Mathematics - 2024-06-11

EE18

01:16

I have three questions spanning almost two pages from Hoffman and Kunze that I read this morning

Would it be OK to post, or better on main site? I know they hate pictures there (understandably) but not sure how to ask without it

EE18

01:26

I tried to compress it to 3 notes. My questions are whether my justifications in the first two places are correct, and also whether my understanding of the proposed procedure for finding g.c.d (in (C3)) is correct

The unhelpful examples I allude to there can be seen below. It seems like they magically know how to combine the different generating polynomials in such a way as to get a monic polynomial which divides all generating polynomials (and then this is the gcd by uniqueness)

Agh I realize the above may be impenetrable without the passages

Fee; free to ignore all of the above. None really urgent in the end, I think I have a decent hang on it. But just not totally sure

Thomas Finley

05:31

0

A: Why is $x^2+x+1$ a factor of the minimal polynomial over $\Bbb R$ just because $x^2+x+1$ is a factor of the characteristic polynomial?

I will prove a much more general statement: Let $T$ be a linear operator on an $n-$dimensional(,$n\geq 1$) vector space $V$. We assume $V$ is vector space over $F=\Bbb R.$ Additionally, let $f(x)$ and $p(x)$ be the characteristic and minimal polynomial of $T.$ Then, $f(x)$ and $p(x)$ have the sam...

2 days ago, by leslie townes

maybe put this problem aside and come back to it in a week

@leslietownes I have added an answer to this. But I need to be sure whether I understood the thing or not. Can you please have a look at it?

I have one doubt in my own solution though, which makes me wonder about it's validity.

The problem is, in my solution I wrote a statement: we may consider $V$ to be a vector space over $\Bbb C$ with the operations of addition and scalar multiplication being the same as the ones that were defined when $V$ was originally considered a vector space over $\Bbb R.$ But can we really make this consideration?

If we can make such a consideration, I wanna know how?

Jakobian

08:23

@ThomasFinley No, for example V = R doesn't extend to a C-vector space

Thomas Finley

08:51

@Jakobian Yeah, I deleted by answer. Thanks for pointing it out!

Thomas Finley

09:13

0

Q: Prove that the canonical mapping from an infinite dimensional vector space to it's double dual is a one-to-one mapping.

What is the canonical correspondence from a vector space V to it's double dual $V^{**}.$ Prove that this correspondence is one-one.($V$ need not be finite dimensional) I tried solving the problem in the following way: Let $V $ be a vector space over $F.$ For any $v\in V$ we define a mapping $\hat...

Need some help with this :/

Thomas Finley

09:46

@SineoftheTime Can you please take a look at this?

psie

11:19

I've read the proof of the inverse function theorem in Spivak's book, however, he states it only with the conclusion that $f^{-1}$ is differentiable, and not continuously differentiable. In the back of the back though, he writes that the continuously differentiability easily follows from the formula $$D(f^{-1})_y=[Df_{f^{-1}(y)}]^{-1}.$$

He remarks that the inversion map is continuous since by Cramer's rule $(A^{-1})_{ji}=(\det A^{ij})/(\det A)$, where $A^{ij}$ is the matrix obtained from $A$ by deleting row $i$ and column $j$. But is this really Cramer's rule? Aren't there a bunch of $\pm 1$ missing?

I don't really see why $f^{-1}$ is $C^1$ if $f$ is. I don't see what is being composed with what. I know the determinant is continuous, that I know.

EDIT: not "back of the back" but "back of the book".

Mats Granvik

11:57

$(1+1-27)-2 (8+1-27)+(27+1-27)-2 (1+8-27)$
$+(8+8-27)+(27+8-27)+(1+27-27)+(8+27-27)$
$-2 (27+27-27)-2 (1+1-8)+(8+1-8)+(27+1-8)$
$+(1+8-8)+(8+8-8)-2 (27+8-8)+(1+27-8)$
$-2 (8+27-8)+(27+27-8)+(1+1-1)+(8+1-1)$
$-2 (27+1-1)+(1+8-1)-2 (8+8-1)+(27+8-1)$
$-2 (1+27-1)+(8+27-1)+(27+27-1)$
$=0$

Except for coefficients $1$ and $-2$ they are all cubes and they sum to zero.

