Mathematics - 2024-03-15

Thorgott

00:06

none comes to mind, except they're ugly

Jakobian

what would this isomorphism mean in terms of algebraic geometry?

Thorgott

that Spec(R) is the affine line over itself

leslie townes

huh, it means the stack on the sheaf of sites is the spectrum of the motive, which means the grothendieck hierarchy at R collapses to pointlessness at grade 2

i really wanted to be the first person to answer that question, thank you thorgott

Jakobian

I do need to come back to reading Liu eventually

Thorgott

interesting point, leslie, you think we can interpret this in the $\infty$-category of motivic spaces?

leslie townes

00:17

thorgott: gosh, i sure hope so

Silly Goose

02:07

Suppose I have a positive semi-definite trace $1$ Hermitian $2 \times 2$ matrix $\rho$. Is it equivalent to maximize the largest eigenvalue of $\rho$ and to minimize the largest eigenvalue of $-\rho$?

I feel like yes, but I am skeptical of my thought. Since online there are many resources on minimizing the largest eigenvalue of a Hermitian matrix, but none on maximizing the largest eigenvalue of a Hermitian matrix.

leslie townes

02:55

silly: there's maybe a background question here, which is, if you just have one matrix, what could you be 'minimizing' over, and maybe the details of the minimization problem are not answered by this side issue, but: the eigenvalues of -rho are the numbers {-k: k is an eigenvalue of rho}. so the largest eigenvalue of rho is the smallest (in the usual sense of the order < on real numbers) eigenvalue of -rho, and not (in general) the largest one.

if you just think in terms of pairs of real numbers (e.g. diagonal hermitian matrices), the entrywise max of (3/4, 1/4) is 3/4, while the entrywise max of (-3/4, -1/4) is -1/4

if by "largest eigenvalue" you mean "eigenvalue of largest absolute value" i guess it is OK

in wanting to push rho's 3/4 closer to 0 you are also wanting to push -rho's -3/4 closer to 0. those two desires are the same.

Silly Goose

So my actual minimization problem is something of the form $\min_{U \in SU(4)} 1 - \text{Tr}([\text{Tr}_2(U\rho U^\dagger)]^2)$ where $\rho$ is Hermitian $\text{Tr}\rho = 1$ and $\rho$ is positive semi-definite. Also I am working over a hilbert space $\mathcal{H} \cong \mathbb{C}^2 \otimes \mathbb{C}^2$, so the partial trace is over the second factor.

leslie townes

i should have said "corresponds to" and not "is" in my first remark above, but hopefully it is clear.

that squaring after taking the partial trace might mess up any expected sign stuff when flipping between rho and -rho. is -rho even meaningful within your model, or is it just something you can do with operators that do fit in the model?

Silly Goose

I am trying to avoid dealing with the proper function above: For the function i mentioned above, I'll call it $\Phi(\rho)$, the absolute minimum value is $0$. So I'd actually like to solve $\Phi(\rho) = 0$. This is true iff the argument of the full trace is rank $1$ (needs to have spectrum $\{0, 1\}$). I am thinking this equivalent problem might be easier to solve; hence, my original question

which is essentially asking about how to maximize the largest eigenvalue of a hermitian matrix (since the argument of the full trace is just a $2 \times 2$ hermitian matrix

albeit a complicated $2 \times 2$ matrix :P

so the equivalence is something like $\Phi(\rho) = 0 \iff \text{Tr}_2(U\rho U^\dagger) \text{ has spectrum $\{0, 1\}$} \iff \max_U \text{spec}(U\rho U^\dagger) = 1$. Where the final iff is because $\text{Tr}\rho = 1$, so $\text{Tr}(\text{Tr}_2(U\rho U^\dagger)) = 1$, so the spectrum is $\{\lambda, 1 - \lambda\}$.

