Mathematics - 2018-03-17 (page 1 of 2)

12:42 AM

According to en.wikipedia.org/wiki/P-adic_order, $\nu_p(a+b) \geq \inf\{\nu_p(a), \nu_p(b)\}$, with equality if $\nu_p(a) \neq \nu_p(b)$. Are any other properties of $\nu_p(a+b)$ known (perhaps an upper bound?), in terms of the prime valuations of $a$ and $b$?

Not necessarily restricted to $\nu_p(a)$ and $\nu_p(b)$.

Xander Henderson

@user76284 well, if $a=-b$, then $a+b = 0$, and so $v_p(a+b) = \infty$

so I am not sure what you are hoping for...

user76284

Yes, but what about a more general case?

Xander Henderson

Again, I don't know what you are hoping for...

For any fixed $a$ and any $N > 0$, we can find some $b$ such that $v_p(a+b) > N$

user76284

Right, but is there any more information we can get about $v_p(a+b)$ in terms of the prime factors of $a$ and $b$?

Xander Henderson

the only prime factor that matters is $p$

user76284

12:57 AM

We know for instance that $v_p(a + b) \geq v_p(a)$.

Xander Henderson

yes... and?

user76284

We also know $v_p(a+b) = \inf\{v_p(a),v_p(b)\}$ exactly, when $v_p(a) \neq v_p(b)$.

Xander Henderson

yes... and?

you can write out the $p$-adic expansions of both $a$ and $b$

user76284

What do you mean "and"? I'm asking if there is anything in the literature that says more about $v_p(a+b)$.

Xander Henderson

the valuation of $a+b$ will be related to the place where the expansions first differ

I don't understand what it is that you want to know...

there is a huge literature about $p$-adic numbers

user76284

1:00 AM

I want to know if we can say anything more about $v_p(a+b)$ than simply it being bounded below by $v_p(a)$ and $v_p(b)$.

Xander Henderson

read Koblitz, or Gouvea, or Roberts, or Artin and Tate

user76284

Do you have a particular reference in mind?

Xander Henderson

the best that you can say, in general, is that $v_p(a+b) \ge \max\{ v_p(a), v_p(b)\}$.

that is the general case

user76284

Of course, in terms of $v_p(a)$ and $v_p(b)$ alone.

Sorry, I should be more clear.

Leaky Nun

p-adic :D

user76284

1:01 AM

What I meant is, what if you know also, say $v_q(a)$ and $v_r(b)$?

Xander Henderson

likely nothing

user76284

For some (or all) primes $q$ and $r$. So not just their $p$-adic valuations.

Leaky Nun

then you can uniquely pinpoint that integer, I think

Xander Henderson

if you know $v_p(a)$ for all primes $p$, then you know exactly what $a$ is

Leaky Nun

sniped :P

Xander Henderson

1:03 AM

but knowing $v_p(a)$ for some $p$ doesn't tell you much about $v_r(a)$ for other primes $r$

Leaky Nun

and then you step into the region of profinite integers

Xander Henderson

mmm... profinite integers

Leaky Nun

which is, in some sense, a generalization of p-adic integers

Xander Henderson

there are also all of the local-to-global results

those are interesting

user76284

Yes, of course. In that case you literally do the addition and factorize it. What I mean is, what if you only know a subset of the prime exponents, or some other property about the prime factorization? Is there some relationship that does not rely on full knowledge of the prime factorization?

I think the appropriate area would be primes in additive number theory.

For instance, $v_2(3^n - 1) = v_2(n)+2$.

But that's just a very specific case. Are there more general results like it?

Leaky Nun

1:05 AM

hmm

but bear in mind, $3^n-1$ and $3^n$ are very far (even under the $3$-adic metric)

user76284

Yes, of course. Actually, I was also thinking about whether anything could be said about the sum in terms of the arithmetic metric.

arxiv.org/abs/0906.0632

Something tighter than the triangle inequality.

I guess I'm looking for general relationships between prime factorizations of two numbers $a$ and $b$ and the prime factorization of their sum $a+b$.

$a$ and $b$ don't have to be positive integers, they could also be positive rationals in which case some of the prime exponents are negative.

