I found a matrix problem yesterday. $A$ and $B$ are two square matrices of same order. If $AB=A$ and $BA=B$ then express $(A+B)^n$ in terms of $k(A+B)$. $k$ is constant number (which depends on $n$). Find $k$. I felt this was quite an interesting problem. :) @DHMO

@RE60K Are you ADG ? (I think I saw you before on this site)

@DHMO I felt it is more complicated. If the rationals are well ordered in the usual ordering, similar to the bounded sequence , then I can describe the irrationals as countable union of open intervals (p,w) with consecutive rationals p,q that go to infinity on both ends. But the rationals are not well ordered in the usual ordering, thus I cannot decompose it this manner. Also, because the rationals are dense, every open interval containing a irrational must contain a rational, thus you canno

describe it in terms of union of any intervals

@Secret you haven't addressed fact 2 at all

@DHMO should it be the complement of arbitrary unions will be the arbitrary intersections of the complements?

@anonymous that's interesting

@Secret that is also correct.

do you mean that fact 2 is wrong?

@DHMO are we excluding the singletons as closed intervals here, if not, the bounded countable sequence explored above will be a counterexample since that is a neither set.

@anonymous wow. $A^2 = ABAB = A(BA)B = ABB = (AB)B = AB = A$

@anonymous I think it should be the number of terms appearing in the binom expansion.

@BalarkaSen but there are $2^n$ terms in the binomial expansion

Right.

beware that $(A+B)^2 \ne A^2 + 2AB + B^2$ in general

@BalarkaSen half that

@DHMO You're on the right track!

@anonymous then by induction $A^n = A$

@DHMO I am aware. In any case, there are $4$ terms appearing there.

similarly $B^n = B$

@arctictern Ah, right.

then use induction to prove that $(A+B)^n = 2^{n-1}(A+B)$

@DHMO Noooo

@DHMO You said the thing I said wrong :P

yup ^

@DHMO why?

@Anonymous Matrices don't commute

In general

@Anonymous because $(A+B)^2=A^2+AB+BA+B^2$ and $AB\ne BA$ for matrices in general

Inductive step: $(A+B)^{k+1} = (A+B)^k (A+B) = 2^{k-1}(A+B)^2 = 2^k(A+B)$

But it's general formula right? Let's assume a,b be constant for sometime

where $(A+B)^2 = A^2 + AB + BA + B^2 = A + A + B + B = 2(A+B)$

since the bounded countable sequence is a union of singletons (which are a special case of closed intervals), and its complement is a countable union of open intervals plus the closed ray based on the limit point of the sequence, which is a neither set

@Anonymous $A$ and $B$ are matrices

Hi chat

Hi @Alessandro

@Secret do I need to perform the inductive step to you?

@AlessandroCodenotti o/

It's also easy to see from binomial expansion, without doing induction.

Hi @Alessandro

There are $2^n$ terms, but because you pair as $k(A + B)$ you divide by $2$

So $k = 2^{n-1}$.

@AlessandroCodenotti buongiorno.

Consider the usual topology on $\Bbb Q$

and the set of rationals with odd denominator

@BalarkaSen Did you try this sum : How to find the maximum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$ are complex numbers.) ? It is definitely one of the most brilliant CN problems I ever saw :) Do try!

@AlessandroCodenotti I think that the set is dense, not open, and not closed.

Am I right?

Here's another one. Assume $BA=qAB$ for a scalar $q$. Find the missing coefficients in the expansion $$(A+B)^n=\sum_{k=0}^n \square A^kB^{n-k}$$

Sounds right

@AlessandroCodenotti I also think that it is connected although I have no proof.

@arctictern assume that $\displaystyle (A+B)^n=\sum_{k=0}^n f(n,k) A^kB^{n-k}$

@anonymous I think I have seen it before, but never tried it. Probably the $z_1, z_2, z_3$ are supposed to be vertices of an equilateral triangle.

Although it's not entirely clear how I would go about showing that...

Then $f(n+1,k+1) = q^{n-k} f(n,k) + f(n,k+1)$...

@BalarkaSen Nooo. It won't be equilateral triangle. Remember that it is not the sum of square of sides of triangle. You are talking about $$\left(|(z_1-z_2)|^2 + |(z_2-z_3)|^2 + |(z_3-z_1)|^2\right)$$ I guess...

But this problem is different

Ah, I guess I was. It's bounded by that though.

Hmm, ok.

