks

00:16

Hey ! :D Did anyone miss me ?

Adeek

hi @KajHansen

Hi everyone

Simple

00:36

Can someone help me do a probability homework problem

Ali Caglayan

sure @Simple

Simple

@AliCaglayan Thanks. The question is Let Y be a Poisson $\lambda$ random variable, and define $X=I_{[Y>]}$, fine $E(|Y-X|)$

and find the conditional mass function of Y given X

Ali Caglayan

@Simple What is $I_{[Y>]}$?

Simple

@AliCaglayan a typo, that should be $I_{[Y>0]}$, an indicator function

Ali Caglayan

When is Y < 0 anyway?

I is only 0 when Y = 0

The probability of Y being 0 is easy to find

Simple

00:48

Y is Poisson, a Poisson distribution only take non-negative value

Ali Caglayan

exactly so X is only 0 when Y is

Simple

When Y is 0, P(X=0) = e^-lambda

Ali Caglayan

So E[Y-X]=(1-e^-lambda)(E[Y]-1)

or something like that

Simple

Should we separate into two cases, since we are looking for the expectation of $|Y-X|$

Ali Caglayan

yes

Simple

00:55

I don't know how to calculate that

Semiclassical

hi chat

Daminark

So after conversing with friends, we realized that the way to homotope a curve to the circle is by $H(t,s) = \frac{\|c(t)\|}{1-s+s\|c(t)\|}$

Simple

@AliCaglayan how do we determine when Y> X and Y<X

user228700

Hello, everyone :-)

user228700

I have a quick question about the existence of the limit of a function.

user228700

01:06

My textbook says that the limit of a function exists if L.H.L=R.H.L=Finite value but it also explicitly notes the following:

user228700

> "Note that we aren't interested in knowing about what happens at that point"

user228700

Why is it that we're not interested about what happens at the point?

Kaj Hansen

@Kaumudi.H, consider $f(x) = 1$ when $x \neq 0$ and $0$ when $x = 0$

Really, the function eval at $x=0$ could be anything

We still have $\lim_{x \rightarrow 0} f(x) = 1$

Now, if we're going to talk about continuity, what happens at that point would matter

As continuity means limit exists and is equal to function eval

user228700

Hmm, OK, that makes sense.

Simple

@Kaumudi.H you can look at $\sin(1/x)$ when $x$ approaches to 0

user228700

01:11

@Simple OK..?

user228700

@Kaj: Thank you :-)

Kaj Hansen

Glad I could help!

Ali Caglayan

01:27

@KajHansen I have an interesting distribution for rank

cloud.sagemath.com/blobs//projects/…

This is the rank of groups of order 128

It peaks at 4

So I wonder if I accumulated this for all groups, would the expected rank be around 3.5?

Kaj Hansen

@Ali, reminded me of a fact I found a while back: something like 95% of groups of order $\leq 1024$ are of order $1024$.

Ted Shifrin

@Daminark: That formula isn't right. You need a vector output, not a scalar.

Daminark

Sorry the norm on top was a mistake

It's $\frac{c(t)}{1-s+s\|c(t)\|}$

Ted Shifrin

Right. And you need to argue that the denominator is never 0.

I think it might be. So this might not be valid.

Maks

@TedShifrin ! I was waiting for you

Ted Shifrin

01:37

Why is that, @Maks? I'm only here for a few moments, actually.

Maks

Do you have a good book to re-learn analysis ?

Ted Shifrin

How hard a book? There's everything from Spivak's Calculus to Rudin.

Maks

derivatives, series,sequences,taylor,integrals, multiple integrals, derivatives on parts, 2D and 3D space, multiple variables function, level curves, and all that

Ted Shifrin

Oh, multivariable, too.

Maks

Mmh, something like basic-medium ?

Kaj Hansen

01:39

For multivariable, there's Ted's book of course. And Hubbard & Hubbard isn't horrible either.

Ted Shifrin

Hmm. People like Abbott as a single-variable source that's easier.

Actually, I think Hubbard & Hubbard is not nearly as good, @Kaj, but I'm biased.

Kaj Hansen

Basic-medium might be covered by, say, Div,Grad,Curl and All That? Won't be as complete as you want though

Maks

@TedShifrin you wrote a book ?

Ted Shifrin

They go off the deep end with idiosyncrasy.

No, he wants analysis, not engineering vector calculus.

