Why does taking all the super diagonal entries of a jordan form as N, and taking all of the diagonal entries of the same jordan normal form as S, giving us $J=N+S$ give us NS=SN? E.g. why do these commute?

Is it something to do with the fact that our diagonal entries are essentially a constant times the identity matrix for each block

and these multiply only with the super diagonal entries of each corresponding block

@evinda Ja, ich werde meine Familie nach der Prüfung besuchen, aber jetzt bleibe ich lieber in Deutschland

Hi guys

I am just reading chapter on group actions

it states the axioms for a vector space over a field F include two axioms that the multiplicative group $F^*$ act on the set V. Thus vector spaces are familiar examples of actions of multiplicative groups of fields where there is even more structure, (In particular V must be abelian group)

how come V is abelian group

with respect to addition they meant to say ?

Yes.

@Alessandro Aha :)

A vector space is, essentially, an abelian group with a compatible action of a field on it.

yeah

Oh I got it, each set of jordan blocks essentially makes up an identity matrix of jordan blocks, and multiplied by its nilpotent matrix companion, it gives us essentially a scalar(the eigenvalue) times the identity matrix * the nilpotent matrix, which will just be a scalar multiple of the nilpotent matrix, e.g. we will just get a scalar multiple of each Jordan block, which is equivalent from left or right

In fact, a vector space is a ring-homomorphism $F \to \text{End}_{\Bbb Z}(A)$ from a field to the ring of abelian-group endomorphisms of an abelian group $A$.

Just as a group action is a group-homomorphism from $G \to \text{Sym}(X)$, the group of all bijective mappings on $X$.

I see

A (left) $R$-module is the same idea, with a ring $R$, instead of a field, and we can similarly consider a monoid acting on a set as a monoid-homomorphism $M \to T(X)$, where $T(X)$ is the monoid of transformations of $X$ (all FUNCTIONS $X \to X$).

@evinda that looks fine

I don't know whats a module

@robjohn Nice, thanks :)

@KarimMansour That's OK, it's just another example. You'll meet some later.

okay good

what happens when you conjugate a nilpotent matrix? or when you conjugate a semisimple matrix?

Wait one sec brb

@DavidWheeler I am just playing with actions a little bit if we consider the following trivial action from $\sigma$ : GxA $\rightarrow$ A where A = G and ga = g + a. Now this action is faithful since we the permutation representation of it will be $\alpha$ : G $\rightarrow$ $S_A$ given by $\alpha(g)$ = $\sigma(g)$

now consider $\alpha(g_1)$ = $\alpha(g_2)$

Hence we will have $g_1 + a$ = $g_2 + a$ hence by cancellation laws we get $g_1$ = $g_2$ hence proved

What is A?

First capital letter of the English alphabet

That is certainly ONE answer

hehehe

@MikeMiller They are still nilpotent and semisimple?

Yes, @Incurrence. And you're done.

A = G where G is some group

@MikeMiller What? How?

My point is, we usually don't use $g+a$, except when we have an abelian group.

@Incurrence: Let's say $P = JAJ^{-1}$ is the Jordan normal form of $A$. You know $P = N + D$. So you know $A = J^{-1}NJ + J^{-1}DJ$, and because a conjugate of a nilpotent is nilpotent, conjugate of semisimple is semisimple, you have the desired decomposition of $A$.

Oh, you're CTAC. I like your new name.

@MikeMiller Thank you haha. How did you notice that?

Checked your profile.

i see

How did you obtain $P=JAJ^{-1}$?

Definition of Jordan form.

@Incurrence You sneaky

@ᴇʏᴇs Why am I sneaky haha?

@Incurrence I was looking for you

Hello!! Is there someone familiar with the Ackermann function??

Let me type it somewhere I can see it prior lol

@MikeM Did you change letters from me?

maybe? the first two things you've got written there are bizarre

Maybe I should write $T=P^{-1}JP$

i'd rather not chase through notation, the idea is just to decompose your Jordan form like you want, and conjugate to get a decomposition of your original matrix

I thought it was an arbitrary matrix $T\in\mathfrak{gl}_n$ is similar to a jordan normal form, so there is some invertible $P$ such that $T=P^{-1}JP$

I'm sure you can implement that

$T=P^{-1}(N+D)P\implies T=P^{-1}NP + P^{-1}DP$

yes

Ahhh and we can see those three matrices to be a single one for both, and we have it okay!

