Mathematics

12:05 AM

hi, i had a question about determinants. i currently find them very unmotivated. of course there is the notion of "signed volume" and the determinant measuring how this volume changes under a linear map. but i find this a bit unsatisfying, firstly because even in the real case if one is not doing e.g. integration i'm not sure why i should be thinking about volumes, and secondly because this definition works for any vector space on any field (so no measure is required)

the other definitions i know of having to do with exterior powers are just as problematic since i don't really understand why one would look at those either

hopefully i'm not coming off as impossible to satisfy!

leslie townes

12:36 AM

det(T) is just the dual of the top wedge power of T. duh.

have you thought of it as a multilinear map, divorced from geometric stuff?

math.stackexchange.com/questions/668/… the answer from benabou is roughly what i have in mind.

everyone has different notions of what constitutes motivation. if you look on that page the majority seem to like the geometrical approach.

people who just like doing stuff with algebra and weird fields and things might like just focusing on alternating multilinear maps. you don't need to introduce exterior algebras to deal with them. but many find this approach too dry

P-addict

yeah! i have been understanding it using the map on the top exterior power. and i agree the geometrical approach is nice, sorry for the subjectivity of the question. maybe this is just a matter of getting used to alternating multilinear maps, but since i am not so experienced with a lot of math i have little experience with even the relatively common ones (e.g. commutators)

so i still find them unmotivated, but it might be one of those "motivated by its consequences" things

leslie townes

if it makes you feel any better, with many classes of operators on infinite dimensional spaces it is possible to prove the nonexistence of anything like a determinant. pick whatever properties you want to extend, they don't go through.

i think one route to intuition is, well, historically it came out of geometry and solving systems of equations, and once you have it, you end up mostly working with through its properties because the formula sucks. and then why not step back and just consider those properties.

like so many other things in math, it may have emerged out of the festering swamp called g--m-try.

P-addict

haha

i think one other thing which makes me curious is the fact that we can speak of determinants and their properties without reference to any choice of basis, purely as properties of linear maps, but this might be more a consequence of the preference for abstraction than anything

leslie townes

12:54 AM

the geometry monster walks the land. he doesn't need a basis, he just wants blood.

Thorgott

1:05 AM

the top exterior power is pretty much just the "vector space of oriented n-dimensional parallelipipeds in V" (where n=dim(V)). this is a one-dimensional vector space and a linear map f:V->V induces a map on this exterior power, which, by one-dimensionality, is necessarily given by multiplication with a scalar. this scalar is the determinant of f, i.e. the determinant is the factor by which the linear map scales parallelipipeds of the top dimension.

this is still geometric, but perhaps it makes it clearer that this does not really require a notion of "volume" or "measure"

P-addict

yeah, this is the definition i'm familiar with. it just seems a bit arbitrary, but this might just be my inexperience talking

copper.hat

1:21 AM

@P-addict you could look at an existing answer, selected entirely randomly math.stackexchange.com/q/2013886/27978

P-addict

definitely i like this answer/motivation, as the multilinear alternating map definition is an interesting algebraic characterization of the determinant (and is beautifully motivated by signed volumes), but maybe my question is more, "why care about multilinear alternating maps especially if we're not doing volumes?", or "why care about exterior powers?"

of course, i know one can, to the annoyance of people trying to help, always play this "why care about XYZ" game, and i'm not trying to do that! i was just wondering if we could go "one step further" than alternating multilinear maps.

of course personally i find linear algebra interesting and beautiful, which is more than enough reason to study it :)

Thorgott

1:36 AM

we care, because it tells us something about the behavior of the map itself

namely, a linear map is invertible if and only if its determinant is

P-addict

yeah, this is an interesting point. maybe there is some "natural" condition we could impose on a homomorphism $\text{End}(V)\rightarrow k$, $k$ the underlying field, which spits out only the determinant? this seems related to the invertibility property

oh, actually, would any such homomorphism preserve invertibility?

never mind that's not true

or maybe it is but the proof i had in mind doesn't work i think

Thorgott

the determinant is not linear, so it'd be a homomorphism $\mathrm{End}(V)\rightarrow k$ of multiplicative monoids