Maks

I would like a book with many examples if possible

Ted Shifrin

01:40

Oh, maybe not.

Kaj Hansen

Fair enough

Maks

Or with exercises and its solutions

I learn much more on the practice than on the theory

Ted Shifrin

You won't find books with solutions to exercises.

Right.

Maks

I need one for vector calculus too

Subspaces, linear transformations and auto vectors...

Kaj Hansen

@Maks, this is going to sound a little sketchy, and you'd be right, but there are truly some decent YouTube lecture series with tons of examples.

Ted Shifrin

01:41

But for multivariable stuff, you might try out my YouTube lectures and see if you like that. There's computations and proofs of theorems.

Maks

@TedShifrin oh... how bad

Ted Shifrin

My lectures have linear algebra and multivariable all mixed up, and you can always ask Kaj for help with exercises :P

Maks

@KajHansen I've checked khan academy's ones, but they are like reaaaaaaally slow and dull

Ted Shifrin

(There aren't exercises in the videos, though ... sadly.)

Kaj Hansen

@Maks, I agree re: Khan Academy. There are a number of alternatives I'm sure.

Ted Shifrin

01:42

I need to head out now, but another decent analysis text (again, no solutions) is by Wade at U Tennessee. He covers single and multivariable analysis.

Maks

Ok then, lets change the question, what do you think is the best way to learn math offschool ?

Daminark

@Ted I don't think the denominator can ever be 0. $s$ only runs from 0 to 1.

Kaj Hansen

Esp. if MathDoctorBob has a series on that. He has some fantastic stuff for algebra and representation theory

Ted Shifrin

But can't you have $\|c(t)\| = (1-s)/s$ for some $s$, @Daminark?

@Daminark: I think you should think about a line segment joining two points.

OK, I'm outta here. Bye, all.

Daminark

No, you can't, $\|c(t)\| > 0$

Alright, see you @Ted

Actually hmm... I'm gonna get chalk and be sure, I might be messing up here

Yeah so if $1-s+s\|c(t)\| = 0$, then $\|c(t)\| = \frac{s-1}{s} \le 0$, which is impossible, so I think this works. Anyway, I'll confer with the others, so see you around, chat!

Martin Sleziak

03:23

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

@SteamyRoot Personally, I think it is reasonable to start with a smaller bounty. Because if you do not get a satisfactory answer and you want to add bounty to the same question again, you have to double the previous bounty until you reach 500.

It's easier to double 50 reputation points and post a new bounty for 100 than, say, start with 200 and then the next bounty will require you to use 400 reputation points.

But I agree with what you said: "Even so, I'd say bounties don't attract that much attention." Meta: How effective are bounties?

Lucas Henrique

Hello there

Can someone help me with a question? I'm trying to prove that the only $r$ that satisfies $r = 2^{r-1},\,r \in \mathbb{Z}^+$ is 1

Martin Sleziak

@LucasHenrique From binomial theorem you have $2^{r-1} \ge 1+(r-1)$ for any positive integer.

The above are the first two terms of the binomial expansions, namely $\binom{r-1}0$ nad $\binom{r-1}1$.

If $r>1$, you will have at least one additional summand there, and the inequality will be strict.

Sorry, if $r>2$ (i.e., $r-1>1$), then you will have more than two summands there.

@LucasHenrique Or maybe easier approach is to show by mathematical induction that $2^{r-1}>r$ for $r\ge3$.

Lucas Henrique

LOL

Martin Sleziak

You can probably find a few similar question on the mains site.

Lucas Henrique

I actually got to that from $2^{r-1} \geq r$

The main problem was to prove that the only number that satisfies $x^y = y^x = n$ is 16

Martin Sleziak

03:39

For example, Proof Using Mathematical Induction that $2^n \geq n + 1$. It appeared in search for $2^{r}>r+1$ in Approach0.

@LucasHenrique Again, there must be many posts about this on the main site. This one seems to have several good answers: $x^y = y^x$ for integers $x$ and $y$

Lucas Henrique

I know that it's easy to prove that, but I wanted to prove that on my own :p (never saw a proof before)

SAWblade

I don't understand what you mean by $x^{y} = y^{x} = n$ is only satisfied by 16.

