Mathematics - 2021-03-09 (page 2 of 2)

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7:00 AM

Buenas Noches @copper.hat and thanks for the help.

So I'm going over the basic idea right now not concerning myself with the $(1,2)$ vector but back to the $y=x$ scenario (ex b, on pg 26).

StudySmarterNotHarder

0

Q: Primality in terms of solutions in the units of $\Bbb{Z}[j]$ are the Gaussian integers?

$p \gt 2$ is a prime number iff the system: $$ x_0 \cdots x_{p-1} = u \\ u\overline{u} = 1 \\ $$ has no solution $x_i \in \{ \pm 1, \pm i\}, i = 1..p$, where $x_i = z_{i} - z_{i+1 \pmod p}$ and $z_i$ are $p$ points in $\Bbb{Z}[j]$, indexed modularly. Can this be proved algebraically or visually? ...

abstract-algebra elementary-number-theory prime-numbers roots-of-unity primality-test

Might be a new primality testing mechanism

What is $j$?

StudySmarterNotHarder

$j^2 = -1$

Because I used up $i$

Huh? So you're in the quaternions?

StudySmarterNotHarder

Nope, I just used a different variable for $i^2 = -1$

Because I used up $i$ as an index already :)

7:07 AM

Then $i$ is now undefined.

These are not indices.

StudySmarterNotHarder

Thanks, fixed in post

$i$ is an index, in my post

$j$ is the complex unit

I failed at twin primes, so now I'm at lower hanging fruit, lol

Don't use $i$ as an index. Keep it as $i$. Use another letter for your index.

StudySmarterNotHarder

@TedShifrin k, fixed in post

@TedShifrin do you think this would hold in $\Bbb{C}$ as well?

I.e. the $p$ points $z_j \in \Bbb{C}$

Doesn't make sense if you leave the land of integers. I am not thinking about the question, though.

StudySmarterNotHarder

7:22 AM

I think it would work in $\Bbb{C}$ as well, but with $x_j \in \{\pm i, \pm 1\}$ still, thinking of a loop of string with unit length segments folding into the grid

I was working with your explanation of what a reflection does. And applied it to the $y=x$ line. Doing that say I have a vector $(x,y)$. The vector orthogonal to it is $(-y,x)$, now according to your description I should be taking the negative of this. But doing that I get $(y,-x)$. WHat did I do wrong?

@TedShifrin

You are not following what I said. Project onto $(1,1)$ and find. what's leftover. That's orthogonal to $( 1,1)$.

7:44 AM

letting $\mathbf{x} = (x,y)$

$ \mathbf{x}^{||} = \frac{x+y}{2}(1,1)$

$\mathbf{x}^{\perp} = \mathbf{x} - \mathbf{x}^{||} = (x,y) - (\frac{x+y}{2},\frac{x+y}{2})$

$ = (\frac{x-y}{2},\frac{y-x}{2})$

7:57 AM

@TedShifrin orthogonal projections.

just to be clear (I don't think that was mentioned, though obviously intended)

yes @robjohn

but for the life of me I'm not getting the correct expression.

8:32 AM

@dc3rd $\left(\frac{x-y}2,\frac{y-x}2\right)$ is orthogonal to $(1,1)$.

8:47 AM

Since I hadn't heard back from you, I went and tried a simpler case to get a better understanding of what was happening. So I worked on what happens when reflecting in the $x$ axis (since I do know the solution, I wanted to just obtain it from the tools I'm learning now)

with that said I falter on something in my algebra, I just worked it out. And I'm close to my "boom" moment of enlightenment...

9:04 AM

Ok going over things I kind of came back to what I typed. So what Ted had mentioned earlier is that the part orthogonal to my line of reflection I would have to take the negative of it. In this case that turns the above into: $\left(\frac{y-x}2,\frac{x-y}2\right)$. I was hoping to obtain $(y,x)$. What went wrong?