Actually the answers to the problems will differ by a wide margin

I don't think so @DHMO

This latter problem is easy

The former one is quite tough :)

Yeah.

@AlessandroCodenotti you're right. Every set is disconnected in $\Bbb Q$.

@DHMO Are you trying the complex number problem?

@anonymous not really. I've seen the solution.

@DHMO Oh...let Balarka do it then :)

I'm not thinking about it right now but I'll try it out at some point.

@BalarkaSen Ah, sure!

@anonymous let's teach you topology lol

It will take some time. It took me 2 days :P

@DHMO Hehe..I wish I could say yes...however, I'm busy with calculus and algebra now !

Hah. It does sound like an interesting problem but I have no immediate insight.

@anonymous I see

@DanielFischer gu'n Morgen

@DHMO Do you know the properties of orthogonal matrices?

@anonymous no

I × I =I for I=unit matrix?

@Anonymous you mean identity matrix

and yes

@Secret have we given up?

Ah, then this is for you: If $A=\frac{1}{3} [(1,2,2);(2,1,2);(x,2,y)]$ is orthogonal then $-x-y$ is? (Orthogonal matrices are those for which $AA^{T}=I$) @DHMO

Yes, $(A+B)^2 ≠ A^2 + 2AB + B^2$ for A,B∈ matrices

Easy enough.

I was replying to the other anonymous :)

@anonymous who got pinged by this message?

@BalarkaSen I am sure you know the properties of such matrices. I want someone to derive it from scratch !

Me✋

$\displaystyle A = \frac13 \begin{pmatrix}1&2&2\\2&1&2\\x&2&y\end{pmatrix}$

$\begin{pmatrix}1&2&2\\2&1&2\\x&2&y\end{pmatrix} \begin{pmatrix}1&2&x\\2&1&2\\2&2&y\end{pmatrix} = \begin{pmatrix}9&0&0\\0&9&0\\0&0&9\end{pmatrix}$

consider the sixth entry

@DHMO Base case S0=[a,b], S0 U [c,d], complement (...,a) U (b,c) U (d,...) is open. Thus S0 is true. Assume Sn=union of n intervals is true, I.e complement Sn is union of open intervals. Inductive case Sn U [ak,ak+1], complement (() U ... U (,ak) U )(ak+1,...) is open. Thus Sn is true for all n.

$2x+2+2y = 0$

$-x-y = 1$

@Secret so where is the contradiction?

@BalarkaSen but it led to the result

Alternatively understand that the columns are all orthogonal in an orthogonal matrix.

what does that mean?

Orthogonal vectors

If the last column is (2, 2, -3) (which doing a cross product calculation gives you) you get that. So the answer is -2 - (-3) = 1.

How you relate orthogonal matrix to orthogonal vector @BalarkaSen? Any rule? From which method you got that?

@Anonymous the matrix product is a bunch of dot products

the diagonal is where the columns dot themselves

and the other entries are where the columns dot other columns

and the other entries are 0

thus the columns are orthogonal

wikimedia.org/api/rest_v1/media/math/render/svg/…

bonus: they all have the same magnitude

wikimedia.org/api/rest_v1/media/math/render/svg/…

@DHMO who?

I think I have uttered some nonsense

@Anonymous Elements of $A \cdot A^T$ are the dot of the columns.

@Anonymous the columns

@DHMO Yeah, the columns are orthonormal, not just orthogonal.

@anonymous are you sure it is orthogonal?

because the magnitude of (2,2,-3) is different

@DHMO By definition orthogonal matrices are $AA^{T}=I$. If that is satisfied then it has to be orthogonal. You are forgetting the factor of 3 outside.

$\displaystyle A = \frac13 \begin{pmatrix}1&2&2\\2&1&2\\-2&2&-3\end{pmatrix}$

@BalarkaSen I didn't understand,can I get an example for 2×2 matrix?

@anonymous no, this doesn't sound right.