Ryder Rude

13:22

hi

leslie townes

@psie looks like cramer's rule to me (the minus signs are buried inside the determinants themselves, should you choose to expand them out)

psie

@leslietownes the above uses slightly different notation, but you see there a lot of factors $1$ and $-1$ in front of the determinants. I think Spivak has omitted those, but maybe doesn't matter

leslie townes

13:38

psie: the adjugate (or i should say "that adjugate," as there are different things floating around by that name) is a matrix. spivak is talking about a matrix entry

it may help to be sensitive to the fact that statements about an "adjugate" may require non literal translation into statements about other matrices formed in ways that are related to determinants

so i would check how spivak defines those things if you are mixing references

(note for purposes of spivak's larger point, whether the (-1)^something is there or not would not affect the claim about the continuity of the resulting formula in the matrix entries)

psie

right, ok, I don't think he defines the adjugate, so I'll just assume he made an omission there that doesn't really matter

looking at that screenshot, Cramer's rule should read $(A^{-1})_{ji}=(-1)^{i+j}(\det A^{ij})/(\det A)$ I think

leslie townes

14:00

'by some horrible but nevertheless explicit formulas, the entries of the inverse are continuous functions of the entries of the matrix' is definitely the best way of reading it

psie

yes :)

leslie townes

great example of textbook remarks that say stuff like 'determinants are more useful theoretically than they are practically'

it barely matters what the formula is, that formula will likely not be your first choice if you ever need to compute something

EE18

14:52

Quick question which I'm hoping to get help with. Am following along with Cantor's construction of the reals but there's a weird negation of a statement as part of a proof that I can't follow. In particular and with regard to the first and last paragraphs in the picture below, how does $r-s, s-r \notin \mathcal P$ imply that there is for each $N$ some $n$ such that $|r_n-s_n| < 1/N$. When I (naively) negate the definining condition for $\mathcal P$ I see that for each $N$ and $n_0$

there are some $n_1,n_2$ such that $r_{n_1-s_{n_1} < 1/N$ and $s_{n_2}-r_{n_2} < 1/N$. But how do I get $n_1,n_2$ to be at the same $n$, as the authors do?

Jakobian

@EE18 broken LaTeX

EE18

Argh OK reposting the last one right now:

there are some $n_1,n_2$ such that $r_{n_1}-s_{n_1} < 1/N$ and $s_{n_2}-r_{n_2} < 1/N$. But how do I get $n_1,n_2$ to be at the same $n$, as the authors do?

Jakobian

15:07

@EE18 we hate pictures as conveying information here too

@EE18 what is the book and what is your work

EE18

@Jakobian Book is Hoffman and Kunze. My work is the first picture from my personal notes

I can't really copy paste all that here without a million messages either

Jakobian

@EE18 I was asking about screenshots in both cases

EE18

Oh, second book is Amann Escher

Yesterday was Hoffman Kunze, today Amann Escher

Jakobian

sigh

that's not what I'm asking

which screenshots represent your work and which screenshots are excerpts from a book

EE18

@Jakobian The very first screenshot i sent, the one with the Latex and grey box and bad writing is my work

The next 3 are Hoffman Kunze

The last one (from today) is Amann Escher

Jakobian

15:14

alright

first comment I have, is that while its reasonable to say "we" when someone is proving something, I find it slightly odd to not refer to yourself simply as "I"

EE18

Ya I guess I just do that because of how math and physics books are usually written

force of habit

Jakobian

what I'm saying is that in those books its reasonable, but here its like listing ingredients you want to buy in a shop

unless there's multiple people buying those ingredients, this is slightly odd

alright so Corollary to Theorem 7 basically says that if $p_1, ..., p_n$ are non-zero polynomials then there exists a unique monic polynomial $d$ with $(p_1, ..., p_n) = (d)$ and if $q|p_i$ for all $i$ then $q|d$

a lot of which is adressed in Theorem 7, so the corollary is really about point (c)

EE18

I agree. So re my comment C2 we then call d the gcd thereof because that q has degree at most d then right?