Thank you, that is awesome!

I have a follow up question that isn't worth making a separate topic, for math.stackexchange.com/questions/1214918/…

Not sure what is meant by "a probability measure on the space of all disturbance sequences"

I could describe a probability measure, though I have not personally heard the term "disturbance sequence."

It's an oddity of the subject. I assume it's just a random variable.

I think the author means that the expectation of the sum is some integral of the infinite summation with respect to a probability measure.

But I can't see why, if that's the case.

*assume it's just a sequence of random variables

hi

Anyone wanna look over my proof ;)

It was a fun one, and I think it is works

what is the difference between the kernel of group action and the stabilizer?

oh I guess

for all b in B

yeah I guess ok makes sense.

sorry for asking a question then answering myself lol

I do that all the time @Karim

haha

I do it to my friends on facebook. Hey how do I do this, and then I answer it in less than a minute. Prior to asking them I spend ages and get nowhere, so weird haha

Can someone help me with a few category theory questions?

yeah me 2 :D

How do I show that all finite dimensional vector spaces form a category 'Vect'? We have learned literally no category theory

Does anyone know how I can prove that if closure A = closure B, then interior of A = interior of B

hi @JulianRachman

i'm trying to recall what the etiquette is regarding 'bumping up' a question from MSE up to MathOverflow.

any thoughts?

@ᴇʏᴇs Hello Bart.

@ᴇʏᴇs Is there a result that says that the complement of the closure is the interior of the complement?

Yes.

Hello @MikeMiller, I am feeling very bad.

The closure of $A$ is the intersection of all closed sets that contain $A$. The interior of $B$ is the union of all open sets that are contained in $B$. If this isn't the definition one starts with, one should prove it equivalent.

I'm sorry, @Jasper.

hi @JasperLoy

@MikeMiller I hope things will improve soon. It's been too long. Sorry to bore everyone, but sometimes I just feel like talking about it in this chat.

I hope so too.

Can someone give me a picture of what $x^2+yz=0$ looks like, I know it's a double cone, but the uni computers have no graphing software, and wolframs image is at a useless angle

I've drawn contours but I can't visualize it in 3d :\

@JasperLoy can you put that in some grapher thing please :)?

@Incurrence I have no software here.

@JasperLoy much sads

@Incurrence It's just a graph, no need to be sad.

I need to draw it for my assignment for some stupid reason, and I have been trying to visualise it for ages lmao

Assignment due in 3 hours and I have to do some category theory thing

Nevermind, I got google's graphing software to work.

Maybe this is too late. But the equality of closures of sets does not guarantee the equality of interiors. For example, the sets Q (rationals) and R (reals) have the same closure in R, namely all of R. But the interior of Q is the empty set while the interior of R is R.

Hi @Sayan

Maybe this is too late too. The exterior of a set equals the interior of the complement. You can either define it that way (as I usually do), or you can prove it from some equivalent definition.And the complement of the closure is the exterior, by definition or by theorem.

@AlanU.Kennington Do you know any category theory?

I try not to know any category theory at all. If I do know anything about it, that's purely accidental. I think I'll get back to my book-writing now. Cheers.

Is this true If G is abelian group then |xy| = lcm(|x|,|y|)?

@KarimMansour No

well if xy is abelian and suppose |xy| = n

then we will have $x^ny^n$ = e

yeah we can't really proceed further

@DiscipleofBarney how come the answer is definitely no

@KarimMansour Have you looked at any examples?

yeah your right

take for example $Z_6$ and take 2 and 3 obviously they don't work

Literally any group not equal to the identity works

yeah

well they will work if we have (|x|,|y|) = 1 but that is obvious since |xy| = (|x|,|y|)lcm(|x|,|y|)

@DiscipleofBarney Do you know category theory?

@Incurrence A little bit, not much. What is the question

@DiscipleofBarney is this true ? |xy| = |(|x|,|y|)| * |[|x|,|y|]| ?

where (x,y) = gcd(x,y)

and [x,y] = lcm(x,y)

What is the other part

no

can you give me a counter example

@DiscipleofBarney Showing associativity of the Vect category, just means showing that $T_1(T_2(T_3(V)))=(T_1(T_2(T_3))(V)$?