Nightmare

hi, i am new to doing proofs for theory of algorithms, and would like some help on how to approach this question: suppose that w exist in T*, prove that (w*w^R)^R = w * w^R

Martin Sleziak

@LucasHenrique I am not sure it is that easy. As one of the answers mentions, it appeared on Putnam contest. So I guess that is not such trivial problem, if it was included in that competition.

Lucas Henrique

There are some Putnam problems that are easy

(If you have a solid background in the subject)

Martin Sleziak

03:46

Well, different people probably call different problems easy.

SAWblade

$r = 2^{r-1}$
$2r = 2^{r}$
$2r = e^{r \log 2}$
$2r * e^{-r \log 2} = 1$
$-r \log 2 * e^{-r \log 2} = \frac{-\log 2}{2}$
$-r \log 2 = - \log 2$
$r = 1$

always nice to find an excuse to use the lambert W function

Martin Sleziak

@Nightmare Perhaps showing these two things would help: (a^R)^R=a and (ab)^R=b^Ra^R. I suppose that you can show that they hold for any word in the given alphabet by induction of the length of the word.

Lucas Henrique

Oh, I have an interesting problem (if you guys want to solve it)

I'll post it in the site

SAWblade

hit me up

i need to procrastinate on my own math anyhow

Lucas Henrique

0

Q: Prove that there are at least 100 pairs of usable boots

A store has 200 boots of size A, 200 boots of size B and 200 boots of size C. In these 600 boots, 300 are of the left foot and 300 are of the right foot. Knowing that usable pairs of boots have the same size and are for different feet, prove that it's possible to find at least 100 pairs of usable...

(I don't know what tags to use, please suggest)

Nightmare

03:56

@MartinSleziak thanks that made it clear :D

SAWblade

combinatorics defo

this is one of those problems that are very easy but hard to put into words

Lucas Henrique

@SAWblade exactly xD

SAWblade

:P

Lucas Henrique

That question was on the book chapter of the pidgeonhole principle

The answer is totally incomprehensible

Martin Sleziak

05:42

@LucasHenrique Since you ask about tags, there is also a tag pigeonhole-principle. And since you said it is from a book, adding the source of the problem to the question would probably be nice. Adding context is mentioned in the faq post on How to ask a good question?

Martin Sleziak

06:02

I have tried to google for "200 boots" "pigeonhole principle". It seems that either the book is not in Google Books or I should have tried a different search query.

Searching for left right boots gets a few more hits, including some hits in Google Books.

So it seems that this problem appears in Mathematical Circles: (Russian Experience) on page 37 with solution on page 225.

Ted Shifrin

06:28

@Daminark: Yeah, I think I messed up a sign. Apologies.

Vrouvrou

06:42

hello, can someone help me

0

Q: question about derivative on $\mathbb{R}^N$

What is equal to this derivative please $$\frac{\partial A^i}{\partial p_j}$$ where $A\equiv \phi(|\nabla u(x)|)\nabla u(x)$ and $A:\mathbb{R}^N\rightarrow \mathbb{R}^N$ and $p=\nabla u(x):\mathbb{R}^N\rightarrow \mathbb{R}^N$ $A^i=\phi(|p|)p_i$ Thank you

functional-analysis derivatives pde partial-derivative

i have a problem with this derivative

please

Daminark

07:21

Ah, don't sweat it @Ted

Balarka Sen

07:59

@PVAL-inactive I am not used to these computations, so just as a sanity check; why is this 0 in $\pi_1$(SO(4))? The clutching function for TS^2 $\oplus$ TS^2 is f(z) = [z^2, 0; 0, z^2], so the nullhomotopy is f(tz) = [tz^2, 0; 0, tz^2], right?

I think there should be an indirect argument. TS^2 is O(2), so TS^2 $\oplus$ TS^2 is $O(2) \oplus O(2) = (O(1) \otimes O(1)) \oplus (O(1) \otimes O(1)) = (O(1) \oplus O(1)) \otimes O(1)$. Now by Euler sequence $O(1) \oplus O(1)$ is $O(2) \oplus \Bbb C$ - which is trivial because it's just $TS^2 \oplus \nu$.

So the whole thing is trivial.

Vrouvrou

please what is equal to $$\frac{\partial p_i}{\partial p_j}$$ where $p=\nabla u(x)$

Socrates

08:37

are such numbers a thing? $a$ satisfies $a=b$ and $a=c$, but $b\not=c$?

Secret

08:59

102

Q: Are there real-life relations which are symmetric and reflexive but not transitive?