I'm not sure if there are good conditions uniquely characterizing it as such

$k$ has a lot of multiplicative endomorphisms, which do not seem to be easily ruled out using some normalization

pritchard

2:13 AM

Hi, if $\frac{1}{2}t_2\leq t_1\leq\frac{1}{2}$, how can we find the bounds for $\frac{2t_1-t_2}{1+t_2}$, i.e. what would $x,y$ be in $x\leq\frac{2t_1-t_2}{1+t_2}\leq y$? Thank you!

copper.hat

not enough information to give bounds.

pritchard

Sorry, also $t_1,t_2\in[0,1]$.

copper.hat

sorry, you had once chance.

jk

pritchard

oops, I forgot to specify sorry!

copper.hat

$x=0, y=2$

you always have $2t_1-t_2 \ge 0$ (and the denominator is always $\ge 1$), you could pick $t_1=t_2=0$ to get this.

pritchard

2:21 AM

Could you please describe how you figured that out?

Or the general method of solving this type of problem?

copper.hat

i don't think there is a general method..

Semiclassical

think about how to make the denominator big and the numerator small, and vice versa

pritchard

@copper.hat Hmm can't $2$ only be obtained if $t_1=1,t_2=0$ but then $t_1\leq\frac{1}{2}$ from the first restriction so this can't be possible?

leslie townes

i agree with copper. unless 'this type of problem' means something very specific

vitamin d

Good morning

leslie townes

2:29 AM

can you believe this guy? he thinks it's morning.

Semiclassical

well, for rational functions and linear constraints there's probably some notion of 'method' in the realm of cylindrical algebraic decomposition

but "make one side as big as possible and one side as possible" remains the basic idea

vitamin d

Good morning leslie, to you too. I will go to sleep now. See you tomorrow

leslie townes

that's my line.

Samyak Marathe

@Semiclassical Yes I tried, 3n + 13, and it spins on 13, 26 and 52

Semiclassical

i wonder what the largest cycle someone has found on an $n\mapsto 3n+b$ type map

Leaky Nun

2:46 AM

is this the Veritasium wave

leslie townes

no, that's the name of my prog rock band

copper.hat

@pritchard sry, should be $y=1$.

Koro

3:24 AM

An $m\times n$ row reduced echelon matrix (rref) $A$ can also be described as either all entries of $A$ are zero or there exists a natural number $r$ ($1\le r\le m$) and $r$ natural numbers $c_1,c_2,\cdots, c_r$ such that 1) $c_1\lt c_2\lt ...\lt c_r$, 2) $a_{ij}=0$ for $i>r$ and $a_{ij}=0$ for $j\lt c_i$, 3)$a_{ic_i}=\delta_{ij}, 1\le i\le r, 1\le j\le r$.

$\delta$ represents Kronecker's delta. I don't understand (3).

The problem with $(3)$ is: $a_{1c_1}=\delta_{1j}$ then $j$ can vary here hence we have on RHS a set of values but LHS is only one value.

leslie townes

the a_{i c_i} is the pivot entry in row i.

the sequence c_j spells out which columns the pivots are in.

Koro

hi leslie!

leslie townes

you have zeros below pivots and anything in a row that comes before the next pivot.

Koro

I know that leslie, the problem is with $(3)$ :-(

leslie townes

putting row echelon form in symbols is insanity. i get why people do it, but whyyyyyyy.

Koro

3:29 AM

I think $(3)$ is wrong.

leslie townes

no it's fine. it says you've got zeros above and below the pivot.

if there's a pivot in row i, column j, then if you run down the entries of the jth column you ought to get 0 except at the pivot.

Koro

But $a_{ic_i}$ for a fixed $i$ is a single value, then how can it equal $\delta_{ij}$ in which j can apparently vary.

leslie townes

let j vary. if it's not i you get zero.

just like if the matrix is in RREF you hop and skip down a column with a pivot in it, you get zero unless you land on the pivot.

Koro

hmm, hard to digest yet. so (3) is replacement for "A is row-reduced".

leslie townes

i agree that it's very hard to read, but unless i'm missing something, it's not wrong.

some people say 'row reduced' to mean only row echelon form and not reduced row echelon form. watch out for that.

it's capturing the RREF aspect.

Koro

3:33 AM

leslie, this definition is from Hoffman &Kunze's section 1.4

@leslietownes here by row reduced, I mean -all columns containing pivots have their non-pivotal entries zero.

leslie townes

ok. then yes. (3) is capturing that.

but beware that in the wild, sometimes "row reduce" only means "put in row echelon form" i.e. zeros to the left of pivots but not necessarily above and below them.

i looked in my lecture notes to see if i had a good slide on this, because i want to compete with ted in promoting my own offerings.

