Hi

Hi

Hi

Hi

Combo breaker^

@EnjoysMath pls

lol

y u do dis

I just had to, the pattern is so obvious. Don't know if it holds for greater though

Didn't you already prove it doesn't?

No such simple proof could work

@EnjoysMath You could have at least done this : Hi

So inconsiderate @EnjoysMath

Plus it is against math chat norms to post questions like this

oh

sry

Yes indeed, we don't do math here

I will work on the $210 = 2 \cdot 3 \cdot 5 \cdot 7$ case to see if it still holds

Strictly off topic please

Rule 3: If you asked your question on the main site, please don't post it on the chat. It will get a lot of exposure without that. If you do choose to post it, title is a better format; it is compact and easier on the eyes.

@EnjoysMath, wtf is up with the plagiarism note on that link above?

@KajHansen I know haha, I went in and read it and no idea

No idea dude. That paper uses math I don't know yet

Link pls

@ᴇʏᴇs it's in my post

LOL

lel

>_>

<_<

it's a math song

Does anyone know where I can find a complete set of the Lectures in Geometry series by Mikhail Postnikov? My university library has the second volume (and ONLY the second volume) and I really enjoy it. But I'd like to start reading the series from the beginning. I'd prefer the English translation, though I can read a bit of Russian, so if someone knows where to get the original Russian set that'd work too.

@Bye_World Have you tried world cat

@DiscipleofBarney I did not even know it existed. That is a useful website, but ideally I'd like to buy the series so I can reference it any time.

@DiscipleofBarney Can you help me prove that the product of two permutation matrices is a permutation matrix? I have been trying, but I can't generalise from the first row

Everything under similarly is not done

@Incurrence Is that what you have done (I havn't read it yet)

Yes

Ignore the very first sum at the top, since I doubled up on using $J$, but it seems I never used that first sum

Actually that is just wrong, maybe you have a better idea of how to approach it?

@Incurrence is that your first linear algebra course?

@KarimMansour No

or second one ?

@KarimMansour It isn't a linear algebra course, I am just using linear algebra for it

I see

It didn't go well though, so I might try using permutations instead

E.g. convert the matrices to permutation form, take the product, and convert back to being a matrix

But I am not yet sure how to do that

By "permutation form", you mean elements of the symmetric group? Because proving the symmetric group is closed (which is equivalent to this) is not hard.

Have you studied group theory yet?

Yes, well I am just dealing with the set that contains all permutation matrices(e.g. only one entry of 1 in each row and column)

I am studying group theory now

@Incurrence I personally prefer to think of matrix multiplication as linear combinations of rows, but anyways. $(AB)_{ij}= \sum_{r=1}^n A_{ir}B_{rj}$, when is that term nonzero? only when we have at some point the both parts are 1 in the sum

Indeed, I used that in the prior part

@Incurrence I guess, what are you having trouble with?

I don't know anymore, I think I need a break - I'll be back after I eat and try it again

What are the nonzero entries in $A$, and use the permutation subscripts for $A$? Same for $B$. Use that permuations notation in the sum

hi @Incurrence ... just stopped in to say g'day to you :P

@Incurrence If you're intent on using matrix multiplication, here's an idea: AB = [Ab_1 Ab_2 ... Ab_n], where b_i is the ith column of B. Ab_i is just the ith column of A. Because each b_i has a 1 in a different row (by definition), each Ab_i is a different column of A. So all you've done is rearranged the columns of A. That won't change the fact that each row and column of A has a single 1 in it and 0s elsewhere.

So it's still a permutation matrix. With some thought, the above could probably be made rigorous.

But not me @TedShifrin ? :(

I'll say hello to you @Kaj! :D

@AlexWertheim ! :)

(The currency of my greetings is undoubtedly lower than that of @Ted's ;A;)

How goes the topology? :)

Look at this edit: math.stackexchange.com/review/suggested-edits/398027, first it changes the meaning and on top of that its like a two year old post (with a perfectly good title), and then for some reason it gets accepted

Oops, Ab_i is the kth column of A (depending on where the 1 was in b_i). The rest of the argument should still hold though.

Not bad at all @AlexWertheim; I'm working on this set right now: math.uga.edu/~pete/4200HW_eight.pdf

Damn, last one huh?

Looks like good stuff though. Are you psyched for your REU this summer? (I think you said you're doing an REU this summer, right?)

Klein bottles and Frechet filters? Lucky you.

