Let [math]T[/math] be an operator on Hilbert space [math]H[/math]. Let [math]A \in H[/math] and [math]\chi_A[/math] its characteristic function.

What does [math]\chi_A(T)[/math] mean?

Juan Allen

Let [math]T[/math] be an operator on Hilbert space [math]H[/math]. Let [math]A \in H[/math] and [math]\chi_A[/math] its characteristic function.

What does [math]\chi_A(T)[/math] mean?

Leo Martin

for x in H

xhi_A (T)(x) = :

1 if T(x) is in A

0 otherwise.

Julian Lopez

Unfortunately no.

Here is a context btw

With your assumption, the dimension doesn't match

Adrian Jackson

bump

Leo Anderson

then don't call chi a fucking characteristic function if it's not one. It's a projector.

Justin Kelly

It is a characteristic function

Andrew Gomez

so what's wrong with the previous response?

T acts on H.

You take an element of H, you apply T to that element.

Then you check if T(x) is in A or not.

and that gives you the result

Maybe I'm misunderstanding something.

In the case of this picthe operator is A.

So you take Ax

you check if it's in the subset B_i

you multiply by ci

you sum over all i

etc

Carter Baker

There is a big difference between [math] f(Tx) [/math] and [math] f(T)x [/math].

Dominic Allen

You are idiot. That's the problem.

Zachary Sullivan

Pleb here. What is a Hilbert space and how do I use it? I've been trying to understand the concept for a while now, but so far I haven't made any progress.

Josiah Lee

bump

Joseph Turner

Roughly speaking a Hilbert space is a (usually infinite dimensional) vector space with a geometric structure similar to that of normal Euclidean space. In fact, Euclidean spaces are exactly finite dimensional Hilbert spaces. Hilbert spaces are used in quantum mechanics. P

Adam Hill

>that's the problem

not explaining the problem

k m8.

Come back when you understand that it doesn't matter here.

Kayden Myers

Nigga, you can't even into muh matching dimensions. What are you talking about? Clean out your eyes and look at the expression.

Parker Davis

I really don't see which fucking dimensions are not matching, I explained how everything matches and you're just here spewing bullshit instead of actually giving information. If I'm wrong, correct me properly or don't say anything.

Jack Young

Ax is a vector. The summand here is supposed to be a vector. In fact, it's a vector from an invariant subspace.

Hudson Long

there's a lot in here that doesn't make sense to my eyes, perhaps because of missing context. i'd guess though, that chi(A) is suposed to be the operator whose existence you get from the spectral theorem (assuming A to be normal). then you can apply any bounded, measurable complex function to your operator.

Adrian Watson

It's defined before introducing spectrum. It should be a limit of polynomial approximation I guess.

Isaiah Torres

therefore the definition of a characteristic function doesn't fit the function used here. Which is my first point. Chi is not a characteristic function as defined in

Jackson Bennett

It's always a good idea to consider a simple example.

Let [math] H = \mathbb{C}^n [/math] and let [math] T: H \to H[/math] be a self-adjoint operator with distinct eigenvalues [math] \lambda_1 , \ldots, \lambda_n [/math].

Then [math] T [/math] can be written as [math] T = \sum_{k=1}^n \lambda_k | \lambda_k \rangle \langle \lambda_k | [/math] and any element [math] | x \rangle \in H [/math] can be written as [math] | x \rangle = \sum_{k=1}^n c_k | \lambda_k \rangle [/math] for some [math](c_k) [/math].

Now you can define a Functional Calculus [math] f(T)x := \sum_{k=1}^n f(\lambda_k) c_k | \lambda_k \rangle [/math]. So if [math]f[/math] is the charactersitc function of some set [math]A [/math] you just put those [math]c_k[/math] that don't belong to an Eigenvalue in [math] A [/math] to [math] 0 [/math] while leaving those that belong to an Eigenvalue in [math] A [/math] alone. This makes it to a projection. If the set [math] A[/math] only includes a single Eigenvalue then it would just be the projection to the corresponding Eigenspace.

Brandon Watson

Yes it is. You can consider it as a limit of a polynomila function.

Brody Ross

>Hilbert spaces are used in quantum mechanics. P

I find your kind of explanation confusing. Just because you explicitly use the word "Hilbert space" when you teach quantum mechanics, doesn't make it untrue that you use Hilbert spaces whenever you do something in R^2.

Same goes with

>usually infinite dimensional

Yes I know what you mean, but.