The magnitude of the first row vector is definitely $1$. Calculate and check. $(1/3)^2+(2/3)^2+(2/3)^2$

=1

I'm talking about the last row vector

I need to check your calculation...if the magnitude of the last row doesn't come out to be 1 then there has been some calc error

you need to check your matrix

what are $x$ and $y$?

real numbers

no, I mean what are their values

Wait a min...let me calculate

The last column would probably be (2, 2, -3) scaled up to unit length...

x=-2 and y=-1 @DHMO

Dot product of 1 st row with 3rd row

=0

and dot product of 2nd row with 3rd row is 0

Then wolframalpha.com/input/?i=2x%2B2-2y%3D0+and+x%2B4%2B2y%3D0

Now the magnitude of last row is indeed 1

@DHMO I don't know. To have no contradictions, it means the irrationals are a countable union of open intervals, but how?, every open interval must contain a rational because the rationals are dense, thus a countable union of any open intervals will not only contain the irrationals, but a countable subset of rationals as well

Oh, wait, I misunderstood your matrix. The columns and rows are switched.

The inductive proof looks fine thus fact 2 must be correct, so something in fact 1 is off

Ok, cross product of (1, 2, -2) and (2, 1, 2)

@BalarkaSen What?

That is (6, -6, -3)

Scale: that's (2, -2, -1).

There you go.

@anonymous I mistook your columns for rows and rows for columns. Nevermind that.

Yep. There is no need of scaling though as 3 is common for all. Just use dot products!

Also, to have no contradictions, it also mean the complement of the bounded countable sequence is also open, thus making the sequence closed. That is, () U () U ... U [,...) is open??

What kind of set are the irrationals. It does not seemed it can be decomposed into open intervals?

@Secret that is fact 1.

@anonymous $\displaystyle A = \frac13 \begin{pmatrix}1&2&2\\2&1&2\\-2&2&-1\end{pmatrix}$?

Yeah.

$\displaystyle A A^T = \frac19 \begin{pmatrix}1&2&2\\2&1&2\\-2&2&-1\end{pmatrix} \begin{pmatrix}1&2&-2\\2&1&2\\2&2&-1\end{pmatrix}$

What I thought was the columns are (1, 2, 2) and (2, 1, 2) and (x, 2, y).

In which case you'd just cross the first two to get the third, upto scale.

@BalarkaSen Hehe :'D Happens to me sometimes :P

@BalarkaSen doesn't matter.

@DHMO It does

alright

@DHMO But I already have a counter argument to fact 1, every open interval must contain countable many rationals, thus the union of them will be a superset of the irrationals. Yet we have just proved inductively that fact 2 is true, so fact 1 has to be false

so instead we have $\displaystyle A A^T = \frac19 \begin{pmatrix}1&2&-2\\2&1&2\\2&2&-1\end{pmatrix} \begin{pmatrix}1&2&2\\2&1&2\\-2&2&-1\end{pmatrix}$

@Secret fact 1 is that the irrationals is not union of open intervals

$AA^{T} \neq A^{T}A$ generally

@anonymous not if they are $I$

$\displaystyle = \frac19 \begin{pmatrix}9&0&8\\0&9&4\\8&4&9\end{pmatrix}$

so you need to check your matrix again @anonymous

Ah, true, left inverse does mean right inverse in matrices.

@Secret specifically because every open interval does have a rational number

@DHMO @BalarkaSen Yeah, you are right. This question was directly from my book. Printing error possibly! I didn't think of it the reverse way until DHMO pointed out

I did check the question just now

$(\dfrac{1}{3} A)^T = \dfrac{1}{3} (A)^T$ ?

yes

For orthogonal matrix , $A^T = A^{-1}$ ?

@anonymous I mean your matrix cannot be orthogonal as row_1 and row_2 are not orthogonal.

yes

(1, 2, 2) and (2, 1, 2) are not orthogonal

Damn these textbooks :P

@BalarkaSen Yes, I saw that just now. Very true!

So the question indeed is garbage.

It is !!

Agreed :D

Good find, @DHMO

So you people didn't checked wheather $A$ is invertible or not,so selfish

@DHMO yeah, congrats :) I would have missed that!

Hi @ATHARVA

@Anonymous Why should we? $AA^{T}=I$. We already know that!

Hi@Anonymous

@DHMO for orthogonal matrices there is no need to check $A^{-1}$ is 0 or not?

Orthogonal matrices are by definition invertible.

^

Geometrically, a set of orthonormal vectors cannot span a subspace of dimension less than their cardinality.