Hence, if I know how to maximize (analytically or numerically) the largest eigenvalue of a $2 \times 2$ hermitian matrix, I can solve this problem. Or, if I can find another equivalent problem to doing this.

okay maybe there is some promising stuff with some results on $2 \times 2$ idempotent matrices

leslie townes

05:48

@Jakobian i wouldn't have thought so either, but it turns out there is a reasonably explicit formula (for example, when n = 100 the smallest distance is 91 sqrt(2) - sqrt(16561), because of course it is). see math.stackexchange.com/questions/4881149/…

Soumik Mukherjee

06:17

Given an infinite subset $S$ of $\Bbb{N}$ and any polynomial $f\in K[t]$, does there always exist a polynomial $g\in K[t]$ such that if $fg=\sum_{i\geq0}a_it^i$ then $a_i=0 \forall i\not\in S$?

Okay so let deg(f(t))=m, We choose $m+1$ monomials $t^{a_1},...,t^{a_{m+1}}$ where a_i's are in S. So we can get m+1 equations of the form $t^{a_i}=f(t)g_i(t)+h_i(t)$ where the h_i's are of degree at most m-1. So we can do some linear algebra stuff to get $c_1t^{a_1}+...+c_{m+1}t^{a_{m+1}}=f(t)G(t)$. Right?

one potato two potato

@BalarkaSen So in some sense, geometric convergence and Hausdorff convergnece are the same, right?

Balarka Sen

07:25

@onepotatotwopotato Yep.

@Thorgott Nope, I don't remember

Pizza

08:50

$$\sum_{n=1}^\infty \frac{\log(n)}{n^3} < \sum_{n=1}^\infty \frac{\sqrt{n}}{n^3}$$

when I use the Direct comparison test, how can I find the term $b_n$ that respects the theorem?

Like in my case above I put $\sqrt{n}$ so the series converged, but I had seen from the internet that this $\sqrt{n}$ was used with the $\log$. How do I know what numbers I should put?

Soumik Mukherjee

@Pizza What is $b_n$?

Pizza

$\sum_{n=1}^\infty \frac{\sqrt{n}}{n^3}$

this

Soumik Mukherjee

Ah ok

Xander Henderson

11:51

@Pizza experience.

Jakobian

12:29

@leslietownes oh... this sucks. I was going to vote to close your question because of lack of focus but you included a more specific one :(

Jakobian

12:47

Joking of course

Thorgott

@BalarkaSen Theres a notion of generalized Whitehead product that generalizes the Whitehead product and proving the bilinearity for that (in the cases where it holds) is quite non-trivial, relying on Whitehead's (the other) theory of nilpotency in groups of homotopy mapping sets and its relation to L-S-category.

I believe the bilinearity for the ordinary Whitehead product admits a direct proof more geometric in nature, but I cant figure it out and the only reference that seems to have it is Whitehead's original 1941 paper and trying to parse its formalism kills my zoomer brain.

Jakobian

@leslietownes now you can think about what happens in $\mathbb{R}^m$

Say $\mathbb{R}^4$

Since $\sqrt{4} = 2$ it seems like we won't be so lucky

Jakobian

14:32

but its a serious joke so you aren't supposed to laugh

psie

16:11

I'm reading an introductory text to probability and the author makes the distinction between events and elementary events. However, the text never states a clear definition of the latter (nor of the former really).

Glancing at Ross' A First Course in Probability, he writes an event is any subset of the sample space $\Omega$. He doesn't seem to talk about elementary events. Feels clearer that way...I do not see why one would like to separate the idea of an event and an elementary event, but again, I have not been given any definition of an elementary event, so I'm not really sure what it is.

Mad

i am reading a proof in my lecture about the negative definitness of a lie algebra and it being compact and semisimple, the proof goes as follows
" if the killing form is negative definit, then g is semisimple....."

is this true generally?( IE negative definit imply semisimple) and how would one prove it

en.wikipedia.org/wiki/Compact_Lie_algebra the claim is even made here

leslie townes

@Jakobian :D

Ryder Rude

@psie hi. an elementary event is a subset of the sample space containing a single element

psie

ah ok, makes sense 👍

Ryder Rude

but it's probably more common to define them to be elements of the sample space

Thorgott

16:25

consulting my old notes, the following things are true:
a) a compact, connected Lie group is semisimple iff its Killing form is negative-definite
b) a Lie algebra over an algebraically closed field of characteristic $0$ is semisimple iff its Killing form is non-degenerate

Ryder Rude

but it's the same thibg

psie

ok

leslie townes

psie one reason that some textbooks might not put too much effort into the distinction (or care about it generally) is that it very much depends on how you choose to model the sample space and not on more "probabilistic" stuff that you can detect via a probability measure or an expectation of something

Thorgott

the point in a) is roughly that, given a compact, connected Lie group, semisimple <=> centerless <=> negative-definite Killing form. either equivalence is non-trivial, relying on the Torus theorems.