Inspired by Halmos (Naive Set Theory) . . . For each of these three possible properties [reflexivity, symmetry, and transitivity], find a relation that does not have that property but does have the other two. One can construct each of these relations and, in particular, a relation that is ...

elementary-set-theory relations

Tobias Kildetoft

10:00

@Socrates Nothing satisfies that unless you redefine equality (which seems like a bad idea)

Socrates

10:31

why not give a number the property of having two distinct values?

wildcard mathematics haha

Tobias Kildetoft

@Socrates that would basically make it no longer a number but a set with two elements

Kasmir Khaan

If 5 -letter ``words'' are formed using the letters A, B, C, D, E, F, G, how many such words are possible for each of the following conditions:
(a) No condition is imposed.

(b) No letter can be repeated in a word.
(c) Each word must begin with the letter A and letters can be repeated.
(d)The letter C must be at the end and letters can be repeated.
(e) The second letter must be a vowel and letters can be repeated.

can anyone help me with this ?

a) 7^5 words

Tobias Kildetoft

(c) and (d) are basically like (a) with the word being shorter

Kasmir Khaan

okay thanks @TobiasKildetoft

b) 7*6*5*4*3

c) 7^4

d) also 7^4

e) 6* 7^4

Can anyone tell me if am thinking correct?

Tobias Kildetoft

10:46

except for (e) yes

how did you do (e)?

Kasmir Khaan

the second letter must be a voyal

there r 6 of them

Tobias Kildetoft

not among the ones you have been given

Kasmir Khaan

omg ><

thanks :)

Tobias Kildetoft

ok, given that the reasoning was correct

Kasmir Khaan

2*7^4

Tobias Kildetoft

10:47

right

Kasmir Khaan

thanks ! :)

am new on combinatorics and i find it harder than analysis

A store is selling 5 types of hard candies: cherry, strawberry, orange, lemon and pineapple. How many ways are there to choose 28 candies?

I this a multichosing problem ?

Tobias Kildetoft

no, stars-and-bars (or number of positive solutions to a linear euqation)

Kasmir Khaan

so i have in total

28 candies +(5-1) dividors

(32,4 ) ?

32 chose 4 ?

Tobias Kildetoft

sounds right

Kasmir Khaan

11:03

thank you !:)

if we insist on chosing 1 of each candy

i think the problem will become the same but with 23 candies left

27 chose 4

Tobias Kildetoft

right

Kasmir Khaan

:D

how to think about this one , 28 candies with at least 5 cherry and at least 2 lemon?

we have to delete 7 pieces right ?

Tobias Kildetoft

right

Kasmir Khaan

but the at least condition is what makes me doubt my answer

(25,4) 25 chose 4

Tobias Kildetoft

well, choosing one of each is the same as having at least 1 of each

("at most" conditions are a lot harder)

note that the problem essentially started with an "at least 0 of each" clause

Kasmir Khaan

11:10

oh okay i see your point :)

if it were at most , then i need to sum up many possiblities

Tobias Kildetoft

yeah, I don't recall if there is a nice way to handle such reqs

Kasmir Khaan

well i think the best way to get hold of this is work all problems on the book

cuz for now i dont have that intuition yet

@TobiasKildetoft my university asked us to use this Grimaldi, Discrete and combinatorial mathematics, 5th edition, Addison-Wesley

Daniel Fischer

11:46

@RandomVariable I think it's possible. But I don't really know what sets on the boundary circle can be convergence/divergence sets. If memory serves, I once read a short overview of what is known, and all simple enough sets can be. But I may be misremembering. If I'm not misremembering too badly, chances are good that it was in an answer by Dave L. Renfro here or on MO.

Socrates

math.stackexchange.com/questions/2101429/… for anyone who likes proofing stuff as an excercise :)

so f(x) looks like $ax^n+bx^{n-1}+...+z$

PVAL-inactive

12:11

@BalarkaSen My point was that $pi_1(SO(4))$ is $\Bbb Z/2Z$ and the clutching function is the concatenation of two functions.

Balarka Sen

Ah, I see.

I was writing down an explicit nullhomotopy of the clutching map

I also gave an alternative proof.

Bhargav

13:17

Hi, I require a little help, I want to simulate a Spring, I have value Velocity, given the velocity I want to know the distance by which the spring will expand (maybe also include extra params like Tautness etc)

DHMO

@Bhargav use F=-kx

Bhargav

k is?

can you explain the params?