@robjohn

9:14 AM

@dc3rd try subtracting twice the part orthogonal from the original

once the part orthogonal will give the projection, twice the part orthogonal will give the reflection

$(x,y)-2\left(\frac{x-y}2,\frac{y-x}2\right)=(y,x)$

hmmmm....intersting......I'm going to ask if my geometrical thinking about this is correct:

I get the orthogonal projection to the vector $(x,y)$. In my picture this vector is pointing in the "southeast" direction. Now what is the motivation behind subtracting it twice?

I should be getting to bed. I don't know what time it is where you are, but it is morning here.......the witching hours to be specific...or the end of them at least

9:34 AM

user image

consider the original point to the lower right

the projection is on the line

the reflection is twice the distance from the original point

so how does the idea of the "negative of the orthogonal projection" fit into this?

@dc3rd subtracting twice the original from the original...

so "twice" may not always be the case then. That is just a scalar factor and there could be situations where I may have to subtract more?

The original point is the projected point plus the perpendicular part. The reflected point is the projected point minus the perpendicular part.

@dc3rd no

I see what you mean...it is the reflected point so based on your explanation on the distance form the original point I understand that part now.

This was all constructive.....I really hadn't been thinking about projections in the right way up to this point. I'm going to leave it at here for now and get some sleep and ask some more questions about it this afternoon. Thanks for the clarification @robjohn

9:52 AM

@dc3rd I've updated the image a bit

I'm still here and just noticed it. :)

sleep well

projected sheep don't jump very high at all

10:51 AM

What would be the range of convergence of the variable $s$ for the sum:
$$\displaystyle \lim_{x \rightarrow \infty} \left( 1 - \sum_{2 \leq a \leq x} \frac{1}{a^{s}} + \underset{ab \leq x}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{1}{(ab)^{s}} \right)$$

I have the feeling I should ask this on main instead, because not everybody in the chat room may feel comfortable discussing it here. On the other everything is recorded also in chat like on main so maybe it is okay to ask it here.

I am hoping for a breakthrough.

11:13 AM

@MatsGranvik what range do you suspect?

@robjohn I expect it to converge in the whole complex plane except at singularities like $s=1$

$s=1+100i$ seems to converge.

well, that would converge since the series for $\zeta(s)$ converges there, too.

12:01 PM

0

Q: Necessary condition to be closed Irreducible subspace

Let $X=\text{Spec}(R)$ where $R$ is a commutative ring with unity. If $Y\subset X$ is a closed subspace that is irreducible, then $Y = V(P)$ for some $P\in X$. Proof. Since $Y$ is closed, $Y = V(I)$ for some ideal $I\subset R$. Need to show $I$ is prime ideal. Let $f,g\notin I$. Then consider $...

abstract-algebra commutative-algebra

Anyone can answer this question?

2 hours later…

1:38 PM

@TedShifrin: I tried the distinct Egyptian fraction trick to expand $5$. It consisted of $99$ terms and the last one was $142548$ digits long.

if E(XY) = E(X)E(Y) what can we say about A and B?

@jeea A comes before B in the alphabet

sorry X and Y lol

you need to specify a bit more

Ah

In general not much

1:41 PM

X and Y are random variable and E is Expected value can we say X Y are independant

@jeea no. If $P(A\cap B)=P(A)P(B)$ then $A$ and $B$ are independent

written in conditional form: $P(A|B)=P(A)$

but if X Y are independant then E(XY) = E(X)E(Y).. so I thought reverse also looks true

It isn't

2:14 PM

If X, Y are independent, then...
$$\begin{align}E(X)E(Y) &= \left(\sum_i x_i P(X = x_i)\right) \left(\sum_j y_j P(Y = y_j)\right)\\
&= \sum_i \sum_j \left(x_iy_j P(x = x_i) P (Y = y_j\right)\\
&= \sum_i \sum_j x_i y_j P(X = x_i, Y = y_j)\\
&= E(XY)\end{align}$$

where am I wrong @Astyx

@jeea yes for independent X,Y then E(XY)=E(X)E(Y)

Sorry I got the mistake

I understand now @Astyx

1 hour later…

3:30 PM

But there's some condition which works

Of similar spirit

Hello

If E[exp(itX)exp(itY)] = E[exp(itX)]E[exp(itY)] for all (real) t, then X, Y are independent

Can I ask if there is a way to see who answered my question, but deleted later?