Indeed @AlexWertheim. I'm doing an REU with Dr. Robert Rumely at UGA. As I understand, we're going to be looking at arithmetic dynamics in $\mathbb{Q}_p$, whatever that means.

We're going to have a crash course in the area once finals are over.

Sweet! Good stuff. I can't say I know what that means either, but anything with the word "arithmetic" in it has to be cool. :)

hahahahaha

so happy

number theory marks are out got in the final 99 and course total 93 :D

How did the fermat's last theorem thing go @KarimMansour

it went really well @DiscipleofBarney I got 18/20 in it I wrote like 20 pages

and proved specific cases

like n = 4,5,7,10,14

Cool

now you know I have all summer I want to like cover elliptic curve and study them

@KarimMansour, I went to a lecture on elliptic curves recently. Truly fascinating stuff.

yeah they are I read about them when I was doing my project

very nice

Elliptic curves are the best! :)

@Kaj, @AlexW ... neither of you were here when I happened by, so stop your sulking :D

@Ted!! :)

Lol, to be fair @Ted, I was sulking because my saying hi isn't as exciting as you saying hi, not because you didn't say hi to me. ;)

I dunno, @AlexW ... relatively few people care about a hi from me ...

I care @TedShifrin

I certainly do! :D

Just saw some amazing young musicians playing with seasoned old hand stars from New York's Lincoln Center ...

On the other hand, @TedShifrin, your hellos often make it to the starred comments ;)

Rarely, I hope, @Kaj ... I see my comment about purely formulaic math got some stars :P

What were they playing, @Ted?

chamber music, @AlexW, my favorite ... Brahms, Martinu, Schoenfeld (modern), a Mark O'Connor jig for violin/viola, and a Dvorak piano quartet.

Amazing 20 yr old violinist and 23 yr old violist ... among other, more seasoned performers

Sounds lovely. I miss going to concerts.

I'm going to miss Athens. Classical music is amazing here. I will have to learn my way around San Diego's classical scene.

I'm sure you'll get some good opportunities in LA and at UCLA.

I hope so. Duke was surprisingly excellent.

I had the good fortune to see, among others, Garrick Ohlsson and Rafal Blechacz. I'll never forget those performances.

(I actually have a picture with the latter. He's tiny, though you would never know it unless you saw him in person!)

We small people resemble that remark.

You're kidding, @Ted. I would have never guessed with you either!

If you do make it past the LA traffic, I suppose I'll get to see for myself.

well, not small in girth, but short, yes :)

Now I am watching your videos again @Ted, looking for indications... :)

You fooled me!

LOL ... Well, I don't exactly tower next to the blackboards :D

Can someone please help me with this: make X the subject in x + xy = y. Please do show working

X ran over the hill

I managed to get X. I tied him up

Now he's alone

;)

I am guessing you mean to get x, by itself, on one side of the equation?

yea, I got it. I just had to focus a little bit

Okay, good

Well..

I don't like fractions: Y = 2 - 1 / x, make X the subject

how do i do that?

Well you want X by itself, start by getting 1/x by itself

so I have to use the whole fraction and not just the x

Huh?

nvm'

x+xy=y?

Then $x(1+y)=y$

no, y = 2 - 1 / x

Oh, OK

I managed to do it but my answer is: x = -1 / y - 2

the actual answer

So $y-2=-\frac1x$ (incidentally, do you have the MathJax rendering? There's a link on the side)

What I wrote was: y-2=-1/x

So 2-y=1/x

And then you just take the reciprocal.

wait, how can I see formulas in chat?

(Note that -1/(y-2) is the same as 1/(2-y)

nvm got it

)

ok so

y-2=-\frac1x

Test: $2^4$

Did you see an exponent?

yea

(Type $y-2=-\frac1x$)

With $s

Do you know why $-(y-2)=2-y$?

I can get upto $y-2=-\frac1x$

'Cause $-(y-2)=-1\cdot(y-2)$. Distribute: $-y+2=2-y$

from there it's like what do I do with the $-\frac1x$

So $-(y-2)=2-y$

Ex: -(5-2)=-3; 2-5=-3

So… you have $y-2=-\frac1x$

yea

Do you see what would happen if we multiply both sides by $-1$?

does it become $y-2+1 = x$

No

what then?

$-1(y-2)=-1(-\frac1x)$. Let's simplify this…

ohhh

you times -1 by both sides

…but tell me if you see how I got that, first

That's what I said.

right, wasn't paying too much attention so anyway

keep going..