Three transformations apply in succession is equal to all of the transformations being made to the operators, and then to the basis(vector space?)?

Nearly every abelian group, every one that has more than two elements @KarimMansour

@Incurrence So what are the morphisms?

Linear mappings

So functions, right?

I suppose so

Yes

Do you know how to show functions are associative

Hmmmm

Let me think for 1 min

No

I know that they need there domain and range to match up

Something like that

and the identity axiom is defined like $f:V_1\mapsto V_1$ how do I write this mapping generally?

can you have two divergent sequences, say $\{a_n\}$ and $\{b_n\}$, where $\{a_nb_n\}$ converges?

@GBeau Yes

Like what? (my text suggested something of the sort)

but I can't think of an example

Two sequences that oscillate between -1 and 1 in the opposite order 1*-1 = -1, 1*-1 =-1

Or I am being dumb, just made that up on the spot

mmmmmmmm

I don't know why that wouldn't work

Nor do I

the example provides insight on how I should think about similar questions, thanks

no problem

because the same example would work for something like $\{a_n+b_n\}$

What do you mean by how to write this map generally? @Incurrence

I don't know anymore I am brain dead

I screwed up since there is no $\mapsto$

only $\maps$

The identity map $f: V_1 \to V_1$ is the identity function so $f(x)=x$

Plus, unless I am mistaken, to show associativity you show that $T_3(T_2T_1)=(T_3T_2)T_1$ @Incurrence

Yes that is probably the case

That was me screwing up on the $T(V)$ thing again

We literally got taught no category theory for this assignment, and I have never seen it before in my life

Well this isn't really category theory, you are showing associativity of linear transformations

Am I?

What about the identity?

What is the question for the assignment

What is that in terms of stuff I have seen before

Okay and you will show that there is an identity

Showing there is an identity linear transformation and that they are associative

Really, is that all it is?

That doesn't seem bad

Its basically a question that says, "describe all the fundamental things in linear algebra you work with every time you do linear algebra"

"If $\{a_n\}$ and $\{a_n+b_n\}$ are both convergent then so is $\{b_n\}$."

(my text just lists a bunch of these at the end of the chapter as things you should figure out and I'm going through them)

@Incurrence Answer me this, what are the objects, morphisms and composition laws?

this one would seem true :<

Objects are all finite dimensional vector spaces and morphisms are linear mappings

@GBeau Looks true for addition, not for multiplication

yeah because for multiplication I can do something like -1,1,-1 and 1/n

Yep

I guess I'll try to write a proof for the addition case

The composition law just requires that for every three objects, e.g. $a,b,c$ there is a binary operation utilizing our morphism(linear mapping) such that we can have $hom(a,b)\times hom(b,c) \to hom(a,c)$.

I'll go through and try to figure out what I suspect on the rest first

"If $\{s_n\}$ is convergent so too is $\{1/s_n\}$." and "$\{s_n\}$ is convergent so too is $\{(s_n)^2\}$" and the last one's converse

So what is it, you told what the axiom is for composition law but you didn't tell me what the law was in this case. @Incurrence

@Dis Linear transformations are associative, which follows directly from the matrix representation of these being matrices, where matrices are associative.

first one not because of 0, I would guess

0,0,0,0,... converges but clearly 1/that doesn't

I don't see a counter example to the second though

(or its converse?)

@Incurrence At some point, in you life, you should show that functional composition is associative,

Second seems true, since we can't have negatives(as an even function)

@DiscipleofBarney functional? where does the 'al' come from?

by its converse I mean "If $\{(s_n)^2\}$ is convergent so too is $\{s_n\}$"

@GBeau That is false

oh, but that has an oscillating counterexample ofc

@Incurrence Just function composition sounds weird to me, maybe it should be that way though,

@DiscipleofBarney What is the difference between a functional and a function, noone has ever said

Oh I see

Vector space to scalar field

So since we can take the linear transfomations as matrices, the identity matrix is our identity mapping @Disc

@Incurrence That isn't what I was talking abou.