I think one should like at this, that single integral alone doesn't determine a function. I think this is similar

Just because in some singular instance expectations agree doesn't mean anything, we'd have to have some kind of parametrized family of those, like above

4:09 PM

i think that moderators, perhaps with a certain level of reputation, can see that. nothing is ever really deleted. but mere mortals do not have access to this information.

4:20 PM

answering and deleting seems less common than asking and deleting (which is annoying). when i've seen answers disappear, it seems that the answerer has discovered some kind of problem with the argument and maybe there's no fix.

i don't know what the "polite" thing to do is in that situation. is setting the whole answer in strikeout and saying "whoops this didn't work" better than deleting? thankfully, i'm never wrong about anything so this is a hypothetical for me.

I recently deleted an answer of mine, because I realized it was wrong shortly after posting it

it sucks to be fallible

4:39 PM

pauca sed matura is my motto. i got it from some guy somewhere. my answers are gems.

i'm also very humble about them, which is rare in mathematics.

gauss is my mathematical great-great-(i think about ten or eleven of these)-grandfather. it runs in the family i guess.

i'm gauss.

@leslietownes

It is fine now. I reached a moderator, and I am just waiting for now. Thanks for giving me an opinion about answering and deleting!

me deleting one of my answers would be like those people who go to museums and try to throw acid on the mona lisa. i'm not going to do it.

well, just one museum, i guess.

i'm fallable to a small degree

i also misspelled that word. to show people how imperfect i am.

@leslietownes Especially if others have devoted time and effort to provide an answer.

@leslietownes some guy... yeah

i guess for a lot of people, question asking is part of an attempt to have other people take an exam for them. so i understand why they do it, it just sucks.

indeed, it is really sad.

4:52 PM

@leslietownes when I see it, I undelete the question and lock it. It is cheating, if it is for homework or an exam

my one non-A in graduate school was a class where there was a take-home exam, and i went to get a burrito, and about a third of my class was taking the exam together at a table. of course they got more problems right. i didn't snitch because who cares. but it was annoying.

@robjohn
Locking a question? How, may I ask??

in the abstract i lean toward the absolutist position that if you can get the information from the cosmos and the argument makes sense, then you win. this does give a leg up to people who just cut and paste, but that's the cold world that we live in. but then i think back to seeing those people taking our take-home exam together and return to reality.

and deleting things that people have responded to is just a terrible move. if you wanna cheat, just do it on your own time. there's certainly enough out there. don't bother people.

What do we get if we multiplied plane's normal vector with a ray direction vector

multiplied in what sense? cross product?

4:55 PM

dot product

if the direction vector lies in the plane, then you'll get zero, because that's what a normal vector is. if the vector simply points to the plane (e.g. because the plane is not through the origin) you can get almost anything.

@soupless I toss a diamond at it and invoke an arcane spell.

Oh. I got it, thanks. Sorry, the head is running slow now

I'm reading a code to the determine intersection of ray with plane, and i'm trying to understand what he's doing so his code is written like this

float Vd = plane.m_vcN * m_vcDir;

// ray parallel to plane
if (_fabs(Vd) < 0.00001f)
return false;

i want to know what VD is

this is from 3d game engine programming book

my first introduction to 3d geometry was other people's code. it was not a good introduction. i'd like to think it is better now. a lot of video cards / video packages have primitives that keep people from reinventing the wheel. i expect that they still do.

5:00 PM

@leslietownes well, I hope it was a tasty burrito

i'm trying to understand the code on my own free time

@leslietownes lots of routines for fast lines and polygons

it wasn't even the good burrito place. it was just the one close to campus. that's ultimately what is so annoying about it.