At most deg d i mean

Jakobian

The uniqueness of $d$ is also part of Theorem 7, so really (c) is the only thing that this corollary adds

So "If $d'$ is ..." is irrelevant, so your work for proving (C1) need not to be checked since it doesn't matter

@EE18 $d = \gcd(p_1, ..., p_n)$ is a simply the least element such that $d$ divides each of $p_1, ..., p_n$. "Least" means in the partial order of divisibility, this transcends the notion of degreees of polynomials and what not

Here we see that if $d'|p_1, ..., p_n$ then $d'|d$, and $d|p_1, ..., d|p_n$. This means that $d$ is the infimum of $p_1, ..., p_n$ in this order

Of course, this is up to some equivalence because this is not quite a partial order, its a preorder, and you have to consider equivalence $x\sim y$ iff $x|y$ and $y|x$ to make it a partial order

As you may now, for polynomials $q_1, q_2$, $q_1\sim q_2$ reduces to simply $q_1 = c\cdot q_2$ for a non-zero constant $c$

This also means $\gcd$ is not defined by a single element, but by a family of elements differing by some non-zero scalar multiple

psie

15:38

I asked a very similar question yesterday. Yesterday I asked about when the map $f':V\subset\mathbb R^n\to M_n(\mathbb R)$ is continuous. I hope I'm not asking the same question now, but when is the map $Df:\mathbb R^n\to\mathcal{L}(\mathbb R^n,\mathbb R^m)$ continuous? There is a theorem in Spivak's, theorem 2-8, that if all the partials exist in an open set containing $a$ and are continuous at $a$, then $Df(a)$ exists, but existence is not the same as continuity, is it not?

Jakobian

@Jakobian this definition works for arbitrary ring $R$, as long as the relation on $R/\sim$ induced by division is a partial order. IIRC to consider those its enough to let $R$ be an integral domain.

but of course there is nothing really stopping you to define it this way even if it isn't a partial order...

either way, what I'm advocating is to think of division related issues in this way (this includes $\gcd$ and similar notions), there being some division-induced order

it might be a good exercise to go through the details of what I just said

EE18

@Jakobian why irrelevant? That’s part of the claim of the corollary right, and HK prove it too, I just want to fill in the gaps that I don’t see

Jakobian

@EE18 because Theorem 7 already says its unique

you don't need to reprove that

EE18

d’ is not monic here though

Jakobian

What they are proving is uniqueness of $d$. This is part of Theorem 7

The whole argument is irrelevant

EE18

15:55

In (C1) I am proving the part of the Corollary that begins with “Any polynomial satisfying…”

Jakobian

Oh I see

You're trying to prove that if $(d') = (p_1, ..., p_n)$ then $d' = c\cdot d$ for $c\neq 0$

EE18

Correct, which would be enough to show what we need given what we’ve shown about d

Jakobian

I don't really like your justification of (C1), because all you need to say is that from $(d) = (d')$ alone it follows that $d' = c\cdot d$ for some non-zero $c$. You know, clearly $d' = q\cdot d$ and also $d = p\cdot d'$, so $d = qp\cdot d$ hence $qp =1$ so $q, p$ are invertible scalars i.e. non-zero ones

now about (C3). The method as I understand it the authors have in mind is as follows. We take some non-zero $p\in M$, and we assume that we can either decide that $p$ divides every element of $M$, or we can find a polynomial $q\in M$ which $p$ doesn't divide. In the latter case, you take the remainder of division of $q$ by $p$. In finite amount of steps of this, you'll be able to decide that the obtained polynomial divides every element of $M$, because the degree gets smaller and smaller

Moreover - this is not only in finite amount of steps, but the amount of steps needed is also bounded by $\deg(p)$

@psie You're asking the same question, yes

$Df$ is just another notation for $f'$

similarly, $L(\mathbb{R}^n, \mathbb{R}^m)$ is, basically, just matrices

Existence and continuity of partial derivatives implies existence of the total derivative, but it also implies continuity (for which, which is reasonable, we also have to require $f'$ exists in a neighhbourhood iirc)

anyway, yes, there is a correspondence between continuity of $f'$ and continuity of $\partial_i f$ for all $i$

Sha Vuklia

16:28

Could someone explain how they arrive at the conclusion that the connected principal coverings are the connected coverings with the largest possible automorphism group? It should follow from what they said before (which I understood), but I don't see how.