Sahaj

hi guys

Thorgott

16:39

ah wait, you can avoid the Torus theorems in one implication

a Lie algebra arising from a compact (connected) Lie group admits an invariant scalar product. it is then Linear Algebra to show that a Lie algebra with invariant scalar product has negative semi-definite Killing form whose radical is the center. in particular, such a Lie algebra is centerless iff its Killing form is negative-definite.

ah, and you can show that a Lie algebra with invariant scalar product has no non-trivial abelian subalgebras (which may or may not be your definition of semisimple) iff it is centerless by studying the corresponding root space decomposition

so it's basically only Linear Algebra after all, except for the part where you have to use Lie Group-Lie Algebra correspondence to establish that a connected Lie group is semisimple iff its Lie algebra is

i havent thought about this stuff in years, hence my disorganization

Jakobian

17:16

I just walked 8 kms like I promised I would

exhausting

I haven't ate anything today, just had coffee

Soumik Mukherjee

@Jakobian great!

Jakobian

I'll do it again tomorrow

Soumik Mukherjee

How much time it took?

Jakobian

I didn't notice when I started but I think I left somewhere around 16 and came back at 18

so 1.5 hour I'd say

biggest problem to me was figuring out a good path to go through

I hate that there's so many people walking now that it got warmer but I guess I'll manage

user 85795

Are you trying to lose weight :^)

Jakobian

17:23

I'm trying to lose weight but also not die from a heart attack when I'm older

everyone neglects the risk of heart attack, but I don't want that to be me

its what most people die of

user 85795

yeah, I've lost close to 50lbs with that in mind

Jakobian

I get really bored when I'm not on a new path but I'll guess I'll manage with that too

@Thorgott can we define embeddings in the category of Galois connections?

I've proved that any Galois connection can be "embedded" into a Galois connection of power sets, but I don't know how to formalize "embedded"

Soumik Mukherjee

@user85795 Just by walking?

Jakobian

By that I mean that I found order embeddings which make a particular diagram commute

Soumik Mukherjee

Recently I have seen very young people dying of heart attack

I don't know if this is a post covid thing.

Jakobian

17:35

No I think its just more access to sugary substances, lack of healthy lifestyle and pc culture

Soumik Mukherjee

probably that, most people are now leading an unhealthy lifestyle

Jakobian

well maybe covid does have something to do with it but I'm not the one to comment

yeah, sedentary

Thorgott

@Jakobian probably

Soumik Mukherjee

what is pc culture? being on the computer all the time?

Jakobian

@SoumikMukherjee I'm not going to define it, but thats part of what I meant yeah

Soumik Mukherjee

17:39

google shows political correctness lol

Thorgott

ncatlab.org/nlab/show/2-category+of+adjunctions

probably the 2nd option here for K = 2-category of posetal categories

Jakobian

@SoumikMukherjee I've used the term wrong then, I didn't know it was a thing already

@Thorgott yep that sounds like it

I mean, that would be the definition of a morphism between Galois connections, I think

but I was asking about embedding because I know its not precisely categorical

I suppose I should take something like the concrete category version of definition of an embedding on wikipedia

but instead of function I grab two functions

but then I wouldn't know why it commutes

okay whatever this is pointless to think about

Thorgott

17:58

the issue is that there's a bunch of choices involved in how strict/lax you want things to be at each level

in your context, you probably want something easier

namely a pair of order-embeddings s.t. one Galois connection restricts to the other

Mad

@Thorgott thank you

user 85795

@SoumikMukherjee I did it by eating a lot less.

(obvious, but it worked for me)

Soumik Mukherjee

Ohh

user 85795

Also, it took literally years.