Leaky Nun

Abc conjecture

The abc conjecture (also known as the Oesterlé–Masser conjecture) is a conjecture in number theory, first proposed by Joseph Oesterlé (1988) and David Masser (1985). It is stated in terms of three positive integers, a, b and c (hence the name) that are relatively prime and satisfy a + b = c. If d denotes the product of the distinct prime factors of abc, the conjecture essentially states that d is usually not much smaller than c. In other words: if a and b are composed from large powers of primes, then c is usually not divisible by large powers of primes. The precise statement is given below. The...

related?

Xander Henderson

I was just looking for a similar link :(

I should have just accepted that Wikipedia is the source of all knowledge :P

Leaky Nun

sniped :P

Xander Henderson

yarp

but I'm okay with that

right... time to go put the child to bed

user76284

1:10 AM

Ok this looks related: en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szemer%C3%A9di_theorem

I suppose a full understanding of this problem would entail solving the Collatz conjecture, twin prime conjecture, and other such conjectures. I have to look for partial results.

By the way, is there any significance to the inner product between two numbers, treating their prime exponent sequences as elements of a module? Do you know if anyone has studied this?

The $L_1$ norm corresponds to the group-theoretic word norm if you treat primes as a generating set.

Xander Henderson

Dude... you're getting algebra into my analysis. :(

$\ddot\frown$

Daminark

Yeah let's do fractals

So let's say you have a fractal. And let's consider irreducible representations of its homeomorphism group... Wait fuck I got algebra mixed in again

Adeek

1:25 AM

hey @Daminark

can I ask a quick question ?

Daminark

Sir

*Sure

Xander Henderson

@Daminark I don't even know what a fractal is, dude...

Daminark

It's a set whose complement is a cofractal

Xander Henderson

...

MatheinBoulomenos

Isn't a fractal just an object with an associated zeta function that has non-real poles

that's obvious, isn't it?

Xander Henderson

1:34 AM

heh

Adeek

I am just replacing some professor to give class on non-abstract differential geometry.

so in ted's videos he defined non-abstract manifolds of dimension k as represented locally as graph of $C^{1}$ functions $\phi : U \rightarrow R^{n - k}$ for U open set in one of the standard coordinates k-planes.

why here we map to $R^{n - k}$ instead of just $R^{n}$ ?

@Daminark

hi @MatheinBoulomenos btw

MatheinBoulomenos

Hi @Adeek

Daminark

@Adeek it'll be best to reason through examples here

Adeek

I think the reason we did in this way in order to be able to define things as like the manifolds of zero set of differentiable function

I think that is probably the reason in order to get meaningful dimension for that probably

Daminark

The best example being, something which is globally the graph of a function

Adeek

1:42 AM

sure

WDUK

hi, i need some help on an integral. I have e^(1/3*x^3) and i tried u subsitution of 1/3x^3 but this gives an integral with both a u and an x in it: e^u/x^2 du so am wondering what other approach i should try

Daminark

Also the level set business is something to keep in mind, think about how the sphere is a level set of a function from R^n to R, and how its dimension is n-1.

Adeek

oh

okay so probably the reason is because of the level set business

Daminark

Well you'll be able to derive the equivalence out of the implicit function theorem

Adeek

ohh I see

I see so it probably comes out of the implicit function theorem

Daminark

1:47 AM

But it's also a good exercise to say okay, let $f:\mathbb{R}^n \to \mathbb{R}^m$ be a smooth function

What is the dimension if its graph as a submanifold of $\mathbb{R}^{n+m}$? Using any definition?

Adeek

wait no

it should be n

n is the amount of degrees of freedom

Daminark

Good, so now substituting what you want, you're good

Adeek

yeah I see that is good

thanks for the discussion

Daminark

No problem!

Also hey @Mathein!

