Mathematics

12:07 AM

Last night dream:

Let $L$ be a linear operator expressed in a vector space with a countable basis set and $a$ be a convergent countable sequence. Then $L$ has the following property:

$$\exists a, La=\lambda a$$

where $\sum_{n=1}^{\infty}a_n = \lambda$

Daminark

You mean a Schauder basis in particular, right?

Oh wait ofc

Sorry

But yeah I'm not clear what $a$ is here, do you mean $\lim a_n = a$ and there exists an eigenvector $b$?

Secret

$a$ is a convergent sequence which is an eigenvector of $L$, which has the property where its eigenvalue is the infinite sum of all its terms

Daminark

Oh so the vector space is a sequence space?

Secret

yup

Secret

12:28 AM

Rarely do I recall the maths in dreams exactly and that it at least does not sounds gibberish

Daminark

Ah okay I'm following now

And fair

Secret

I have not learnt functional analysis yet, thus further investigation of it has to be done later

But my gut feeling is that that equation might restrict the possible Ls alot due to the eigenvalue depends on all terms in the eigenvector

Daminark

Actually in thinking about the Hamel vs Schauder basis thing I've realized you can prove that a Banach space cannot have a countable basis via BCT

It's really nifty, so you can show that a proper subspace of a normed space has empty interior

Because if you take a ball $B(x,r) \subset S$ (the subspace), then note that for any $z\in V$, you have $x + \frac{r}{2\|z\|}z \in S$

But then $z$ is a scalar multiple of that vector minus $x$, so $z\in S$

Now if you have a Banach space, assume $\{x_n\}$ form a Hamel basis

Then $\textrm{Span}(x_1,\ldots,x_n) = X_n$

Each $X_n$ is closed and has empty interior

But then their union can't be everything by BCT

Which is fun

Jasper

12:54 AM

@Daminark Next time, in your exam, write nifty as well, lol.

Daminark

"Yo Professor [REDACTED] check out this sick lemma"

I'm not sure which of the two fits better

Jasper

lmao

I have a joke, so listen.

You can only lmao once in your life, because once you do it, you no longer have an ass.

Daminark

Heh

Jasper

@LeakyNun I seriously doubt the authenticity of that news.

@Daminark What would happen if you said 'Yo Professor, you are sick!' LOL

Secret

Grrr, this proof is immune to visualisation...

hmm...

The subspace $S$ is spanned by any $B(x,r) \subset S \subset V$ ?

thus it has no choice but to be V itself?

Hmm...

How about this:

In a normed space, $z \in V$ can always be rescaled because it can be converted into a unit vector. This result in any $z$ to lie within some $S$ where $B(x,r) \subset S$ for any $B \subset V$. Since we can start with picking a $z \in V -S$, it then follows that $V-S$ is empty

Daminark

1:19 AM

I mean the finite dimensional picture is that a ball "pokes out" into each dimension

Secret

1:32 AM

Another thing I noticed is that any $z \in V$ will end up be $\in S$, thus $S$ cannot be proper if it contains balls

@Daminark Btw, since the above theorem holds for any normed space, moving back to $\Bbb{R}^3$ with the eucledian norm, and picking a plane passing through the origin (thus a 2 dimensional subspace) does it mean that every point on this plane is the boundary in the topological sense?

Daminark

1:54 AM

Yup

Secret

Ah, now it makes sense, because back in the times I am a layman (or at least back in linear algebra 1st year undergrad), all the points inside the plane will be intuitively like the interior to me

Daminark

2:06 AM

Yeah, the trick is what dimensional ball

Secret

2:24 AM

Let $X = \bigcup_{n \in \Bbb{N}}X_n$, where $X_n$ is defined as above the span of the hamel basis. Since by BCT $X \subset V$, is there a name for the space $V-X$?

One thing I am interested is how going from countable to uncountable will allow the space to reach $V-X$

Secret

2:38 AM

hmm...

Actually, suppose I have a set of 2-balls in $\Bbb{R}^3$

then it logically follows that if I want the union of all 2-balls to be $\Bbb{R}^3$ itself, I will need uncountably many of them?

But maybe the analogy is not correct, as a countable union of finite sets is countable

Daminark

Closed 2-balls, yeah, by BCT

Secret

I am trying to understand the hamel basis in countable normed space. Since in such space, each vector has countably many components, then since a countable union of finite dimensional sets is countable, then it may seemed that countably many suitably chosen balls that spans 2 or 3 dimensions each should be able to fill in the whole space, but it does not as the property that the only subspace that has an interior is the whole space itself, and thus by BCT they cannot union to fill in the whole

space

So the question is then what do the missing points in $V-X$ look like, such that it takes uncountably many balls to reach them

Daminark

When you're in a Banach space, I dunno if you wanna think about components

Unless you're specifically looking at a sequence space

Secret

2:55 AM

sequence spaces $\ell^p$ ($\in \Bbb{N} \cup \{\infty\}$) seemed to be the most simple and one of the most ubiquitous example of a countable banach space, thus I think we can start with that as an example

Daminark

Define "countable" Banach space

One with a Schauder basis?

Secret

Yeah I think so, as otherwise we cannot talk about each component of the vectors I think?

I think what I have in mind is the following:

In sequence space, we can index each element of the sequence with the natural numbers, thus each sequence has coutably many elements

I think I am trying to generalise that to some kind of formal sequence which is any countable set and these countable sets form a normed vector space

Daminark

Okay, so one way of saying it is a complete normed subspace of R^(infinity)

Secret

yeah that sounds right

Daminark

I mean the best thing is to explicitly say you're working in l^p

Secret

3:00 AM

right, so that means sequence space is the word I am looking for

Daminark

Other Banach spaces with a Schauder basis have a coordinatization of the similar sort but thinking about linear operators through that tends to be clunky

I guess unless you're in L^p and doing Fourier but I dunno about that so I'll refrain from opening that can of worms

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$Mathematics$ Mathematics