Jakobian

@EE18 this seems a reasonable concern to me. I believe the answer is that this is not a contrapositive, but the usage that those are Cauchy sequences. Namely, we can let $|r_{n_1}-r_{n_2}|<1/N$ and $|s_{n_1}-s_{n_2}|<1/N$ and so $r_{n_1}-s_{n_1} > r_{n_2}-s_{n_2}-2/N > -3/N$. Hence $|r_{n_1}-s_{n_1}| < 3/N$

Sha Vuklia

I don't understand what "coverings with the largest possible automorphism group" really means

EE18

17:54

@Jakobian am at work now so can’t quite respond but I will take a look when home this evening. Thank you so much for taking all that time

psie

20:25

@Jakobian how can one prove this, the continuity part? The existence of $Df(a)$ when the partials exist in an open set around $a$ and are continuous at $a$ is a theorem in Spivak's and that's clear. But I'm unsure about the continuity of $a\mapsto Df(a)$ when the same assumptions hold.

Jakobian

20:37

@psie one way would be to notice that $Df(x)$ for $x$ in domain of $f$ is a matrix consisting of $\partial_i f(x)$

now if $Df$ exists in a neighbourhood of $a$, and $\partial_i f$ are continuous at $a$, then this follows from your basic theorems about continuity of functions into products

psie

right, and we want to look at $Df(x+h)-Df(x)$ as $h\to0$

Jakobian

I mean, what I had in mind is that $x\mapsto Df(x)$ is a map into a space of matrices, which topologically is a product space $\mathbb{R}^{m\cdot n}$

psie

ok, interesting, and what did you mean by "continuity of functions into products"?

Jakobian

and compositions with the projections $\pi_{ij}:\begin{bmatrix} a_{11} & ... & a_{1m} \\ ... & ... & ... \\ a_{n1} & ... & a_{nm}\end{bmatrix} \mapsto a_{ij}$ are just your standard partial derivatives

leslie townes

if the map has 'components' its continuity can be assessed in terms of them

psie: jakobian began answering that just before you asked it :)

psie

20:44

ok, I see :)

Jakobian

@psie $Df$ is continuous at $a$ iff $Df\circ \pi_{ij}$ are continuous at $a$

and $Df\circ \pi_{ij} = \partial_i f_j$ or $\partial_j f_i$, one of those

while I'm not sure which one, it doesn't really matter

this has some generalization to Banach spaces

But you still break up $Df$ into finite amount of parts, just maybe not scalar-like, like partial derivatives

For multivariable calculus those chunks are just scalar functions

Juliamisto

22:18

Hello. I have a question for native speakers of languages that are not English. As per the alphabet of what language do you read Latin letters (or, to be more precise, 1-letter names of variables) in mathematical expressions in your native tongues? My native tongue is Chinese, and I, like many other Chinese people, read "a", "b", "c", "d", ... as "a", "bee", "cee", "dee", .... (Numerals such as 0, 1, 2, ... are not read as English words but Chinese words.)

leslie townes

i have to ask, is "z" "zee" or "zed"? :)

Juliamisto

Both variants seem to be widely used in China.

s.harp

Zed is objectively superior, but I personally only use zee

Jakobian

22:35

@Juliamisto "zet"

The alphabet is Polish alphabet which is Latin alphabet with few additional letters and without x and q

Xander Henderson

@s.harp "Zebra" is the best alternative.

Jakobian

We don't do the "ee" our "e" is like in the word ... well, "zed" lets say

a, be, ce, de, e, ef, gye, ...

the "ee" would be our "i" (sort of)

Juliamisto

I see. Thanks.

ee is /iː/ (using IPA).

Jakobian

Our i is /i/

leslie townes

my daughter is in a school that teaches her in spanish but we use english at home and it is hell on spelling. "no, that's spelled with an E. an ENGLISH E, not the spanish i. you know the spanish [sounds like A but means E]? that one. E."

Jakobian

22:47

So without the... elongating, is how I'd describe it based on feelings alone

Juliamisto

"a", "e", "i", "o", "u" in Pinyin (the most common romanization system for Standard Chinese) are /a/, /ɤ/, /i/, /ɔ/, /u/, respectively.

Jakobian

a, e, i, o, u for us is /ä/, /ε/, /i/, /ɔ/, /u/

Juliamisto

The foreign language most widely used in China is perhaps English. There are many people who know the English alphabet more than Pinyin.

That may be one of the reasons for which we read Latin letters as per the alphabet of English...

@leslietownes You live in the US, I presume?

Jakobian

23:09

@leslietownes that's because of the complex way the English speakers pronounce words, I'd say

There is no simple this letter is pronounced that, rather the pronouncuation of specific letters changes almost arbitrarily depending on the word

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