@fido9dido Vd looks like it is supposed to be the dot product of the surface normal with m_vcDir

the fabs stuff is basically testing if Vd is zero, within some tolerance, because nothing floating point is ever going to actually be zero. i'm not sure of what the code is trying to do.

5:09 PM

i know i just don't understand the meaning of the dot product! what does this value represent , for cross product we get a perpendicular vector but what does this value represent

fabs is the c++ version of absolute

Use your $\vec x\cdot\vec y = \|\vec x\|\|\vec y\|\cos\theta$ to see what dot product means.

if a vector lies in the plane, or is representable as being parallel to the plane, its dot product with the normal is going to be zero.

in floating point, it might be some junk at the end that isn't quite zero.

If $\|\vec y\|=1$, it gives you the (signed) length of the projection of $\vec x$ onto $\vec y$. Nice triangle picture to draw.

don't draw any pictures at any time. it's a trap.

once you start, you can't stop, and then you wind up a geometer. the strong boy will "cut it out"

@leslie Maybe you're overdoing this over and over ...

What is cute once gets very tiresome.

5:13 PM

my wife has made this point. we strongly diverge on whether repetition makes things funnier. but noted and thank you.

ah thanks

I speak as one of the room mob bosses.

it's hard to come up with bits to do. i just want to do one of my, like, four or five bits. don't make me rehearse a whole new hour of material.

@leslietownes I am a strong proponent of drawing pictures.

Is it really drawing if the computer does it for you?

2

@leslietownes or worse, a physicist ...

5:17 PM

yes, that is the end of civilization.

Ten rounds of applause for Astyx :)

proof is a social process...

i remember talking to a physicist once about a functional analysis problem. he was using intuition imported from lie groups, where the topologies are simpler. i asked him, do you realize what we even have classes on functional analysis for? the unbounded operators class? all of those are classes precisely because you can't do what you're doing. he didn't care.

there was a picture to go along with it, too.

and he probably got the right answer.

the problem with therapy is that there aren't a lot of them who can dig deep into the weak-star topology on the dual of a vector space. and that's where a lot of my issues arise.

if the underlying model is changing quickly, it doesn't make sense to invest in formalising early on.

Mike Tyson put it differently "Everyone has a plan 'til they get punched in the mouth".

my outside member met with me to talk about my thesis. he was a physicist. he asked a few very uncomfortable questions about operator theory. then he told me that the government was using 9/11 to manufacture fear. they were testing emergency sirens at the time. then he told me about world war two and signed my thesis.

5:23 PM

Mike Tyson's plan was therefore not to get punched in the mouth.

Very rational guy

mike tyson in his heyday was something else.

@Astyx Does this make it better?

just that a physicist has a different focus...

@robjohn robjohn is fond of the software that wiggles his computer drawings.

You just applied an XKCD-like style didn't you :p

5:27 PM

@Astyx indeed

it looks so real!

needs a little more randomness :-)

i talked to the xkcd guy once on irc. he was looking for examples of math stuff that could be expressed elementarily but was difficult. i suggested denominators of partial sums of the harmonic series. never made it into a comic, which makes sense, because if you think about it there's nothing funny about them.

Nothing worse than when a note is just a bit off from the harmonic

Well, some of us don't have good relative pitch.

(Having been raised in a family of musicians, mine is painfully good.)

5:33 PM

@TedShifrin if it's a little high, they'll walk

even if you're playing with a relatively clean tone, a pinch harmonic automatically makes what you're playing sound like heavy metal, and is always awesome.

@leslietownes That's fundamentally repressive.

if i ever write a book, i'm going to use that as a quote on the back. "Fundamentally repressive." Robjohn.

i love that.

my daughter bought me xkcd 0 a few years ago. excellent. lots of math stuff.

including the mathematics of cunnilingus.

@copper.hat I love XKCD

@copper.hat I'd never seen that one before.

5:39 PM

@robjohn i always wonder what it would be like to spend an evening drinking with folks like monroe, larson, etc.