Stuff can't degenerate into a subspace of lower dimension.

googling every word

The issue here is that our matrix wasn't orthogonal in the first place.

so a matrix is orthogonal iff it represents a set of vectors mutually perpendicular and having magnitude 1

yeah

OK, I am going to chicken out of the JEE problem session :P It's too hard for me, and I have stuff to do too

@BalarkaSen "too hard for me"...oh stop bluffing me :'D

@DHMO ok in that case, fact 1 will contradict fact 2. Even though induction requires some kind of well ordering, the intervals we state in the proof are arbitrary thus fact 2 should also hold even when the set is dense

@Secret but we have proved fact 2 via induction

Yes, and we also showed fact 1 is also true. Fact 2 say complement of any union of closed intervals are union of open intervals, and we have proved it is true. Fact 1 say the complement of a countable union of closed intervals the rationals are the irrationals, which by fact 2 should be union of open intervals. But using dense property of rationals we have shown the irrationals cannot be a union of open intervals, hence factc1 contradict fact 2

@Secret so how to resolve the contradiction?

Hallo

@ChristianAndrews welcome

Thanx, I'm just watching how this chat works

This chat works in mysterious ways

I suspect the key has some to do with that the rationals are dense in order to bypass this apparent contradiction, but I don't have any idea. Point is, I cannot even decompose the set of irrationals into unions of any intervals except an uncountable union of singletons

@Secret of course you cannot. any ordinary interval must contain an irrational

do you want the answer?

Yeah I can see it, I'll be back later

Do you have one more hint?

Have a nice day

@Secret apparently fact 1 is true and fact 2 does not apply to the rationals

so think about why we cannot apply fact 2 to the rationals

We have just showed fact 1 is true, that is, the set of irrationals cannot be decomposed into any union of any open intervals, hence open sets. So the base case of the induction is empty and thus the whole induction falls apart, making fact 2 fail for rationals and its complement

@Secret "the base case of the induction is empty"?

enumerate the rationals

add the singletons of the rationals one by one

the complement will always be open

by the induction

@Secret

But the rationals are dense, there is no one after the next in the usual ordering?

I can still enumerate them

1/1,-1/1,2/1,-2/1,1/2,-1/2,1/3,-1/3,3/1,-3/1,4/1,-4/1,3/2,-3/2,2/3,-2/3,1/4,-1/4 etc.

I.e moving in some sort of zigzag according to cantor

yes

But if the next term is larger than the previous, this is not the usual ordering

why do we need to use the usual ordering?

the induction doesn't depend on the ordering

(Actually what I just said about usual ordering has nothing to do with the induction argument, as you mention induction is independent of ordering)

What do they exactly mean that linear maps are compatible with scalar multiplication and vector addiction?

My book explicitly uses the word "compatible"

I can see similarity, sure

But I don't understand the meaning of compatibility here?

@ShaVuklia A(x+y) = A(x)+A(y)

A(kx) = kA(x)

I understand that, but how would you define compatibility then?

what I have just said is compatibility

Because as for now, I wouldn't know when I can and cannot use it

Oh right, but that's pretty specific. Isn't there a broader definition?

Is it that they obey the same rules?

intuitively, doing A first then B, and doing B first then A, give you the same result

A is the linear map

B is addition

@Astyx bonjour

Hi

Have we not greeted already this morning ?

sure

Ok, I'll stick to that intuitive notion then

thanks

Oh, I get it now. Definition compatibility: "the ability [...] to work successfully with other machines or programs (or in our case, operations)". So they don't 'interfere' with each other.

heh

What does your 'heh' mean? I'm interpreting it as a 'yes' and a 'huh?' at the same time

no, it's a laugh

if $E$ and $E'$ are field extensions of $F$ and have the same finite degree over $F$ then any $F$-homomorphism is an $F$-isomorphism

why so?

I mean for example what stops me from mapping all the elements in $E$ but not in $F$ to $0$ in $E'$?

@Secret have we given up?

dense: for all x<y, exists x<z<y.Now, {x} U {y} closed, complement (...,x) U (x,y) U (y,...) open. Remove z gives (...,x) U (x,z) U (z,y) U (y,...) open. Remove k1,k2 gives (...,x) U (x,k1) U (k1,z) U (z,k2) U (k2,y) U (y,...) open. This can be repeated forever, thus the intervals are get smaller and smaller. As I throw away countable number of rationals in the middle of each open interval, I get $2^{\aleph_0}=\aleph_1$ open intervals in the union. We have not proved by transfininite induction,

@Secret that is far from the dense, but I'll let the first sentence pass

thus fact 2 may fail for uncountable unions.

Anyway I need to go onto the plane now, chat later

@Secret and $2^{\aleph_0}$ is not $\aleph_1$

@Secret safe flight

@Secret actually removing $n$ numbers give you $n+1$ intervals

so it is definitely countable

@Astyx tu sais topologie?