MatheinBoulomenos

hey @Daminark

Adeek

1:55 AM

Btw @MatheinBoulomenos now I bought all the books that I am gonna use for next 2 years

I am gonna stop buying anything anymore :d

MatheinBoulomenos

ha, I'm buying too much math books, too

Daminark

Springer Sale was truly the best business decision

Adeek

I bought Algebraic Geometry and Arithmetic Curves, Several Complex Variables with Connections to Algebraic Geometry and Lie Groups ,Algebraic Number Theory,and Serg lang complex analysis.

those Plus the ones that I have should keep me busy for few years

@MatheinBoulomenos it is an infection haha

@MatheinBoulomenos This book seems very interesting "Several Complex Variables with Connections to Algebraic Geometry and Lie Groups"

MatheinBoulomenos

sounds good from the title, yeah

Daminark

Hopefully they're not lying about what they cover

Adeek

2:01 AM

yeah I sure hope not

it has 5 stars

Daminark

Oh it was a shitty joke about the title

"Lie" groups

MatheinBoulomenos

that's not pronounced like that

Daminark

True but imposing these kinds of conditions only limits what you can do

0celo7

2:27 AM

shit, springer sale rn?

Karl Kronenfeld

Surely a collection of people formed a "lie" group at some point in the past, for some reason or other.

Balarka Sen

this is a sophus ing dumb joke

Karl Kronenfeld

The Sophists in Ancient Greece!

Balarka Sen

Hi/bye @Karl Kronenberg

Karl Kronenfeld

Hey yo @Balarka

LegionMammal978

2:37 AM

Just curious, is there any method of FSM (with initial and terminal states) reduction which guarantees that originally accepting inputs will still accept, while disregarding originally rejecting inputs?

Balarka Sen

@KarlKronenfeld How's life

Karl Kronenfeld

@BalarkaSen things are going fine. How are you?

Balarka Sen

Good to know. I'm not terrible, but a bit pressed with exams

Karl Kronenfeld

Naturally. Hopefully they are going well nevertheless.

Balarka Sen

They haven't started yet but yeah I'm working on it. Thanks.

0celo7

3:20 AM

@BalarkaSen someone suggested I write a list/summary of all proofs of the Hodge theorem

seems a bit unproductive but would definitely be cool

Balarka Sen

I would read it

0celo7

I'm at like 6

one semi-original in the sense that I haven't seen it written or hinted anywhere

but not unexpected if you know this stuff

Madhuchhanda Mandal

4:17 AM

Hey

I have a number theory (probably) question

MatheinBoulomenos

@MadhuchhandaMandal just ask!

Madhuchhanda Mandal

Let {x} denote the fractional part of x. Now , I'm trying to develop an algorithm which finds integers n (in increasing order) such that {nπ} goes on decreasing. Here's an sample output of what I'm trying to achieve :

MatheinBoulomenos

hmm, no idea, sorry

Balarka Sen

Ideally one would start by finding a continued fraction expansion of \pi

Madhuchhanda Mandal

At present what Im doing is simply iterating through all integers and checking is if {nπ} is lesser than the least output printed

But my question is,

Is it possible Mathematically to predict for what integer n the fraction part of the product may be less than a particular value ?

Balarka Sen

4:35 AM

It's not hard to prove that there is such a sequence $n_k$ of integers such that $\{n_k \pi\}$ is decreasing, because all the real numbers of the form $\{n \pi\}$ for some integer $n$ are equidistributed in [0, 1]

Coming up with one would require to study a continued fraction of $\pi$, like I said

This looks like a good place to start

Madhuchhanda Mandal

@BalarkaSen Thanks a lot. I will Google Continued fraction

Is continued fraction of every decimal number possible ?

Balarka Sen

For sure, just do a Euclidean algorithm.

Madhuchhanda Mandal

Okay. Looks interesting. Lemme Google

Thanks

Balarka Sen

The one I wrote is not quite a continued fraction but a generalized continued fraction because the numerators aren't all 1's

Madhuchhanda Mandal

I see

@BalarkaSen Equidistributed means?

Balarka Sen

4:46 AM

Ah right so that means for any subinterval $[a, b] \subset [0, 1]$, the density of the sequence $\{n\pi\}_{n = 0}^\infty$ is $b - a$, length of the interval $[a, b]$.

So in particular for any subinterval of $[0, 1]$ there are lots (hence at least one) of real number of the form $\{n\pi\}$ for some $n$.

In particular for every $\epsilon > 0$, there is a $n_\epsilon$ such that $\{n_\epsilon \pi\} \in [\epsilon/2, \epsilon]$.