I got offered "Thing explainer", which is a book by him

@leslietownes I doubt that my Blues namesake ever used a pinch harmonic, though.

I missed his birthday by one day!

i don't think he did, no.

there are a few recordings of django reinhardt on electric where he gets close, but doesn't do it. i think the first appearance on an album was the ZZ top guy.

@leslietownes That late? hmm

django's best electric recording is "night and day."

it does seem late. i mean maybe it happened earlier but wasn't purposeful.

5:54 PM

math puzzle: can you find an n so that you can partition the n+1 binomial coefficients \binom{n}{i} into four subsets with equal sums?

i think a lot of guitarists purposefully avoided it. jerry garcia for example always soloed with an incredibly clear tone.

and duane allman. they would not have wanted that in the sound.

anush, can i, or can my computer? because that's what your asking. :)

you're. goodness. my father would kill me.

Pursuant to my discussion yesterday of ancient posts magically becoming active now, here is another example. @robjohn: Can you explain why this is suddenly active?

6:11 PM

that's weird

@TedShifrin It was bumped by the system: "This question has answers that may be good or bad; the system has marked it active so that they can be reviewed."

The system is nuts. This is happening with way too much frequency now.

If the OP didn't accept an answer or even see fit to comment 3 years ago, isn't it a bit futile to try to rectify that now?

Hey guys, I'm trying to prove this equation by induction: $\sum_{i=1}^{n} ix^i = \frac{nx^{n+2} - (n+1)x^{n+1} + x}{(1-x)^2}$

I got stuck in inductive step:
$$p(k) \Longrightarrow p(k+1)$$

if i just vote on one of these answers, will it go away?

@leslie Good question. It's for this sort of reason that I tell OP's with whom I'm communicating that they should accept an answer when they're satisfied. I don't know if it takes an acceptance. Perhaps robjohn knows.

6:14 PM

i sometimes upvote decent answers that don't seem great because i understand it makes it harder for the asker to delete a question.

maybe the system is just prompting us to do that.

I've stumbled on several of these in the last few days, but hadn't looked for these issues.

It's going to be a little bit of algebra, @Matheus. Did you factor out the $x$ to start with?

Please do not spam us. That is rude.

@TedShifrin I'm not sure exactly what the algorithm is.

sorry, I tried to edit my question

So why did you get stuck?

the inductive step focuses upon the (k+1)th term, where we have some notion of what the sum of the previous k terms is, and a good guess at what the sum will be. what goes wrong?

6:19 PM

I get this equation, but this isn't prove the inductive step:

$\frac{kx^{k+2} - (k+1)x^{k+1} [-(1-x)^2 + 1] + x}{(1-x)^2}$

Well, now you have to work out the algebra.

Ok

That's how all these proofs work.

Ok, this makes sense

thanks

You got it?

6:21 PM

not yet. I'm trying

Ahh, thanks

I assume you got it.

I couldn't see the result only seeing. Sorry, I call you for nothing kkk.

No problem.

4 hours later…

10:46 PM

What's a good book as an intro to Riemann surfaces?

you can try Miranda

it's more algebraic in nature, which I assume you'd want

lol that would indeed be preferrable

preferable?

idk English isn't my thing

Thanks :P

hey, @Thorgott - idk if you mess with much homotopy stuff, but have you ever seen the notation $\Delta(n,p)$ used? has something to do with simplices, and I think $p \leq n$ for it to make sense

nvm found it

11:02 PM

I haven't, but good you found it

@EdwardEvans Do you really mean Riemann surfaces or do you mean algebraic curves?

Forster is in German, but it's analytic. I know some standard sources, but English. Fulton's Alg. curves might be best for you.

what's the difference :P

I mean, the course name is "Riemann surfaces" and the research group focusses on various incarnations of modular forms

so probably Riemann surfaces more than algebraic curves

So that's analytic.

right? idk

I'll check out Forster too

11:10 PM

Griffiths has a beautiful elementary book, Intro to Alg Curves.