Koro

@leslietownes :)

@leslietownes i understand that. some people also use "pivots" and some use "leading non zero entry" to mean the same thing :)

leslie townes

i am among these people.

Koro

I also say pivots since i watched professor Gilbert Strang's lectures :)

15 mins ago, by Koro

An $m\times n$ row reduced echelon matrix (rref) $A$ can also be described as either all entries of $A$ are zero or there exists a natural number $r$ ($1\le r\le m$) and $r$ natural numbers $c_1,c_2,\cdots, c_r$ such that 1) $c_1\lt c_2\lt ...\lt c_r$, 2) $a_{ij}=0$ for $i>r$ and $a_{ij}=0$ for $j\lt c_i$, 3)$a_{ic_i}=\delta_{ij}, 1\le i\le r, 1\le j\le r$.

But still, I am having a hard time digesting how can a singleton value equal a set

leslie townes

so my notes spend several slides on row echelon form, with blanks where we would have done examples. then there's one slide saying what RREF is, two blanks for two examples, and then the slide says "The book calls this 'Gauss-Jordan elimination.' You can use it, or not. I will not separately test this topic and if you can put a matrix in row echelon form and solve systems of linear equations without it that is fine with me."

i shouldn't have put that in print.

Koro

3:40 AM

@leslietownes I understand this is the idea but it's not visible to me from the equations alone.

copper.hat

this is one of those things that should not be put into symbols.

Koro

@leslietownes haha, solving by elimination and not using RREF was fine with you :)

My college professor would have deducted grades if he asked about finding rank of a matrix in exam :(

leslie townes

it should be a_{j c_i} = delta_{ij}, should it.

Koro

no

copper.hat

somethings are best described pictorially

leslie townes

3:45 AM

my attitude was, the arithmetic mistakes you do with backsubstitution are really no different from what you do with matrix operations, so i'm not going to ask you to go to RREF unless you want.

koro one thing to keep in mind is that people tend not to formally reason with matrices in RREF, so it is not really that helpful to symbolically express what RREF is. it's a calculational tool.

there's the fun fact that the product of RREF matrices is RREF, but who RREF'in cares, as far as i'm concerned.

Koro

@leslietownes hmm but still it is painful for me to not be in a position to either confirm the definition as right or wrong. :-(

EM4

@leslietownes hiiii!

leslie townes

how about a_{ij} = delta_{i c_j}? that feels better.

Koro

@leslietownes Leslie, the copy I had had this $j$ made very small so I confused that with $i$ so the problem is solved.

leslie townes

oh.

shame on paper books, they don't let you zoom in.

Koro

3:58 AM

So we have $a_{ic_j}$

leslie townes

i wonder how many times they use that formalism later in the textbook.

Koro

@leslietownes it was a softcopy so it let me zoomed in. Sorry about causing confusion :(

copper.hat

it is really never used in $\mathbb{R}$ numerical work.

leslie townes

i was more befuddled than confused.

Koro

After zooming in a lot, I saw it was $j$ and not $i$

leslie townes

4:00 AM

depending on the age of the book, typesetting of subscripts was really bad about things like that. i vs. j.

this is one of the things that motivated knuth to develop tex

Koro

how did they type before tex/latex?

copper.hat

my non numerical exposure is limited, so my perspective slightly warped, but i do not understand the huge amount of space given to rref.

Koro

i think it's because it saves space (else you'll have to write so many +, - signs)

leslie townes

knuth wrote a good (free!) article about it. projecteuclid.org/download/pdf_1/euclid.bams/1183544082

he takes one journal and gives examples of how its typesetting evolved over the years, pointing out the good and the bad. very fun reading if you're into it.

also, tex hadn't fully matured yet, so there's a fun work-in-progress flavor to it

copper i don't know if you saw it, but when i taught linear algebra it got one slide that ended in "you can use it, or not."

Koro

Also, I have seen in some books written before 1950, they write "shew that" instead of "show that"

copper.hat

4:04 AM

the selectrix had a key for doing it

other mortals had to use a type-it

leslie townes

"shew" was hopefully gone by 1950. unless people were putting on an affectation.

the US patent office is horrible at typesetting math. they will fail to subscript and superscript, they will miss fraction bars, everything. usually because the attorneys submitting the work regard all of it as noise anyway and nobody reads it.