Picking a sequence of $\epsilon$'s tending to $0$, this gives a sequence of $n_\epsilon$'s such that $\{n_\epsilon \pi\}$ is monotonically decreasing to $0$.

It's easier to prove density than equidistribution, I just used the strongest theorem available. Look up "equidistribution modulo 1" for more on this.

In general there is nothing special about $\pi$, it holds for any irrational number.

I don't know how to construct such a sequence of $n$'s, though. This just proves existence.

Madhuchhanda Mandal

@BalarkaSen Okay okay.. I seem to get it now. Isn't it like uniform distribution of mass of a rod along its length and then claiming between points [a,b] mass is (b-a)* M/(L) ?

Where mass denotes number of points ?

Balarka Sen

Yes, exactly.

It means the sequence $\{n\pi\}_{n = 0}^\infty$ is uniformly distributed in $[0, 1]$, that's the correct intuition.

Madhuchhanda Mandal

Yes .. so it proves the existence of such a sequence

Balarka Sen

Quite right.

Madhuchhanda Mandal

I'm thinking how to develop an algorithm...🤔🤔

To extract the n's

Balarka Sen

4:57 AM

This I do not know. Interesting question, though.

Madhuchhanda Mandal

Yeah. Ofcourse. It's very interesting.

Perturbative

Yo everyone

Balarka Sen

Mystical greetings

Perturbative

If we take out the part in the definition of a group that requires inverses to commute, do we still get a group?

I'm trying to find a counterexample, but can't think of one right now

Daminark

Wait do you mean you only assume one-sided inverse?

For any $a$ there's some $b$ such that $ab = e$ and a priori not necessarily $ba = e$? If so then yeah that's enough for a group. Even one-sided identity I think

Perturbative

5:11 AM

@Daminark Yeah that's what I meant

Balarka Sen

Think about aba

Alex

~dancing queen~ youtube.com/watch?v=xFrGuyw1V8s

Balarka Sen

Scratch that. You have to think about a c such that bc = e.

c = (ab)c = a(bc) = a

Notice that you need left identity = right identity in the proof.

Perturbative

Thanks Balarka!

So one-sided identity isn't enough for groups, correct?

But one-sided inverses will do just fine

Balarka Sen

I think it's enough but I can't be arsed to write down a proof that one-sided identity implies two-sided identity and both of them are the same

3algebra5me

If you think a little you should be able to write it down

@Alex Ugh fucken 70's pop

The worst

Semiclassical

5:26 AM

lol

Balarka Sen

I only listen to real music

Semiclassical

Bee Gees - Stayin' Alive (1977)

►

Alex

@BalarkaSen You said to think about abba

The only song I know from abba

Daminark

So let's say $e$ is a left-sided identity and that you have left-sided inverses. $ee = e$, so $a^{-1}ae = a^{-1} a$, so $(a^{-1})^{-1} a^{-1} a e = (a^{-1})^{-1} a^{-1} a$. Then the left hand side, by associativity and left inverse is gonna be $ae$, and the right side is $a$, so $e$ is a right-sided inverse

Semiclassical

i know a few abba songs off the top of my head for better or worse

Waterloo, SOS, Knowing Me Knowing You, Super Trooper

plus i'll take 70's pop over contemporary pop any day

Daminark

5:30 AM

Now we have a two-sided identity (which we know is unique lest that be a problem) so let's prove we have two-sided inverse.

Alex

I'll take suicidal depressive black metal over 70's pop any day

Balarka Sen

^^

Daminark

(Also double check me for errors, I did this argument a while back because on my first algebra pset we got docked a quarter of the points because we usually just established one-sided inverses (one of the problems was fucking commutative anyway) but this was a while back)

Balarka Sen

3edgy5you

Semiclassical

lol

Alex

5:31 AM

:')

Semiclassical

its all relative

Alex

Make A Chance ... Kill Yourself - du er alene

►

good^

Balarka Sen

Merzbow - Pulse Demon (Full Album)

►

good^

Alex

Noise metal?

Balarka Sen

Just noise. Raw noise.

skibi pop pop

Daminark

5:33 AM

$a^{-1} a = e$, so $a^{-1} a a^{-1} = e a^{-1} = a^{-1}$, so $(a^{-1})^{-1} a^{-1} a a^{-1} = (a^{-1})^{-1} a^{-1} = e$, while the left hand side simplifies to $aa^{-1}$, so that's a right-sided inverse.