Medium between alg and analytic.

nise

and I guess the relevant prereqs for an intro are complex analysis and point-set topology?

Hey @TedShifrin, I finished the problem I was working on last night. I went beyond just "solving for a solution". I wrote down a "stream of conciousness" of how I thought things out....I'm going to paste it here. I would like some feedback to know if I'm approaching these things in the right way because I don't want to treat using the projection as some black box algorithm, but instead as a necessary tool applicable when needed....Anyways....here are my "thoughts":

You mentioned that "part of a vector parallel to $(1,2)$ stays the same, and the part orthogonal to that reflects, i.e., turns into its negative.". I superficially understood it, but didin't have the deeper understanding of what was going on. So instead of working on the $(1,2)$ case. I started off from reflecting in the $x$-axis.

So I drew a vector (you're common $(x,y)$ vector going in the north east direction and asked myself the question, "What happens to this vector when you reflect it in the $x$-axis?" Well to do that I had to actually get a mirror and reflect it. From that a vector …

This is where you're suggestion of the projection comes into the equation. So I take my vector $(x,y)$ and project it onto the $x$ - axis. Now from this I can get the orthogonal projection. With this vector I can now describe the reflected vector $(x,-y)$. Which will end up being $-\mathbf{x}^{\perp} = (x,-y) = \mathbf{x} - 2\mathbf{x}^{\perp}$. This idea I got from the drawing and you mentioning in one of your lectures of to think of vector subtraction as "what needs to be done to get back to one vector from another". Well then to get to the reflection I would end up having to subtract aga…

Of course in higher dimensions it would be the same ideas, but on a larger scale.

A question I had from this was trying to understand why we applied the projection? What I gathered from this was I was fundamentally not actually appreciating a vector being decomposed into its projection and orthogonal pieces. I then questioned "why is this needed?". From what I gather once we can express a vector into those two pieces we can "provide directions" to any other vector.

Is this the right way to think about these things?

You want to go back to section 1.2. If you want to project across the line spanned by $\mathbf y$, you write $\mathbf x = \mathbf x^\| + \mathbf x^\perp$. The reflection will be $R(\mathbf x) = \mathbf x^\| - \mathbf x^\perp = \mathbf x - 2\mathbf x^\perp$. Maybe it's more useful to write this as $\mathbf x - 2(\mathbf x - \mathbf x^\|) = 2\mathbf x^\| - \mathbf x$.

That way you don't need to separately compute the perpendicular part. But it's all the same geometry, just different algebra.

This works in higher dimensions, and you'll see that later.

Is my understanding of why projection is used as a tool correct as well?

11:26 PM

Yes, the orthogonal decomposition is the key idea. This is what works productively in higher dimensions (this will come up in Chapter 5).

And the normal equations — which are crucial with variance in statistics — are one of the ways of understanding this (that's what's in Chapter 5 because of least squares being calculus as well as linear algebra).

That's interesting and intriguing. I say that because we have all these "fanciful" concepts we develop but at the core foundation of it is "moving vectors" around

See I've encountered normal equations in the studying I've been doing with regards to linear regressions and them setting the equations up in matrix form, but now this is giving me a way to actually "see" what is happeneing instead of just the algebra

Well, we're looking for lots of examples of functions. These are among the geometrically interesting linear functions (projections and reflections across subspaces).

You'll see later that the normal equations arise by trying to solve inconsistent systems of equations (see Chapter 4) by finding the closest point for which you can solve (i.e., consistent).

This is a projection.

o/

@TedShifrin projection as minimization is one of those concepts which seems like it can't be as fruitful as it actually is

it's so simple but absurdly powerful

Well, the Pythagorean Theorem is powerful, isn't it? :D

lol

you're not wrong

11:33 PM

Don't tell leslie. It'll get him all wound up again.

lol

I suppose the adjacent leg is a projection isn't it?....

Yup, and other leg is the $\mathbf x^\perp$ we've been talking about.

Yesssss............because a vector can be shifted over since it is a distance....................the power of mathematics... :)

So using a projection also provides us with a "set of axis" as well no?

one concept within here is the notion of random variables forming a vector space.

