Mathematics - 2017-03-04 (page 3 of 3)

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Semiclassical

23:00

(namely, what's that by definition of Ker?)

Maks

@Semiclassical You mean the image of the kernel of T ?

Semiclassical

Right.

Vrouvrou

@AlessandroCodenotti yes

Maks

Isnt that T ?

Semiclassical

uh, no.

Maks

23:01

I mean, all the vectors in ker(T)

Alessandro Codenotti

Are those sets in $\theta$? Are they in $\tau$?

Vrouvrou

it is in $\theta$ by not in $\tau$ @AlessandroCodenotti

Semiclassical

You're not thinking sensibly. What does it mean to be in the kernel of T?

Maks

The image of the kernel of T is 0..

Semiclassical

Right.

That's all I was after.

Maks

23:02

Yeah, I was kindda brainstorming before hahah

Alessandro Codenotti

@Vrouvrou so $\tau$ is not finer than $\theta$

Maks

Then $Nu(T) \subset Im(T)$

Vrouvrou

so $\theta$ is finer

Alessandro Codenotti

not necessarily

Semiclassical

No.

Alessandro Codenotti

23:03

they might be incomparable

Maks

In this case

Semiclassical

the kernel is still a subset of the domain, and the image is still a subset of the range.

Alessandro Codenotti

Do all subsets of $\Bbb R$ not containing $[x_0]$ have a finite complement?

Vrouvrou

Alessandro Codenotti

so those sets are in $\tau$ but not in $\theta$

Maks

23:04

If my range and domain belong to different fields then I cant compare them ?

Alessandro Codenotti

vector spaces rather than fields but yes

Vrouvrou

thank you @AlessandroCodenotti

Alessandro Codenotti

you're welcome

Semiclassical

A can't be a subset of B if A and B contain different elements.

Maks

Then in this case $Nu(T) \not\subset Im(T)$ and $Im(T) \not\subset Nu(T)$

And is $T$ isomorphic ?

$T$ is isomorphic is its bijective

Alessandro Codenotti

23:07

vector spaces are isomorphic, maps can be isomorphisms

Maks

In this case I say it is isomorphic

But I can only tell it by intuition

Alessandro Codenotti

An isomorphism of vector spaces is a bijective linear map, right? Is your map linear? Injective? Surjective?

Maks

So, if my map is linear then its isomorphic ?

Alessandro Codenotti

it makes no sense in english to say that a map is isomorphic

Maks

And its linear if
$T(0) = 0$
$T(\alpha + beta) = T(\alpha) + T(beta)$
$T(c\alpha) = cT(\alpha)$
right ?

@AlessandroCodenotti It does in spanish hahaha

Alessandro Codenotti

23:13

"isomorphic" is an adjective that refers to vector spaces, not functions

Semiclassical

In english, you'd say that the map is an isomorphism and that the range/domain are isomorphic.

Maks

What about, Isomorphism ?

@Semiclassical yeah, that's it

user400188

@jublikon I left chat shortly after that first message, (so I'm not sure if it was already answered by another) but as to the answer to your second question: http://chat.stackexchange.com/transcript/message/35823948#35823948
$\overline {x_1}x_2 + x_1 \overline {x_2}$ simplifies to $x_1\oplus x_2$, where $\oplus$ is the XOR operation. It does not simplify to true.

Maks

I sometimes struggle translating concepts

@Maks If I prove this, then T is an isomorphism ?

Alessandro Codenotti

so I have to go now, but if you want to check whether your function is an isomorphism you should check whether it's linear, surjective and injective (Actually the last one is not needed, a linear map between finite dimensional vector spaces is injective iff it is surjective, you should prove this though)

Simple

23:19

hello, chat

Semiclassical

hi @Simple

Still doing g.f. stuff?

Simple

I am writing a paper about knot theory

Semiclassical

Ah.

Vrouvrou

You are here @AlessandroCodenotti?

Henning Makholm

Is it just me, or is there a sudden influx of (otherwise unrelated) questions that insist on writing definite integrals with \limits in question titles?