~2.5lbs per year

Soumik Mukherjee

18:14

:O

Pizza

18:32

How can I understand which of the two terms goes to infinity faster?$$\lim_{n\to\infty} \frac{(\log(n))^3}{n}$$

Jakobian

@Pizza think about how slowly logarithm goes to infinity. In reality it goes much slower

third power doesn't change it

Pizza

$$\lim_{n\to\infty} \frac{\log(n!)}{n}$$

And I have noticed that this is not always true, as in this example

Jakobian

but its not logarithm

its logarithm of factorial

Pizza

Yes

Jakobian

yeah. So it makes sense it could behave differently

If I take $\log(10^n)$

well its also just $n$ but thats not the point

when I say logarithm I mean $\log(n)$

Pizza

18:39

And how do I understand when one thing goes faster than another?

Jakobian

you have to understand that $n!$ goes to infinity very rapidly

Pizza

How did I understand that $\log(n!)$ went to infinity faster than $n$?

user 85795

Plot them.

Jakobian

when you look at expression like $\frac{\log(n!)}{n}$ its not immediately clear what its behaviour is

but something like $\frac{(\log(n))^3}{n}$ is a different story

Pizza

My internet was crashed ... Now I'm reading

Jakobian

18:44

since $n!$ goes to infinity almost as fast as the exponential, we can't claim anything about limit of $\frac{\log(n!)}{n}$ without further argument

Pizza

@Jakobian If I use hopital can I understand it?

Jakobian

I wouldn't call that understanding

Balarka Sen

L'Hopital doesn't even apply

user 85795

Understanding will come from plotting them...

Jakobian

you can calculate the limit but its different from what you're asking

I suppose understanding would come from asymptotics of $n!$

to obtain rate with which $n!$ diverges to infinity we use the so called Stirling's formula

$n!\sim \sqrt{2\pi n}(n/e)^n$

Balarka Sen

18:51

That's backwards, because Stirling's is usually derived from looking at $\log(n!)/n$ and comparing that with the integral of $\log(x)$.

Jakobian

the formula itself comes from taking logarithm of factorial, $\ln(n!)$, and approximating the obtained sum $\sum_{k=1}^n \ln(k)$ by bounds with integrals

namely, $\ln(n!)\approx n\cdot \int_1^n \frac{1}{x} dx$

I know that Balarka. If you were to let me speak I'd say that sooner or later

Balarka Sen

Regardless of if you know that or not, we don't obtain the rate at which $n!$ diverges to infinity by using the so-called Stirling's formula. The so-called Stirling's formula is derived by trying to understand the rate at which $n!$ diverges to infinity.

Stirling's is a much more precise estimate. I think we can understand $\log(n!)/n$ more easily: $\sqrt[n]{n!} \geq n/(1 + 1/2 + \cdots + 1/n) \geq n/(1 + \log(n))$ by GM-HM and an elementary bound on the harmonic series. Taking log (using monotonicity), we get $\log(n!)/n \geq \log(n/(1 + \log(n))$, and the lower bound goes off to infinity.

Pizza

But if I use hopital on $\lim_{n\to\infty} \frac{\log(n!)}{n}$, so I get $+\infty$ , I can say that the above part goes to $\infty$ faster than the one under ?

Jakobian

also note: Just because Stirling formula helps you to understand what the limit is, doesn't mean its the most efficient way of calculating it.

Balarka Sen

L'Hoptial does not apply, @Pizza. $n!$ is not a differentiable function, it's a function only defined for positive integer $n$.

What do you mean by applying L'Hopital?

Jakobian

19:02

@Pizza whats the derivative of $\ln(x!)$?

Thorgott

$\log(n!)\sim n\log(n)$

Jakobian

you can use Stolz-Cesaro theorem but not l'Hospital

Pizza

@BalarkaSen Do the derivatives

Balarka Sen

@Pizza And that is nonsense for $n!$, because what does $(1/2)!$ mean?

Pizza

@Jakobian To be honest, I don't know

@BalarkaSen i dont know ...

Jakobian

19:07

definition of $x!$ for a real number $x$ is an advanced topic and you won't know of it

Balarka Sen

Does it mean anything?

There is no definition of $x!$ for a real number $x$.

Jakobian

but since l'Hospital is not available to you, you can't use it

Balarka Sen

Best not to conflate the gamma function with factorial.