@Incurrence here

Oh ok

Composition of functions if you prefer

Does my matrix logic work though :)

If you already have that matrices are associative (normally people have functions are associative before learning about matrices)

I have proved the function thing before, maybe 7 months ago

And probably again 1.8 years ago

I think I remember answering one of your proof verification questions about composition of surjective and injective functions

You did indeed math.stackexchange.com/questions/986980/…

Good afternoon everyone.

Hello

Hello

Hello

my fund-raising efforts have been fruitful today

That's good to hear

My assignment solving efforts have been fruitful also

Although I have learnt little of category theory

I am not as scared by it

did you learn what a product and a co-product are?

I did not unfortunately, but it got changed to a bonus question for some reason(as did 5ii which is apparently impossible at our level under the time constraints)

It is something along the lines of a unique mapping from a product of two objects toward only one of the objects

The idea is to characterize a product/co-product in terms of mappings, instead of specific constructions

It's most transparent when you consider objects as sets, and arrows as functions

But mine is in regard to linear mappings(matrices) and vector spaces, does this make it more difficult or just more opaque?

no, you just replace the sets with vector spaces, and the arrows (functions) with linear maps

Although it's possible to do the construction with an indexed family of objects, it's easier to see what is going on with just two "factors"

So, how can we characterize the cartesian product of two sets with just functions?

See, normally we use "elements": $A \times B = \{(a,b): a \in A,b \in B\}$

But there's two "natural functions" (called projections) $A \times B \to A, A\times B \to B$.

namely: $\pi_1(a,b) = a$, and $\pi_2(a,b) = b$.

OH I see

So, now let's introduce our "players": $A,B,A\times B$ and $\pi_1,\pi_2$ to a new set, $C$

Yep

Suppose we have two functions: $f:C \to A$ and $g: C \to B$.

Yep

How can we "combine" them into one function: $h: C \to A\times B$?

Wait is that the same C as above?

yes, same $C$.

$h=f_1\times f_2$

Yes, or put another way: $h(c) = (f(c),g(c))$

Ahhh I see

That is: $\pi_1\circ h = f$, and $\pi_2 \circ h = g$.

The function $h$ is UNIQUE among functions $C \to A \times B$ with the properties I just listed.

That makes sense

Is your C my $\operatorname{Vect}$?

$C$ is an object. In this case, the collection of all objects is $\mathbf{Set}$.

In $\mathbf{Vect}$, the objects are vector spaces.

So, in $\mathbf{Vect}$ we characterize the direct product $U \times V$ of two vector spaces by two linear maps $p_1:U\times V \to U$, and $p_2:U \times V \to V$.

These maps (and the product) have the property, called a universal mapping property, that if $W$ is ANY other vector space, with two linear maps $f:W \to U$, and $g:W \to V$, there is a UNIQUE linear map $L: W \to U \times V$ with:

$p_1\circ L = f$, and $p_2\circ L = g$.

This is often said as "$L$ factors through the projections".

By uniqueness, we are justified in calling this map $L$ by $f \times g$.

Note I didn't actually mention ANY vectors, just linear maps, and vector spaces.

Yes, that was my mistake earlier

The product (in this case the direct product) is only unique up to isomorphism. For example, $U \times V$ and $V \times U$ are BOTH "products".

As real vector spaces, both $\Bbb R \times \Bbb C$, and $\Bbb R^2 \times \Bbb R$ are direct products of $\Bbb R^2$ and $\Bbb R$, even though they are not "identical".

In fact, any vector space of dimension $\dim(U) + \dim(V)$ can serve as a direct product of $U$ and $V$ (although defining the projections might take some ingenuity).

So it's more technically correct to say that $U \times V$ is "a" direct product.

And you can prove it is, since the functions: $p_1(u,v) = u$, and $p_2(u,v) = v$ are in fact linear maps with the desired properties.

The thing is, the same "definition" works in several different categories: sets, monoids, groups, rings, topological spaces, modules, vector spaces, algebras.

So using the "mapping properties" allows for a "portable definition", instead of one tied to the specific structure you are working with.

This is very powerful, as it allows great economy of effort: prove once, apply many.

@DavidWheeler I am stuck with an integral

Can you help me

no idea if i can or not

can somebody assist me in understanding something my professor wrote about a problem?