DHMO

a constant depending on the spring

F is the force

x is the displacement

Bhargav

yea I understand how to get by the force, all I have here is velocity

DHMO

in that case you use energy

Bhargav

13:19

should I break down force to ma again?

DHMO

the kinetic energy is 0.5 mv^2

you find the kinetic energy

and the potential energy given by 0.5kx^2

Bhargav

so how do I get the displacement?

ah so the kinetic energy and potential energy are equal

DHMO

no

the sum of the kinetic energy and the potential energy is constant

Bhargav

ah ok

DHMO

you need to have another parameter

velocity alone isn't enough

Bhargav

13:23

i can assume mass and tautness of spring

try out different values see how the graph looks

i.e displacement vs time

DHMO

is the velocity taken when x=0?

Bhargav

yes

DHMO

then you have x at that point

at that point the potential energy is 0

so as you said, kinetic energy loss = potential energy gain

Bhargav

what would be the correct values for K?

@DHMO

Balarka Sen

@Bhargav velocity of what, exactly?

sarcopsy

13:39

Hello people

I have a question about finite fields

If $E$ is an extension of $F$, Gallian defines a primitive element of $E$ to be $a\in E$ such that $F(a)\cong E$

Socrates

are there sets A,B, such that

$|f(A)\cap f(B)|>|f(A\cap B)|$

f is a function

Balarka Sen

$f(A \cap B)$ is a subset of $f(A)$ and $f(B)$, hence of $f(A) \cap f(B)$.

sarcopsy

The multiplicative group $GF(p^n)^* := GF(p^n) - {0}$ is shown to be cyclic. If we consider $GF(p^n)$ as a extension of $GF(p)$, does a primitive element $a$ in $GF(p^n)$ necessarily generate $GF(p^n)^*$?

Socrates

yeah, but I somehow fail to see an example

Balarka Sen

What example?

Socrates

13:47

such that the inequality holds

sarcopsy

@Socrates $f(0) = 0, f(1) = 0$. Then set $A = \{0\}$ and $B = \{1\}$

Socrates

ah, I saw that too complicated

Balarka Sen

It holds for all sets and functions. Take your favorite function.

Socrates

for A=B it doesnt hold

Balarka Sen

Ah strict inequality. Sure.

Alessandro Codenotti

13:49

@sarcopsy how do you define primitive?

Socrates

@sarcopsy thanks

sarcopsy

@AlessandroCodenotti Gallian defines primitive as $F(a)\cong E$ iff $a$ primitive where $E$ is an extension of $F$ and $a$ in $E$

@Socrates np

@AlessandroCodenotti But there seems to be many variants of 'primitive' floating around so its a bit confusing and I wanted to clear it up as well

Alessandro Codenotti

Yeah, as far as I know primitive is defined as being a generator of the multiplicative group

DHMO

@Bhargav k is a constant that depends on the material of the spring

sarcopsy

@AlessandroCodenotti Yeah, if that was the case it immediately follows, but with this definition of primitive I'm not sure if it works

Socrates

13:55

does $f^2(x)$ equal $f(f(x))$?

Alessandro Codenotti

But your definition is weird, you can have $GF(p^n)~\Bbb F_p[\alpha]$ with $\alpha$ not a generator of the multiplicative group if I remember correctly

I'm quite sure I've seen an explicit example of that in the abstract algebra course, I can look it up in my notes later if you're interested

DHMO

@Socrates depends on the context

sarcopsy

@AlessandroCodenotti It would be very nice if you could! Thanks

Socrates

@DHMO that certainly can only be, if the whole codomain is a subset of the domain

DHMO

@Socrates I don't even have your whole question

sarcopsy

13:59

@AlessandroCodenotti These kind of things annoy me to no end haha

Socrates

@DHMO it stems from this question math.stackexchange.com/questions/2101537/…

DHMO

@Socrates I fail to see how it cannot be simply [f(x)]^2

sarcopsy

@DHMO If your codomain of $f$ doesn't have multiplication defined then it's going to fail, I guess

Balarka Sen

@Socrates eg sin^2(x) is (sin(x))^2

Socrates

because $f^{-1}(x)$ is also not the reciprocal of $f(x)$ (in general)

Alessandro Codenotti

14:01

I'll look it up tonight because I'm uni now, I hope I'll remember, ping me otherwise @sarcopsy

DHMO

@sarcopsy our codomain is R there

@Socrates depends on notation

sarcopsy

@AlessandroCodenotti Thanks, I'll remind you!