Pizza

Now I'm going home and turning on the PC, because I haven't read everything you wrote

user 85795

Please do.

Pizza

19:08

I'll let you know.

user 85795

cya

robjohn

19:21

@leslietownes: have you heard from Ted since Saturday?

leslie townes

i have not. i was wondering.

robjohn

He seems to still be posting on main, but I haven't seen him say anything in chat since Saturday.

leslie townes

ah, so he is OK. maybe taking a well deserved break. although, should he ever want our answers to questions that nobody asked, or unsolicited advice on any aspect of his life, we'll be here. i hope he knows that.

i'm going through his videos now, working up a series of notes on how he can improve his teaching

come to think of it, maybe my notes would work best as my own series of reaction videos

robjohn

I'm sure that will help

Xander Henderson

19:36

@BalarkaSen I mean, you could replace $n!$ with $\Gamma(n-1)$ (or is it $n+1$? I can never remember...) and you get something with a derivative. </sarcasm>

Oh, shoot. Y'all already had that discussion. That's what I get for being late to the party. :(

user 85795

@leslietownes please post your reaction vids here for further input.

leslie townes

let a hundred flowers bloom

user 85795

from one drop of water

Balarka Sen

@XanderHenderson :-)

Pizza

@Jakobian ok but if it was like $(2n)!$ , how did i use that asymptotic estimate?

@Thorgott how did you do...

Xander Henderson

19:42

@Pizza Sometimes it helps to make really dumb estimates. For example, look at $\log(n!) = \sum_{k=1}^{n} \log(k)$. A silly upper bound is to note that $\log(k) \le \log(n)$ for any $k = 1,2, \dotsc n$. So $\log(n!) \le n \log(n)$.

A silly lower bound is to observe that half the terms in $\sum \log(k)$ are bigger than $\log(n/2)$. Hence $$\sum_{k=1}^{n} \log(k) > \sum_{k=n/2}^{n} \log(k) > \sum_{k=n/2}^{n} \log(n/2) = \frac{n}{2} \log\left( \frac{n}{2}\right). $$ (Ignoring fence-post errors)

Hence $$\frac{n}{2} \log\left( \frac{n}{2} \right) \le \log(n!) \le n \log(n), $$ which means that $\log(n!)$ is roughly $n \log(n)$ (for large enough $n$).

Again, these are very rough estimates, but sometimes you don't need a sharp estimate---you just need something that will get you into the right league (doesn't even have to be in the same ballpark).

It is only when the really dumb estimates fail that you go looking for better ones.

Balarka Sen

Oh I love that

That's even simpler than GM-HM

Thorgott

yeah, Xander's argument is also what I had in mind

Pizza

@XanderHenderson but this can also be applied when $$\sum^{\infty}_{k=1}$$ ???

Xander Henderson

@Pizza How do you mean?

Pizza

@XanderHenderson above sum there is not $n$ but $\infty$

Xander Henderson

19:55

@Pizza Why would the sum be infinite?

I am using the fact that $\log(ab) = \log(a) + \log(b)$ and that $n! = n(n-1)\dotsb(2)(1)$ to rewrite $\log(n!)$.

Pizza

@XanderHenderson my exercise was a numerical sequence $\sum_{n=1}^\infty \frac{\log(n!)}{n}$

Xander Henderson

@Pizza I made no comment on that sequence. I was commenting only on $\log(n!)$.

But if $\log(n!)$ is bounded below by $(n/2) \log(n/2)$, what does that say about your sum?

Hint: it does not look all that great, vis-à-vis convergence...

Pizza

@XanderHenderson the series diverges

Xander Henderson

@Pizza Why?

Pizza

because the limit is +infinity and the series has positive terms

Xander Henderson

20:03

(No one really cares if the series diverges or not---you have a 50/50 shot at guessing right on that. What an instructor really wants to know is if you can articulate how you know that the series behaves the way it does.)

@Pizza What is "the limit"? There are at least two limits in the problem which may be relevant... Don't use pronouns.