Try to simplify my last equation.

$-y + 2 = x$

No, $-y+2=\frac1x$.

errr... how come it's like that? didn't you multiply out the 1 from the fraction?

On the right, $-1(-\frac1x)=-(-\frac1x)$ — double negative — $=\frac1x$

Sorry, typo

Fixed

right. -1 * -1 = 1

You're confusing numerator and denominator, I think…

…stuff would only cancel if I multiplied by the denominator, x.

> Since $x\frac1x=1$

but you didn't multiply instead you got rid of the negative

I multiplied by $-1$ to get rid of the negative, yes.

$-y+2=\frac1x$ now. Should we continue, or are you still confused?

this might sound really basic but whenever there is a fraction with a negative in the middle e.g. $-\frac1x$

> Note that $-\frac1x=\frac{-1}x=\frac1{-x}$

so when there is a negative in the middle of the fraction, it applies to both numerator and denominator

No

Yeah, but remember that $\frac{-1}{-x}=\frac1x$. If you have two, they cancel.

so it applies to just the numerator

It is not at the same time

What isn't?

7 marks, haha

$-\frac1x=\frac1{-x}$, right?

@Rememberme

wait, can you show me the full working for this with steps: $y=2-\frac1x$

Yes but minus cannot be both in numerator and denominator at the same time

that way I can understand and let you know

Like I said.

@Rememberme

I didnt read the full transcript@columbus8myhw

Steps do far:

$y=2-\frac1x$

$y-2=-\frac1x$

$-(y-2)=-(-\frac1x)$

Take minus and reciprocal

$-y+2=\frac1x$

You OK so far, @F4z?

yes, keep going

$2-y=\frac1x$. See why?

@F4z take the reciprocal now

x=1/(2-y)

> $-y+2=(-y)+2=2+(-y)=2-y$, since order doesn't matter in addition

there we go

Addition is associative as well as commutative (abelian)@F4z

So, $2-y=\frac1x$. Taking reciprocals:

$\frac1{2-y}=x$

@F4z, you OK?

sorta, I'l trying to keep up

fractions and powers really screw me over with rearranging but i'll get there

Its just simple manipulation@F4z

@DiscipleofBarney Hahahahaha

Reciprocal basically means "flip upside down the fraction". Like, turning $\frac23$ into $\frac32$

i know that

hi!!!!!!!@Incurrence

Since $x=\frac x1$, it's reciprocal is $\frac1x$

@Rememberme How are you

And vice versa

Great!!!!!!!Just proved some very difficult theorems@Incurrence

@Rememberme That's awesome!!!

@Incurrence I think my next name will be "7 marks" or "Opportunity for 7 marks" (although I have been thing "The Geometers Agenda" )

@DiscipleofBarney Oh you plan to change it periodically xD?

@Incurrence but i am still having problems with abstract algebra

@DiscipleofBarney Does your real name have initials PP for real?

@Rememberme So am I

@Incurrence Not sure. Yes it is, I am also thinking of changing it back to my actual name (that is the name I use on MO)

Oh I guess it didn't change on MO, because you can only change so often, let me see if it has been long enough

I dont know when i will be able to start homomorphisms @Incurrence

My username is "Columbus Ate My Homework," though someone once told me he thought the hw meant highway for a sec

It has been long enough...

Columbus ate my highway. Omnomnom

@Rememberme You have been doing them for a long time I imagine

@Rememberme $e^{x+y} = e^x e^y$

what is one thing to remember other than "what you do on one side, you have to do on the other" when making the subject of an eqauation?

Well, remembering laws of fractions and negatives (and the logics behind them) is always good

But I guess one thing to remember is, get the term(s) involving $x$ all on one side.

Alone on one side, with no other terms.

@DiscipleofBarney I have a map $\phi(P_1)$ where $P_1$ is a permutation that does the following:

(The $x+xy=y$ one was trickier, since you have two $x$ terms…

…but it became easier when we factored out the $x$.)