Koro

haha

leslie townes

they are pretty good with typesetting sequences of nucleic acids. ACGCTAGTCACCGCTTACGATCGATCCTGGA

copper.hat

AAAAGGGGGGG

leslie townes

put some uracil in there. AAAAUUUUUGGGGGG

Koro

4:07 AM

copper, T and C are also important I think

not sure though

copper.hat

i'm surprised levis doesn't have a pair with some ACGTU derived name

leslie townes

there are companies that will print out any sequence you want. you could actually get a bottle full of AAAAGGGGGGGG if you wanted it.

that would be a weird birthday gift.

copper.hat

its all greek to me

actually no, i understand a smattering of greek

vitamin d

good morning leslie

copper.hat

why just the lawyer?

returning in a week for a few weeks to brush up on my accent

love_sodam

4:48 AM

0

Q: Definition of the module of compatible $\mathcal{U}$ families of $\Gamma$.

In Spanier AT (p.325), there is a notion $\Gamma(\mathcal{U})$ called the module of compatible $\mathcal{U}$ families of $\Gamma$ where $\mathcal{U} = \{U\}$ is a collection of open sets and $\Gamma$ is a presheaf of modules on $X$. Definition. A compatible $\mathcal{U}$ family of $\Gamma$ is an...

leslie townes

guten morgen

vitamin d

5:04 AM

Genosse

copper.hat

6:20 AM

the totally tropical taste

leslie townes

did we lose the cold war? are people still saying genosse?

one time i was in stockholm and i was with some extreme left-wing friends, and a taxicab driver refused to drive us to the bar we wanted to go to. i didn't know at the time, but it was some kind of hotbed of swedish communsim.

probably in my FBI file now.

we were going there for the pricing of the refreshments, not the politics. i didn't plan on reading any marx while i was there.

whatever.

copper.hat

i think it is 'amusing' how many in the us view sweden as some sort of social nirvana

leslie townes

6:37 AM

i liked it when i was there although i don't think i would want to live there full-time.

my friend lived in the equivalent of a housing project. it was cool but he had like 750 sq ft to his name. i need a tiny bit more.

copper.hat

my main memory was people were not interactive and alcohol was prohibitive. returned quickly to copenhagen on the ferry where alcohol was less prohibitive.

leslie townes

copenhagen for me was like a time travel back to the 1970s. which i had not experienced.

they had smoking zones in the airport. like a little rope around where people could smoke.

that reminded me of restaurants in california circa 1985.

copper.hat

it had many similarities to ireland at the time except people were not interactive

leslie townes

except it wasn't 1985.

danes are distant up to a point and then beyond it there's no filter. there is no middle ground.

i have a gradient between 'i don't know you, go away' and 'here's the worst thing in my mind right now.'

copper.hat

yep, that would be my impression

leslie townes

6:50 AM

i inherently sympathize with any culture that focuses on the worst thing in one's mind

copper.hat

i will confess my innermost secrets to a random person but reveal nothing to those closest.

leslie townes

is that cultural?

i have this problem too.

copper.hat

i suspect something to do with potatoes

leslie townes

i met someone at a party once for about 2 hours, and dumped like half of my life's story on her. we're friends on linkedin and still check in from time to time.

she knows stuff about me that my wife doesn't know.

and i know an enormous amount of stuff about her life that her husband doesn't know. this is continents apart so there is nothing pervy about any of it.

it's just easier to talk to people if you know there won't be some kind of immediate interruption in your social circle.

user178758

I think it has something to do with the presence of social media and potatoes.

leslie townes

7:02 AM

that is a good observation.

Shobhit

Hello. I have a question. So in a game, if you win in the $i'th$ round you are paid $2^i$ dollars and to play the $i'th$ round you have to pay $2^{i-1}$. The probability of you winning in any round is 0.5. Whats the expected loss player incurs before profiting from the game. So if I win in round $i$ my win trumps all my losses, so I made a profit.

leslie townes

someone should do something about the potatoes.

Shobhit

So let X denote the money lost in round $i$ (such that I have never won before) and p(X) be its probability. So the expected money lost upto round $i$ would be $\frac{2^{i}-1}{2^{i}}$ and I took its limit to infinty. Am i correct?

Mary Star

7:53 AM

Hello!!

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$Mathematics$ Mathematics