Jakobian

@leslietownes LOL

Pizza

@XanderHenderson $\lim_{n\to\infty} \frac{n}{2} \log(\frac{n}{2}) = +\infty$

$\sum_{n=1}^\infty \frac{n}{2} \log(\frac{n}{2}) $ has positive terms

for the Direct comparison test $\sum_{n=1}^\infty \frac{\log(n!)}{n}$ diverges

since $\sum_{n=1}^\infty \frac{n}{2} \log(\frac{n}{2})$ diverges

Xander Henderson

Much better.

Thorgott

you don't even need a test to say that a series whose terms don't tend to $0$ diverges

Xander Henderson

But you should be looking at $$ \sum \frac{\log(n!)}{n} \ge \sum \frac{(n/2)\log(n/2)}{n} = \frac{1}{2} \sum \log(n/2) = \infty. $$

Or, even more simply (as pointed out by @Thorgott), if $a_n \not\to 0$, then $\sum a_n$ diverges. Period.

Pizza

20:15

ah , right it was divided by n because it wasn't just $log(n!)$ , but it was divided by $n$

but in this case why shouldn't I use that test??

Xander Henderson

@Pizza Why use a nuke when a fly swatter will work?

There is no reason that you can't use the direct comparison test, but you don't need it. If the general term of a series doesn't go to zero, then the series does not converge. Why make your life harder than it needs to be?

Pizza

I reduced the starting series with another one, therefore $a_n$<$b_n$ (where $b_n$ is the starting series), making the limit of $a_n$ and seeing that it had positive terms. Now I don't have to say that $b_n$ diverges because it meets that test?

Xander Henderson

Alternatively, you can note that $\log(n!)/n \ge \log(n/2)/2 \not\to 0$, hence $$\sum \frac{\log(n!)}{n}$$ diverges.

user 85795

Playing with nukes is fun!

Xander Henderson

@user85795 Careful! You'll put your eye out!

Pizza

20:20

yes, but does the starting series diverge because it respects that test, or am I doing something wrong?

Xander Henderson

A Christmas Story | You’ll Shoot Your Eye Out Kid | 4K UHD | Warner Bros. Entertainment

►

user 85795

Thnx

Xander Henderson

@Pizza The series diverges because it doesn't converge.

There are many ways of showing that it diverges.

One way to show that it diverges is to compare it to another series. A more elementary trick is to show that the general term does not go to zero.

The latter is probably simpler (indeed, how do you know that $\sum \log(n/2)$ diverges?).

Thorgott

citing a more complicated reason for why something is true when a simpler reason exists (particularly in a case where the complicated reason relies on the simpler one implicitly) is generally considered bad mathematical practice, unless you're trying to make a specific point out of it

Pizza

@XanderHenderson because the limit is infinite and the series has positive terms.

Xander Henderson

20:31

@Pizza Sure, but can't you say exactly the same thing about $\sum \log(n!)/n$?

Pizza

here I didn't know which term went to infinity the fastest

Xander Henderson

13 mins ago, by Xander Henderson

Alternatively, you can note that $\log(n!)/n \ge \log(n/2)/2 \not\to 0$, hence $$\sum \frac{\log(n!)}{n}$$ diverges.

Pizza

but in this case I would calculate the limit of $\log(n/2)/2 $ which is a smaller series than the starting one. Since we have seen that it diverges, now to say that the starting one also diverges I don't have to say that for this test if a minor series diverges then the one that has been minored also diverges?

@XanderHenderson

Xander Henderson

20:49

@Pizza If I can show that $a_n\not\to 0$, then that is sufficient to show that $\sum a_n$ diverges.

$\log(n!)/n \ge \log(n/2)/2 \not\to 0$.

Done.

I don't need to compare the series to something smaller, because I can compare the terms of the series to something smaller.

Again, you can use the direct comparison test for series. Indeed, we saw above how that argument goes. But that argument is less elegant.

Anyway, I have to get to the airport. Later.

user 85795

Cya

Pizza

Okay, thanks so much!

EE18

Is this true? $\sum_{i=0}^n \binom{m+i-1}{i}= \binom{(m+1)+n - 1}{n}$

Doesn't appear in my book but it seems to get used

Jakobian

Pascal's rule

In mathematics, Pascal's rule (or Pascal's formula) is a combinatorial identity about binomial coefficients. It states that for positive natural numbers n and k, where ( n k ) {\displaystyle {\tbinom {n}{k}}} is a binomial coefficient; one interpretation of the coefficient of the xk term in the expansion of (1 + x)n. There is no restriction...