Alex

It's recommending me Macintosh plus three times in the recommended list

Daminark

@Perturbative this should do it unless I'm drunk on the flu

Alex

I wonder why

Balarka Sen

hahahah

get v a p o r w a v e d

@Semiclassical You know what I like about 70's pop

Alex

I haven't actually listened to the whole album, I'ma hit that now

Balarka Sen

5:33 AM

They can be slowed down and turned into masterpieces

good ^

Ｉ　ＣＡＮ　ＭＡＫＥ　ＩＴ　ＢＥＴＴＥＲ　ＦＯＲ　ＹＯＵ

Like I said, I only listen to real /mu/sic

Perturbative

@Daminark Don't you mean $e$ is a right-sided identity? (In the post before Semi said the abba songs he knows)

Balarka Sen

@Alex I actually don't dislike Macintosh Plus

It's kinda dope, in a sense

D O P E

Alex

@BalarkaSen I only like the canonical song from their album

From what I've heard

Daminark

5:40 AM

Oh maybe I flubbed that, I'm tired

Trust yourself more than you trust me at the moment

Balarka Sen

@Alex Have you tried Tom and Jerry wave?

ＬＯＮＥＬＹ

►

Alex

Actually yes hahaha

I think you showed me ages ago

Balarka Sen

I love the comments below

Perturbative

@Daminark Just went through it, it looks good to me

Balarka Sen

@Daminark Sorry to hear about the flu

Hope a speedy recovery

Daminark

5:48 AM

Thanks

Perturbative

Hope you have a quick recovery Dami!

Secret

<S_h|P>

I need to figure out what this means...

It looks nothing like a wavefunction

1 hour ago, by Madhuchhanda Mandal

Is continued fraction of every decimal number possible ?

Hmm... is Chaitins's constant a decimal given how we cannot even compute its digits...?

Balarka Sen

Yes it is, like any real number. The digits just cannot be computable by a Turing machine.

Secret

interesting

Secret

6:40 AM

This notion of transcending the normal bounds of something is very common in our society. For example, how a person's idea, if it is considered as good/bad/whatever enough, it get sucked into the education system, and then will start to influence the younger generation

Thus, in a minimalistic sense, education is a bridge between individuals to the society in terms of influences. It converts a local influence into a potentially global one

We tried to capture this minimalistic notion by coming up with the concept of interfinite, something that bridges the finite (individual) to the infinite (society at large in the perspective of individuals)

note how annoying it is that this concept is not even logically consistent with first order logic

Mar 4 at 7:29, by Secret

Proposition: Let $A$ be a Non-Archimedean semiring with a linear order. Then interfinite elements does not exist

another lesson of this and similar blog posts, is that don't think that every public blog post you wrote has no consequence to the society. When any of these high level entities such as the education system start putting them into their plans, it can have a global impact. The internet makes this transition even easier as internet memes spread very far and quickly like wildfire, and all that is need is enough people to share about it

Perhaps, a more accurate way to formalise this idea is before the internet, society is really infinite in the perspective of individuals, but ever since the internet, society becomes finite. That, is much easier to model and is also consistent

schrodinger_16

7:03 AM

0

Q: Two Touching Ellipses - Tangents, Centres and Collinearity

Consider two ellipses, touching each other at a point (i.e. they've a common tangent at that point). It is given that they also have two more external common tangents, which when extended meet at P. The centres of the ellipses are A and B. Prove that A, B, P are collinear. I've attache...

In this question

If we want to use the affine transformation

Can we change both ellipses to circles?

or just one?

And why so?

Secret

From the way the ellipses are oriented, it seems they semimajor axes (resp major axes) are orthogonal to each other, that means, any sheering that sheer the ellipses along one of these axes will make the other more elliptical as one becomes circular, and it does not seemed sheering in the diagonal direction will work either, as the resulting ellipses will be slanted

I don't recall what other kinds of affine transformation that are not sheering nor translation (which does not do anything) are, I need to check

ok checked, so there is no affine transformation that can make both ellipses circular, and the best you can do is make one of the circular

(I don't know how to prove this algebraically however...)