Je connais un petit peu oui

@Astyx en francais?

Comment ça ?

topologie en francais

Je ne comprends pas ce que tu veux dire

tu sais les glossaires de topologie en francais?

Oui

donne-moi un sous-ensemble de $\Bbb R$ qui est ouvert, dense, et disconnexe

$\Bbb R^*$ ?

c'est pas un sous-ensemble de $\Bbb R$

Euh ...

oh

Si ?

$\Bbb R\setminus \{0\}$

merveilleux.

quel est $\Bbb R^*$?

$\Bbb R^* = \Bbb R \setminus \{0\}$ where I live

oh, desole

Pas de quoi

sont tous les ensembles fermes, un reunion des intervalles fermes? @Astyx

Non, par exemple l'ensemble de Cantor

je demande pas reunion fini

Dans ce cas pour tout sous-ensemble $A$ de $\Bbb R$ on a $A = \bigcup_{x\in A}\{x\}$

Enfin pas que de $\Bbb R$ d'ailleurs

I already posted this in the topology and calculus chat rooms, but they are inactive

@Astyx oui

je viens de decouvrir que ma question est stupide

Et si tu veux une réunion dénombrable il ne me semble pas

il ne te semble pas que quoi?

Que ce soit vrai

es tu francais?

Que tout ensemble fermé de $\Bbb R$ soit une réunion dénombrable d'intervalles fermés

Oui

:o

Pourquoi ?

pourquoi on utilise le subjonctif ici?

Il ne me semble pas que ce soit vrai

Je ne suis plus sûr des règles

forum.wordreference.com/threads/…

il dit l'indicatif ici

Mais je suis presque certain que ce que je dis est juste :)

french.about.com/od/grammar/qt/subjunctive_ilmesemble.htm

donc c'est le subjonctif en negation

@Astyx considere la topologie de la base [x,y).

@Astyx est (1/n) convergente?

On dit plûtot : "Est-ce que (1/n) est convergente ?"

ou "(1/n) est-il convergente"

"elle" du coup

Une suite

While it is nicer to your fellow chatters to speak in the language that most people here speak, you don't always have to. I've had conversations in Hebrew on chat.se.

@Mithrandir shalom

Hello!

Someone was complaining that the conversation was not in English.

@Mithrandir It's more difficult to discuss French language in English :p

When ? Where ?

In moderator flags.

@DHMO Is it not the case that both the assumption that $2^{\aleph_0}=\aleph_1$ is consistent and the assumption that $2^{\aleph_0}\ne\aleph_1$ is consistent?

Ah right

That is, neither raises a contradiction by itself.

@robjohn yes but we can't assume it to be so

but that's irrelevant to the discussion anyway

@DHMO we can assume it and go from there

okay

@robjohn and please do not spoil the answer

I'm having Secret think about it

How do you define convergence of a sequence topologically @DHMO ?

@Astyx it has a limit point

A unique one

not necessarily

it can converge to all values

in some topologies

Ie $\bigcap \overline A_{n}$ is a singleton where $A_n = \{x_k, k\ge n\}$ right ?

just not empty

I must say I haven't done much topology outside metric spaces

@Astyx Consider $A = [0,1] \cap \Bbb Q$.

Can you give me a sequence $(b_n)_0^\omega$ of open intervals whose lengths sum to a value strictly less than 1

but which fully covers $A$?

With the usual topology ?

this question is independent of topology

Sorry I wasn't paying attention

Well you asked for open intervals

yes

not open sets

Mmm

That doesn't seem possible

notice that the sequence extends to infinity

countably infinity

Well, it is a sequence

there are finite sequences

Kinda

but it cannot be uncountable

why?

Because that's the definition of a sequence

It's an application from $\Bbb N$ to a set $A$

feel free to change the word "sequence" to "set"

why can't it be uncountable?

I don't think I understand your question

because you can't sum an uncountable amount of items...

anyway, let's go back to the question

Yes you can

But what does a sum have to do with this ?

"lengths sum to a value strictly less than 1"

@AkivaWeinberger hola

Hello

Oh I see

The sum of an uncountable amount of positive numbers always diverges

alguien estaba reportado por que hablo en un otro idioma :o

¿Reportando a ti?

@Akiva The sum of an uncountable amount of nonzero numbers always diverges

no a mi, no soy un moderator