@EE18

if you know Pascals formula you can prove this identity

its a telescoping sum

robjohn

21:30

@EE18 $$\binom{m+i-1}{i}=\binom{m+i}{i}-\binom{m+i-1}{i-1}$$

This is just a rewrite of Pascal's Rule, and it telescopes

Pizza

Does anyone have any pdfs on geometry and algebra exercises??

Jakobian

21:45

@Pizza amazon.com/dp/3834806765

why not both?

Pizza

but what exercises should someone usually start with?

diagonalizable matrix, maybe it's the toughest topic?

Jakobian

You start with prevarieties and work your way up to schemes

I'd say

(@Pizza its a joke)

Balarka Sen

@Pizza Read David A. Brannan's book "Geometry". Tons of exercises, some worked out.

Useful for learning mathematics in general.

Pizza

@Jakobian ...I was still trying to understand what "prevarieties" meant

Jakobian

22:03

thats also a joke

EE18

But robjohn if it telescopes shouldn’t I end up with two terms?

Jakobian

yes, but one of them is zero

(my answer was exactly the same)

EE18

Appreciated, sorry I am on mobile right now so missed some of this

Jakobian

I meant that I can also comment on it since it was mine as well

Pizza

22:24

is this definition of a system of generators of a vector subspace , good?

it's a set of vectors, each in the subspace, of which every vector in the subspace is a linear combination

Jakobian

yep

EE18

Ended up getting it, thanks again Jakobian and robjohn

Pizza

@EE18 did you write all this?

Jakobian

@EE18 too complicated

there is no need to split $\binom{m-1}{0}$ from the rest

EE18

22:55

I had to because my version of pascals identity requires i>0

Negative numbers not as yet defined

Thorgott

that is an unnecessary restriction

Joe

23:21

I am in desperate need of advice on how to draw an $\mathfrak a$.

Pizza

$$\begin{pmatrix}1 & 0 & 1\\\ 0 & 1 & 1\\\ 0 & 0 & 0\end{pmatrix}$$

Is this an example in $\Bbb R^3$ that is not a basis?

From what I understand, the third vector can be expressed as the sum of the first two vectors. So, these vectors are not linearly independent?

But I have a doubt, Can I use the matrix like this in the representation? Or do I have to separate each column?

Joe

@Pizza: Yes, that is not a basis, for the reason you have given. More precisely, the set of vectors $\{(1,0,0),(0,1,0),(1,1,0)\}$ is not a basis of $\mathbb R^3$ (it doesn't make sense to assert that the matrix you wrote down is a basis, since the matrix you wrote down is quite clearly not a set of vectors in your vector space).

It might help to look up the precise definition of a basis again.

By definition, if $V$ is a real vector space, then a basis of $V$ is a subset $E$ of $V$ which is linearly independent and whose linear span is equal to $V$.

On the other hand, in practice, when we are trying to work out whether $\{v_1,\dots,v_n\}$ is a basis of $V$, we might arrange the $v_i$ into columns of a matrix.

Thorgott

23:48

@Joe you've just done it!

Joe

Handwrite, I mean.

Pizza

@Joe So if I want to represent the basis of a vector space, don't I have to write everything inside a single matrix?I mean so it writes something like this: (),(),() then

Joe

@Pizza: No, you don't have to write it in a single matrix. But what is a basis, to you?

Thorgott

tbh, I just draw a square and then extend the right vertical side slightly further to the bottom

Pizza

@Joe 1) None of the vectors in the basis can be expressed as a linear combination of the other vectors in the basis 2) Each vector in vector space can be uniquely expressed as a linear combination of the vectors in the basis.

So practically, a basis is a set of vectors that satisfies these two properties?

Joe

23:55

@Pizza: Yes.

@Pizza: You need to remember what kind of mathematical object a definition refers to, not just the properties it satisfies.

Pizza

Mm yes

So how can these examples be represented graphically??

Because then it is wrong as I wrote

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$Mathematics$ Mathematics