11:36 PM

"since it is a distance"? Nah, we just agree that we can use what physicists like to call "free vectors" when we draw pictures of vector addition.

(not sure if that's the concept here)

You're right "free vector" was the term I was looking to describe this concept.

won't forget it now.

I never make a big deal of this, but some authors do. I also refuse to use different notations for points and vectors, but even some calculus books insist on that.

So I'm not nearly so pedantic as some people think.

well

But you can afford to be because you fully understand each of the concepts. I on the other hand am not at that luxurious position yet.....

11:38 PM

i think it's inoffensive to have a different notation for the point, e.g., point A. it's having different notation for the coordinates of point A vs. the vector pointing from the origin to A

as a pedant, I would like to distance myself from people who demand different notations for points and vectors

I actually think there is a lot of pedagogical insight in that distinction. But... I drop it very quickly to avoid cumbersome notation.

what they're doing is less pedantry and more futility

A derivative is a limit of a difference of points, which is a vector, whether or not you think of these as the same.

No, you're not :) I was pretty careful in the book and in lectures to say that when we write $\mathbf x$ we think of it as an arrow from the origin to the point with coordinates $\mathbf x$. Buy when we do addition we can shift the tail to the head of the vector to get the parallelogram picture.

LOL @Thor

@MikeM: I agree that it's important to understand the affine structure — that we can subtract points and add a vector to a point. But I'm not going to go there unless forced :P

11:40 PM

To me a vector is ultimately an instruction. If you start somewhere, it tells you how to move. This is why vector addition makes sense. You just do one instruction after the other.

So that's totally the free notion.

"The Committee which was set up in Rome for the unification of vector notation did not have the slightest success, which was only to have been expected." -F. Klein, Elementary Mathematics from an Advanced Standpoint, 1925

And the way to draw this is the "free vector" stuff. The reason arrows which may be translated to one another represent the same vector is because they're doing the same damn thing.

LOL @Thor

Yes, and some people want to start pontificating about equivalence relations.

i'm so spoiled by easy physics cases

(easy in terms of geometry, i mean.)

11:41 PM

There's a reason we introduce vectors and not just arithmetical manipulations of coordinates of points. One is geometrically meaningless and the only operations we ever bother with are the geometrically meaningful ones.

And it's good to sanity check that an operation you've come up with isn't junk.

Coordinatewise multiplication to get a product of vectors? Junk.

And now you're going to talk about tensors and pseudovectors :P

But if one just thinks of coordinates of points and these are random operations we come up with, why not just multiply coordinatewise?

Yes, I like to emphasize the going back and forth between geometry and algebra at the beginning of these courses.

Let's keep this at multivariate calculus level boys.....the insights are intriguing...

@TedShifrin yeah, vectors/pseudovectors is where the physics language starts to get inconsistent

11:43 PM

Yes, precisely. That's a great question. What things are basis-independent?

Everything I'm saying I say at the start of multi.

A bit less compressed.

is a vector something you add in combinations, or is it an object that behaves in a certain way under transformations

@MikeMiller, it was that sort of thinking that landed me in this mess....I had zero appreciation for any geometrical intuition and completely disavowed it

I mean, multiplying coordinate-wise makes sense

Day 1 and day 2 of the course are different from day 25.

11:44 PM

it's the standard ring structure on k^n, it just isn't geometry

Right. Agree.

It's not going to transform to anything when you change basis, though.

@Thorgott yeah, hadamard product is a thing

@Thorgott 😒

LOL

11:45 PM

And I can also take Kronecker/tensor products of matrices but I don't teach that in a linear course.

eh

Which reminds me. Let me link a question that ought to have a nice answer, but I don't think it does other than the standard one.

i take issue with that omission, somewhat, because so much of my use of linear algebra is in the QM context, and tensor product is absolutely important there

but i grant that that's specific to me

Here.

