/mg/ - Math general - Perfectoid edition

what are you talking about?

Actual /mg/ thread here

But /mg/ isn't a reddit thread? So this seems to be the legitimate one.

Who ENS here?

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>But /mg/ isn't a reddit thread? So this seems to be the legitimate one.
Scholze is the personification of redd*t

You.

It doesn't matter what you think. Reddit threads encouraging off-topic discussion and redditry such as despising the anime culture of this website will not be tolerated.

What are some applications of vector spaces which do not have a basis?

Please be my friend.

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>vector spaces which do not have a basis
There is no such thing silly user.

Appart from the memes, functional analysis deals with this shit, i.e. linear spaces with eveb an ubcoubtable basis such as function spaces. It's used in fourier analysis and harmonic analysis which is useful for PDE, but yea
All vector spaces have a basis no matter how much the retards here yell.

My answer will depend on what exactly you study.

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Don't listen to that physicist

You seem to know what he's talking about. Can you share this information with me?

>Appart from the memes, functional analysis deals with this shit, i.e. linear spaces with eveb an ubcoubtable basis such as function spaces.
I said no basis, not uncountable basis.

No such thing sweetie

Most vector spaces do not have a basis unless you assume it does

>Most
What do you mean by this? Can you provide one example?

>What do you mean by this? Can you provide one example?
The real vector space of functions [math] \mathbb{R} \to \mathbb{R} [/math]

He's just some retarded troll ignore him.

>He's just some retarded troll ignore him.
I'm not a "he".

Yea, you are just a faggot

Prove that it doesn't have a basis.

It definitely has a generator set

>It definitely has a generator set
That is not a basis.

>faggot
Why the homophobia?

>Prove that it doesn't have a basis.
The burden of proof is on you.

It has a subset that is linearly inependent and that spans the whole space.

>It has a subset that is linearly inependent and that spans the whole space.
The burden of proof is on you.

isnt (1) the basis??

Reminder to report and ignore the retard.

>isnt (1) the basis??
What function "(1)" is meant to denote? And why does it generate the whole space?

proofwiki.org/wiki/Vector_Space_has_Basis yea I know, it's a "popsci" link, hilarious, but just to prevent any newb falling for you bullshit.

see

>And why does it generate the whole space?
hmm because you can multiply 1 for every number and you get the whole R

You are also retarded my dude. The space is all the functions, not all the real numbers.

[math]\forall x\in \mathbb{R}[/math] let [math]V_x: \mathbb{R}\rightarrow \mathbb{R}, V_x(y)=\chi _{\{ x\} }(\{y\})[/math], these functions are clearly linearly independent, and [math]\forall f:\mathbb{R}\rightarrow\mathbb{R}, f=\int_{\mathbb{R}}f(x)\cdot V_x(x)dx[/math]

>The space is all the functions, not all the real numbers.
They are isomorphic vector spaces.

Not in any way or form.

Sorry, there isn't supposed to be an (x) after V_x at the end there.

Commit sudoku

Alright, I'll look for my puzzle magazine first thing in the morning.

interval calculus :$

>They are isomorphic vector spaces.
The burden of proof is on you.

>∀x∈R let Vx:R→R,Vx(y)=χ{x}({y}), these functions are clearly linearly independent, and ∀f:R→R,f=∫Rf(x)⋅Vx(x)dx
That's not necessarily a linear combination.

Next?

>The space is all the functions
uhm... so if we multiply every function by the costant function 1 we will get all the space

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explain to me what i did wrong

>explain to me what i did wrong
Everything.

ok, so you have no argument lmao

>ok, so you have no argument lmao
There is no notion of multiplying functions in this vector space.

then how the fuck are you supposed to generate the space if you can't have a linear comb

The field are the reals, multiplying by the identity function is not a fucking linear combination, is a whole new operation.
By reading definitions carefuly

Which one? T'es un feuj ?

>then how the fuck are you supposed to generate the space if you can't have a linear comb
You can have linear combinations.

>multiplying by the identity function is not a fucking linear combination
>multiplication is not linear
Are we reaching new level of retardness or what?

"the identity function" is not a scalar in the vector space.

so what is a scalar in that vector space?

But the space of functions doesn't form a field you inbred retard, so you cannot form a vector space of the space of those functions over itself.
The real, as the set of real numbers which are not the same as the spacw of real valued functions.

>so what is a scalar in that vector space?
Any real number.

[eqn]
x^2 - 5 = 2\\
\frac{d(x^2 - 5)}{dx} = \frac{d2}{dx}\\
2x = 0\\
x = 0
[/eqn]

>infinitely many summands
>linear combination

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real not reddit thread here

en.wikipedia.org/wiki/Hilbert_space#Orthonormal_bases
>In the infinite-dimensional case, an orthonormal basis will not be a basis in the sense of linear algebra; to distinguish the two, the latter basis is also called a Hamel basis. That the span of the basis vectors is dense implies that every vector in the space can be written as the sum of an infinite series, and the orthogonality implies that this decomposition is unique.

>an orthonormal basis will not be a basis

I'm not , but perhaps you should've clarified ahead of time that you meant a basis in the general sense and not in the specialized sense that "basis" is often used to mean in the context of discussing a Hilbert space like R → R

I said "vector space" and "basis", not "Hilbert space" and "orthonormal basis".

well, ok

Is he actually this stupid, or is he just doing it for attention?

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>well, ok
Which inner product can you put on this space of functions to get a Hilbert space?

oh right, oops

He's a pretty decent researcher apparently.
Just very autistic about foundational stuff I guess.

to be fair the rise of set theory in mathematics is basically an endorsement of autism

"Autistic" is a bizarre way of describing it, given that being literally autistic myself has probably been one of the biggest factors in me actually learning & understanding foundations

Recapping some basic alg topo as a bedtime book.

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FRUMP

may your dream sequences be exact

how hard would it be to learn multivariable calc on my own? ive just finished integration at uni.

How much linear algebra do you know?

not much. should i learn that first? is there anything i should look into before starting linear algebra/multi-variable calculus

*anything else?

Intro-level multivariable calculus is like 80% calculus + linear algebra. I may be forgetting something, but I think you should be able to at least learn some important basics as long as you're comfortable with single-variable derivatives and with vectors and linear transformations.

Hmm, you don't really did that much libear algebra, or you can learn it as you need it, also what is your goal? I.e. are you a mathematician?

What bedtime book are you using?

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How does this impact anything? You're a single autist.

cool! not really sure what linear transformations are either yet. i probably know them, just not by name.

i hope to be one! im a first year math major

>im a first year math major
>not really sure what linear transformations are
How does this even happen?

A linear transformation is something that maps vectors to other vectors and which is "linear", which has several equivalent definitions. The standard one in the context of formal math—albeit maybe not the most enlightening one—is that a function [math]f[/math] from one vector space to another is called "linear" if
1. for any vectors [math]v, w[/math] in the source space, [math]f(v + w) = f(v) + f(w)[/math], and
2. for any vector [math]v[/math] in the source space and scalar [math]a[/math], [math]f(av) = af(v)[/math].

There are better explanations, but it'd take a while to explain, and a proper source would probably do it better than me anyway.

A few simple examples—rotations about the origin are linear, as are reflections through lines/planes that pass through the origin. Uniform stretching along one or more axes is also linear.

>?
that's so cute