NotThatGuy

7:43 AM

Are "where am I going wrong in my calculation" questions on topic on Math.SE? Can I ask them in chat?

Alternatively, does anyone have a nice and simple explanation of using Markov Chains to calculate the probability of reaching each terminating state for cross-dependent probabilities (as in P(A) = f(P(B)), P(B) = f(P(C)), P(C) = f(P(A))) somewhere? The explanations I've seen so far are either a bit math-heavy or goes off on some long tangent about things I don't care about.

The whole "ignore all previous states and only consider the most recent one" worked* for one calculation, but then it failed for another one.

schrodinger_16

8:08 AM

@Secret, Yeah, alright. I understood that now.

That done, the proof using analytic geometry is still not easy

It may be straightforward, but is lengthy

Have you tried it?

Could you post it if you have?

A method using only geometry works fine too

NotThatGuy

8:48 AM

Never mind, I figured it out (to some extent). Random fact: with a probability p of winning a point in tennis, the probability of winning the game from deuce is: p^2/(p^2 + (1-p)^2)

Secret

9:06 AM

ugh, this is not going anywhere...

I don't know the relationship between the angles of an ellipse and its tangent

Evinda

9:21 AM

Hey

does it hold that each 2x2 symmetric matrix with positive diagonal elements is positive defined?

Hi @LeakyNun
Do you maybe have ann idea?

Secret

take e.g. diag (0,1,0), then it is easy to see that if the vector is e.g. 1,0,0, you get zero for its quadratic form

Evinda

But in our case we have a 2x2 matrix :/

So which matrix could we pick?

Secret

oops I forgot to read the 2x2 bit, hmm... my first intuition is the determinant will be important, but let me check closely just in case. Luckily you have a symmetric matrix here, thus the maths is a bit easier to handle

Ok, turns out the rules are quite simple: You pick any 2x2 symmetric matrices with eigenvalues all positive, that is enough to guarentee positive definiteness

Mary Star

Hello!

Does someone of you have an idea about my question: https://math.stackexchange.com/questions/2694783/convergence-of-iteration-method-criterion-using-iteration-matrix ?

Do we check the converegence of an iteration method only the spectral radius? Or also the infinity norm of the iteration matrix?

Evinda

A ok, thank you :) @Secret

Drew Brady

9:45 AM

Anyone here good at combinatorics?

Shobhit

if $f$ is a homeomorphism between $(X,\sigma) \to (Y,\delta)$, is it true that if $U \in$ $\sigma$, then $f(U) \in \sigma$. I know the definition of a homeomorphic function, i have tried manipulating its properties, but could not conclude if its true or not.

Astyx

9:59 AM

@Shobhit The inverse function of $f$ is continuous thus $f^{-1}(f(U)) \in \sigma \implies f(U)\in \delta$. And $f^{-1}(f(U)) = U$ because $f$ is bijective (unless I'm missing something obvious).

Alessandro Codenotti

@Shobhit What's your definition of a homeomorphism?

Shobhit

@Astyx i may be interpreting your answer wrong, but could you please read the question again, its " is it true that if $U \in$ $\sigma$, then $f(U) \in \sigma$" (note they both belong in $sigma$).

Alessandro Codenotti

Are you sure it's not $f(U)\in\delta$? Because $\sigma$ has nothing to do with the codomain of $f$

Shobhit

@AlessandroCodenotti $f$ is bijective with $f$ and its inverse continous.

@AlessandroCodenotti i am not sure, thats why i asked, i have read wiki and searched MSE for this, all to find what i already know, i read this from a friends notebook, he could very well may have written wrong, i skipped class that day. So wanted to confirm if its true or he copied wrong the from the blackboard

Alessandro Codenotti

I think that should be $\delta$, it seems to be part of a standard exercise, namely showing that $f$ is an homeomorphism iff it is continuous, bijective and open

Shobhit

10:10 AM

ok, ty for your help

Astyx

Sorry I was away

What Alessandro said

You can't say anything about the image of $f$ and $\sigma$ (and more importantly it doesn't make any sense if $X \ne Y$)

Shobhit

oh ok

Evinda

10:41 AM

@Secret And something else... if there is an upper triagonal matrix R with positive diagonal elements such that $A=R^T R$ does it imply that A is positive definite?