$P_1 = \begin{pmatrix}1&2&\cdots&n\\\pi(1)&\pi(2)&\cdots&\pi(n)\end{pmatrix}\to M_\pi$ Where $M$ is a permutation matrix with $1$'s at positions only $i,j=\pi(i)$

I want to show this is homomorphic, which I can do easily with any given example

But I don't know how to compose $P_1 = \begin{pmatrix}1&2&\cdots&n\\\pi_1(1)&\pi_1(2)&\cdots&\pi_1(n)\end{pmatrix}, P_2 = \begin{pmatrix}1&2&\cdots&n\\\pi_2(1)&\pi_2(2)&\cdots&\pi_2(n)\end{pmatrix}$

Given that I am not sure how to say which element of $\pi_1(i)$ will go into which mapping in $P_2$

Basically there is no nice way with that notation, but since this is a permutation you can have the top part of $P_2$ to be $\pi_i(i)$'s and then the bottom rows to be $ \pi_2 ( \pi_1 (i))$'s @Incurrence

Hmmm

But the top part dictates the row

Then the composition (if I am doing it in the desired direction) would be $i \mapsto \pi_2( \pi_1(i))$

Oh wait

Yes, that seems right

Oh y es

That seems good, thank you

And showing this map is surjective seems really strange, pretty much I just take a $M_\pi$ and say it has values $1$ at all $i,\pi(i)$ and zeroes elsewhere, and then I can just make my $P\in S_n$ from this

Is that acceptable? It pretty much just comes from the definition

The subscript $i$ in $\pi_i (i)$ should be a $1$

Basically yes

So you did end up showing it was a homomorphism?

Else I could use my already made injectivity proof

@DiscipleofBarney For this part? Yes I can now

Thanks for your help again

No problem

Why did you want to know if my initials were really PP? @Incurrence

@DiscipleofBarney Just out of curiosity, since if you are changing your names periodically, it may have been a fake name

It is sort of a funny name @Incurrence

@DiscipleofBarney Lol you saw that did you xD. I just meant it has double P

Which seems unlikely lol

I mean no offense about your name lol

@Incurrence Yah. No offence taken

lmao

Are your real initials AD?

[you're in my class!]

Haha, Maybe... :P

I wonder if you are my tutor and you are judging me for all of my questions xD

Are you going to mark me down?

@Incurrence By 7 marks

I knew it!

Do you know much about normalisers? @Disc

@Incurrence I know of them, I haven't done a ton with them though

I am not sure which, but I think I have to show that permutation matrices are a subset(or maybe a subgroup) of the normaliser of the diagonal matrices group

So $N_G(T)=\{g\in G| gTg^{-1} =T\}$

So all the elements that don't change any diagonal matrix(??)

Oh wait no

I misunderstood

$T$ is the subgroup of diagonal matrices

$G=GL_n(\mathbb{R})$

So I need to find all $g\in G$ that don't change $T$ under conjugation

Sounds like it

and then show that the permutation matrices are a subset of this group

Well I guess the normaliser might just be a set

So subset of the normaliser I will say

I guess if that is the problem...

Well it doesn't say, and it does have the subset symbol without saying either are a group

(and sometimes lecturers leave it out by mistake)

So yah the problem is asking you to show if the set contains the permuations matrices (or it is asking you to prove it does)

Hmmm well

Taking $pTp^{-1}=T$ anyway

Since it just permutes back

sooooo

Wait is that all it is?

wut

Does it

?

Doesn't it? Taking the inverse permutation?

$PTP^T$

I'll prove it!!

assuming its right

Damn my claim is very false

I am doing some stuff related to small cancellation theory and it sort of looks like I am drawing stick wieners :D

Picture for proof?

xD

Sort of, I am just working through an example

Oh, I meant send me a picture for proof of said stick wieners lol

xD too legit

I wonder if I will get flagged...

Nah

That^ might get flagged lmao

Haha

So if $P\subset N_G(T)$

I just need to show that all $P$ congugate $T$

But I have proven this false...

How did you do that? What is an example?

$\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$ against $\begin{bmatrix}a\\&b\\&&c\end{bmatrix}$

What did it come out to be?

Underconjugation gives $\begin{bmatrix}a\\&c\\&&b\end{bmatrix}$

So if the P=P^{-1}(when it is its own transpose)

It swaps elements

Oh then that isn't a counter example, I think you are misunderstanding what the normalizer is

Probably

Its not saying that it leaves the elements unchanged, its just saying that it needs to keep the element in the same set.

That conjugation is still a diagonal matrix

It looks like I have a $P\in G$ that can't $PTP^{-1}=T$ so that means that $P$ isn't in $N_G(T)$

Oh

Ohhhh

Well that's weird notation

So $gTg^{-1}\in T$

Is what it means

The $T$ is the set of diagonal matrices, not a matrix itself

$N_G(T)=\{g\in G| gTg^{-1} \in T\}$

So it still means subset, not element

Oh I see

My diagonal matrix is still in the set