@DHMO Oh, ok. That being said I've always thought of the trig functions $sin^2(x)$ as the exception to the rule rather than the other way around

DHMO

@Socrates please ask in the comments the meaning of f^2(x+y)

Socrates

@DHMO appereantly your interpretation was the right one

DHMO

@Socrates by substituting x=y=0 we can see that f(0)=0

then by substituting just x=0 we can see that f(y) = f(-y)

Adeek

15:02

hi @BalarkaSen here ?

Kasmir Khaan

hi chat

i need help with setu up of tripple integrals

y = x^2+z^2 , y = 8-x^2-z^2

what is the region on the xz plane ?

i know that x^2+z^2 < y< 8-x^2-z^2 in order for the region to make sense ( finite)

but on xz plane y = 0

that gives me x^2+z^2 = 0 a point

and 8-x^2-z^2 = 0 circle of radius sqrt8

am thinking correct here?

DHMO

yes

Kasmir Khaan

this does not give me correct answer i set up the integrals many times

the region on xz plane how to find that

DHMO

what is the question asking for?

Kasmir Khaan

its a tripple integal

DHMO

15:09

what is the integral?

Kasmir Khaan

My problem has been in setting up the boudries but ill post it one second

tripple integral of sqrt ( x^2+z^2 ) dx dy dz

region R : y= x^2+z^2 , y = 8-x^2 -z^2

Ali Caglayan

@kaj related to yesterday

DHMO

@KasmirKhaan the region would be a double-cone

Kasmir Khaan

what is the projection on the xz plane ?

DHMO

a circle

Kasmir Khaan

15:12

i have found the boudries for y

DHMO

it would be useful to imagine the region

Kasmir Khaan

okay but how to find that?

how to find the radius of the cricle ?

DHMO

if you can imagine the region, you will see that it is a double-cone

it has rotational symmetry of infinite order around the y-axis

reflection symmetry about the plane y=4

Adeek

anybody wants to discuss the following problem.

DHMO

you don't need to project the region on the xz plae

y=4 cuts the regions into two cones

Kasmir Khaan

15:16

how to set up that?

because litterly that what i dont understand on these problems

DHMO

firstly you need to be able to imagine the region!

Kasmir Khaan

okay thank you

DHMO

then you will see two cones

divide the integral into y=0..4 and y=4..8

when y=0..4, you only care about the cone y=x^2+z^2

(hint: polar substitution)

when y=4..8, you only care about the cone y=8-x^2-z^2

Balarka Sen

@Adeek Now I am

Kasmir Khaan

@DHMO its not a cone its paraboloid y=8-x^2-z^2

so is y = x^2+z^2

DHMO

15:25

@KasmirKhaan right

but my statements are still true

Adeek

@BalarkaSen I am thinking about the following problem. It would be nice if you could give me a hint, but not a full solution(No spoilers). Suppose R is a ring. Consider $M_n(R)$. Describe all left and right ideals. Find all two sided ideals.

Kasmir Khaan

Okay thanks @DHMO

Balarka Sen

@Adeek Err. I can't really give you a reasonable hint, though I know what the ideals are.

Adeek

ok don't tell me I will think about this for sometime.

Balarka Sen

There's a more-or-less obvious guess. But I won't spoil the fun.

Adeek

15:30

I mean first thought were just lower triangle matrices.

Alessandro Codenotti

@Adeek start describing ideals in $M_n(K)$ with $K$ a field

Adeek

but that isn't everything I think.

Alessandro Codenotti

And see what happens in this case

Adeek

ok @AlessandroCodenotti

Alessandro Codenotti

Hmmm actually I don't know about left and right ideals, but I have a guess knowing the answer for the 2 sided case

Adeek

15:32

well part of the 2 sided ideals are diagonals.

But that isn't everything.

If I is two sided ideal in R. Then, $M_n(I)$ is two sided ideal.

I don't think that is everything though.

Balarka Sen

It is everything.

Oh, actually I know a proof. Think of $M_n(R)$ as a $R$-module

Adeek

oh

Semiclassical

15:48

hi chat

SteamyRoot

ohi

Balarka Sen

Hi @SemiC

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$Mathematics$ Mathematics