Secret

but positive diagonals does not imply positive eigenvalues, right?

Leaky Nun

@Evinda $$\begin{bmatrix}1&1\end{bmatrix} \begin{bmatrix}1&-2\\-2&1\end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix} = -2$$

@Evinda Let $A=R^TR$. Then, $\langle x, Ax \rangle = \langle Rx, Rx \rangle = \|Rx\|^2 \ge 0$, so it is positive semidefinite

But $R$ is non-singular, so $Rx = 0 \to x = 0$, so $A$ is positive definite

Secret

huh, never thought about using inner product properties...

Astyx

inner product and positiveness are very tightly linked

Leaky Nun

well, I could have phrased it as easily as $x^TAx = x^T R^T R x = \|Rx\|^2$

Evinda

10:48 AM

@LeakyNun $\begin{bmatrix}1&-2\\-2&1\end{bmatrix} $ cannot be written in the form $R^TR$... or am I wrong?

Leaky Nun

@Evinda I was replying to an earlier message

Secret

yeah, guess I have gone rusty 3 years after my 2nd year linear algebra class (which I actually did in my 3rd year)

also, that matrix has a negative eigenvalue, thus it is not positive definite and hence Leaky have gave the relevant counterexample

Evinda

@LeakyNun Ah yes, I see...

Mary Star

It holds for the spectral radius of a matrix A that $\rho (A)=\max_i \{|\lambda_i|\}$ and $\rho (A)\leq \|A\|$, right? Does this hold for each norm? Also $\rho (A)\leq \|A\|_1$, i.e. does it hold that if $\|A\|_{1}=0.01$ then there is an eigenvalue such that $|\lambda|\leq 0.01$ ?

Evinda

So $A$ is positive definite since $\langle x, Ax \rangle \geq 0$ and $\langle x, Ax \rangle =0 \iff x=0$, right? @LeakyNun

Leaky Nun

10:59 AM

yes

Evinda

@LeakyNun Nice, thank you :)

Mary Star

Hello @LeakyNun !! Do you have an idea about my question above?

Leaky Nun

no

Secret

the only thing I have in mind is it seems to involve spectral theorem of linear operators, which seemed to be functional analysis stuff thus I don't think Leaky nor I nor Evinda will have idea

For example, despite I heard that term all the time, I still have no idea what a spectral radius is

Alessandro Codenotti

biggest absolute value of an eigenvalue

It arises naturally in numerical analysis as well

Mary Star

11:04 AM

but is it less or equal to any norm of the matrix?@AlessandroCodenotti

Alessandro Codenotti

It's true for matrix norms induced by vector norms, I don't think it holds in general

Mary Star

So, does it hold for 1-norm of the matrix? @AlessandroCodenotti

i.e., $\rho (A)\leq \|A\|_1$

Alessandro Codenotti

Yes, the $||\cdot||_1$ norm is induced by a vector norm

Mary Star

Ah ok! So, if $\|A\|_{1}=0.01$ then there is an eigenvalue such that $|\lambda|\leq 0.01$, right? @AlessandroCodenotti

Alessandro Codenotti

Not just one, all of them

Mary Star

11:09 AM

Ah yes, since the max of the eigenvaulues is less than 0.01, then all of them are.

Thank you!

Leaky Nun

what is the 1-norm?

Secret

|x|+|y|+|z|+...|

Mary Star

It is defined as the max of the sums of the absolute values of the columns: $\max_j \sum_i |a_{ij}|$

Alessandro Codenotti

Which is also the operator norm if you think about $A$ as an operator $\Bbb C^n\to\Bbb C^m$ and use the $||\cdot||_1$ norm to turn those into normed vector spaces

Niing

12:10 PM

Since $span(\emptyset)={0}$, could I say "$0$ is a linear combination of nothing"?

The context is linear algebra.

Secret

12:45 PM

19

Q: Why $\mathbf{0}$ vector has dimension zero?

According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace $V$ is defined to be the minimal number of vectors required to generate $V$ (pp. 4). Then why does the $\mathbf{0}$ vector have dimension zero instead of one? Shouldn't it be true that only the empty...