Kronecker product is useful

11:47 PM

Of course tensor product is important. But we already cannot cover everything that belongs in an introductory course. That isn't important enough for the common student to force out things that are.

Thanks Ted.

fair

I'm out of patience on this one.

Was that a sarcastic thanks? I'm never sure :P

Nah.

It's what I was typing.

11:48 PM

Anyhow, Semiclassic, since you want to think tensorially, look at the question I just linked. It's all about turning the linear map $\sum a_{ij}(e_i^*\otimes e_j - e_j^*\otimes e_i)$ into a $2$-vector.

LOL, oh.

So in this discussion of points and vectors, when is it of benefit to think of things in terms of points and why?....because reading again over the comments and actually thinking about everything I have learned and translating it to vectors it does "bring a picture" to my mind..

the other thing there is that "linear algebra" is a moving target. linear algebra for a high school student looks substantially different than a grad course

Because classical geometry is arguably about points in the affine plane, nothing to do with vectors.

I've never taught a grad course in linear algebra. I've taught tensors and exterior algebra briefly in my diff geo to use them.

Representation theory is probably graduate linear algebra.

In geometric problems points appear naturally. If I tell you to find the distance from a point to a line the natural thing to do is think about vectors PQ where P is your point and Q is on your line and try to minimize the length of these vectors.

You can rephrase this by replacing P with the vector OP and Q with the vector OQ but the picture gets all muddled.

Now you've got this random origin you're drawing arrows from for some reason.

@TedShifrin does "the standard one" mean the sledgehammer method?

i.e. just calculate it out

11:52 PM

Yeah, I should explain to @dc3rd that "the affine plane" means you have the plane without a distinguished origin. To turn points into vectors you need to specify an origin. But geometry is independent of the origin you choose.

The standard one is to go to the normal form with $2\times 2$ blocks, @Thor.

Day 1 for me. To talk about coordinates you make choices. Origin, what is right, what is up.

Vectors require less choice.

Yes that makes sense @MikeMiller

I have found after all the years that I have to spend more effort on subtraction than I thought when i started teaching. So many students randomly get the direction wrong on their answer. So I have to remind them of elementary school and doing $6-2$.

I was gonna make an AC joke, but your edit stopped me

(Coordinatizing vectors requires one to pin down "right" and "up" still but best to sweep this under the rug until change of basis in linear algebra.)

11:54 PM

@TedShifrin so by establishing an origin we can now "build up" a machinery to describe phenomena i.e by using vectors

By establishing an origin you can introduce algebra into it, @dc3rd.

Which makes leslie happy.

@TedShifrin Yup I draw a line too. What is 1+1, really? 1 is really "the place you get moving 1 right from 0" which is more easily described as an instruction: move one right. To get to 1+1 we move right once more. All can be drawn as little arrows above the number line. Same with subtraction and scaling.

People are familiar with vector pictures since they were kids. They just forgot.

Definition 5.1. The simplex category $\Delta$ has natural numbers as objects, and morphisms
$$
\boldsymbol{\Delta}(n, m)=\operatorname{Cat}([n],[m])
$$
A simplicial set is a presheaf on $\Delta$.

this gives me HoTT ptsd

its a fancy way to state order preserving maps

@MikeMiller and then to make it "esoteric" we give it a name like the taxicab metirc........

After the first day, I never draw $x-y = x+(-y)$. It's always answering $? + y = x$.

And still students will get it wrong until they die.

11:56 PM

figured it was simplical sets, but I don't mess with those

@TedShifrin and me

Yes, @Thor, of course "and" you.

You are more multifaceted than you like to let on.

since order preserving map = functor on ordered set

s

I'm like a nontrivial homology class, many facets but nothing inside

@user2103480 that is completely natural

Seems I'm going down the math history hole because it appears "introducing the algebra" is done to provide a symbolic way to describe the geometry.....@TedShifrin, so without the geometry there would be no algebra............leslie might not like that train of thought.....

@Thorgott no ITS A FUNCTOR

11:59 PM

@dc3rd Descartes

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