(I'm just not following his logic)

He writes this:

in the example he gives, how does the $t_n=t_n/s_n\cdot s_n$ work with the preceding statement so show $\{t_n\}$ diverges to infinity?

proof by comparison test?

@KajHansen Do 6.16!

Hi, anyone here to chat with me?

Are you here @ᴇʏᴇs?

Hi @DanielFischer, how is your health?

@JasperLoy So-so. Had a thing with the stomach which I'm recovering from. And a runny nose. But nothing serious.

@DanielFischer I am still struggling with mental problems. =(

Greetings

@Chris'ssis Hi, was waiting for you to come =)

@JasperLoy Hi :-) How are you doing?

@Chris'ssis Not too good. Some new things happened the past two weeks, now I need to deal with them. The thing about anxiety is that it spreads easily. But the meds help to calm me down a bit.

@JasperLoy Have you seen this one?

youtube.com/user/TheMiro0r

@Chris'ssis Thank you so much. I am going to take a nap, see you later.

@JasperLoy OK

@robjohn the problem with the logarithms (the one I posted it somedays ago here) is so so so amazing ...

@Chris'ssis Hello !

@Hippalectryon Hello

@Chris'ssis Have you ever done sums like $\sum_{k\in\Bbb{N}}\dfrac{k}{a\uparrow^k b}$ ?

@Hippalectryon Sure.

For instance, what does the one above give ?

@Hippalectryon Now I'm pretty busy with my book ...

Oh ok :-)

That's a good thing !

@Hippalectryon it's a telescoping sum ...

Hi @Hippalectryon

@ᴇʏᴇs o/

@Hippalectryon By the way, did you manage to solve the problem with the nested radical I gave you some months (or just weeks?) ago?

@Chris'ssis I unfortunately didn't have time to spend on it, but don't worry, It's stored on my computer. I'll be rather busy until August.

@Hippalectryon ahh, OK

@Chris'ssis is substitution the only method to solve indefinite integrals

I am getting unbelievable huge answers

@SayanChattopadhyay did you try the integration by parts?

@Chris'ssis Heya

@N3buchadnezzar Hi

My dog (the big one) just came to my door I let me know he wants to eat something delicious. So, I'm out to the store. Actually, we'll eat together today. :-)

I found this

mathim.com

It's like a private place to chat with math

@columbus8myhw The SE chat is still better though

ye

Back.

But I ca't see why this statement is true for $\{1,2,4\}$

@Silent $B=\{\}$ and $C=\{2,4\}$. Does that count?

Is there a kind soul with some knowledge of algebraic number theory who could review this answer of mine? It is fairly short an doesn't use much other than unique factorization of fractional ideals.

Oh

Whoops

@Silent Forget that. $B=\{1\}$, $C=\{1,2,4\}$

@ᴇʏᴇs: Did you sort out your closure/interior question? Hint: It's not true.

@columbus8myhw, but they are not disjoint!

gah

I can't read, apparently

I don't think that this is possible, then.

I doubt if the question is true

:(

@Silent Your question asks for distinct subsets, not disjoint...

@A.P., oh!

@columbus8myhw, sorry!

So... neither of us can read?

By the way, I trust you're all familiar with the fibonacci numbers $F_0=0,F_1=1,F_2=1,F_3=2,\dots$?

@columbus8myhw More or less... why?

An interesting thing is d'Ocagne's Identity

$$F_mF_{n+1}-F_nF_{m+1}=(-1)^nF_{m-n}$$

Nice ^^

But wait! There's more!

(By the way, if you want to go ahead and try to prove it, go ahead; it's not hard.)

So, if you divide by $F_mF_n$, you get

$\displaystyle\frac{F_{n+1}}{F_n} -\frac{F_{m+1}}{F_m} =(-1)^n\frac{F_{m-n}}{F_mF_n}$

Which looks kinda nice

except that the RHS is kinda ugly.

However, if you set $m=2n$, stuff starts canceling out!

$\displaystyle\frac{F_{n+1}}{F_n} -\frac{F_{2n +1}}{F_{2n }}=(-1)^ n\frac{F_n}{F_{2n}F_n}=(-1)^n\frac{1 }{F_{2n }}$