Jose Reed

The problem is that it was introduced before introducing spectrum

Jack Howard

alley cat, yee :)

Hudson King

Okay familia here's how it works. As stated before, Hilbert spaces are an abstraction of our usual Euclidean space into an arbitrary number of dimensions. This abstraction allows us two important extensions: defining the Hilbert space [math] \mathcal{H} [/math] over the complex field [math] \mathbf{C} [/math] and defining an inner product [math] \langle \cdot , \cdot \rangle [/math] in the same [math] \ell^2 [/math] sense as in typical Euclidean space (basically, the Pythagorean theorem).

Point 1 is sorta trivial in the context of explaining the concept, so I'll skip that (you just add imaginary numbers to the real numbers, throw in some c.c.'s, and so forth).

Let me address point 2 by first addressing your comment

>doesn't make it untrue that you use Hilbert spaces whenever you do something in R^2

Generally speaking, you're right. Here's the thing though -- when you speak just of [math] \mathbf{R}^2 [/math], you're just talking about the Cartesian product between 2 real fields. There is no structure within it: no *metric* which tells me how vectors within [math] \mathbf{R}^2 [/math] are additively measured from one another, no *inner product* to tell me how they can multiplicatively interact with one another. We take for granted that when we speak of the n-dimensional real field [math] \mathbf{R}^n [/math], we actually mean the metric space [math] E = (\mathbf{R}^n , d) [/math], where the metric [math] d : \mathbf{R}^n \times \mathbf{R}^n \to \mathbf{R} [/math] is defined by [math] d(x,y) = \sqrt{\sum_{j=1}^n (x_j-y_j)^2} [/math] for [math] x = (x_1, x_2, \dots , x_n),\ y = (y_1, y_2, \dots , y_n) [/math].

So it's not the field over which you define your space that matters -- it's really all about the metric, from which you have a sense of distance and length between elements of your space.

Camden Thompson

[math] [/math]

So let's bring that all back to the Hilbert space. We equip the Hilbert space, not with a metric, but an inner product: [math] \mathcal{H} = ( \mathbf{R}^n , \langle\cdot,\cdot\rangle ) [/math] (I'm sticking with real Hilbert spaces for now). The inner product [math] \langle\cdot,\cdot\rangle : \mathbf{R}^n \times \mathbf{R}^n \to \mathbf{R} [/math] is defined as [math] \langle x , y \rangle = \sum_{j=1}^n x_jy_j [/math] (using the same vectors as in the previous post). Indeed, it is the usual inner/scalar/dot product we're so very familiar with in Euclidean space: that's because all Euclidean spaces are Hilbert spaces. (Note that you can define the metric/norm from the inner product and vice versa.)

The divergence, of course, happens when you consider [math]n \to \infty [/math] -- now, thinking of Euclidean space makes no sense! But what are infinite dimensional vectors but functions -- or rather, the sequence of values that a function takes over the domain of the space. We have to redefine some notions of length and direction with these infinite dimensional vectors, of course. We move from the finite vectors [math] x\in\mathbf{R}^n,\ y\in\mathbf{R}^n [/math] to functions [math] f\in L^2,\ g\in L^2 [/math], where [math] L^2 [/math] is the space of square-integrable functions on [math]\mathbf{R}[/math]. We now define the inner product as [math] \langle f , g \rangle = \int f g [/math] with some measure of the integration (Riemannian is usually good enough for physics purposes, I believe). Just as when [math]x [/math] and [math] y [/math] orthogonal functions, we say [math]f [/math] and [math] g [/math] are orthogonal functions if [math] \langle f , g \rangle = 0 [/math]. Thus we have a sense of projection and direction between functions (infinite dimensional vectors), just as we did with the usual vectors.

In a sentence: (inf. dim.) Hilbert spaces allow us to extend the geometry of Euclidean spaces into function spaces.

Liam Hughes

What's about the actual question?

Aaron Parker

Have you ever worked with fourier series? Congratulations, you've worked with the concept of an infinite dimensional Hilbert Space. Each function [math]\sin(\frac{n\pi x}{L}),\ \cos(\frac{n\pi x}{L})[/math] is orthogonal to all others over the range [math]-L

Carson Carter

So what's the correct answer?

Kayden Myers

>They then form a basis in a Hilbert space

No, one of the first things you should learn about Hilbert spaces is that they never have a countable basis. This is usually proven with Baire category theorem.

Zachary Ross

bump

Julian Wright

bump

Matthew Johnson

bump

Connor Evans

So what'S the correct answer?

Aiden Hernandez

x

Isaac Adams

Whoah, thanks user. If only all posts on this shithole of a board could be this insightful.

Gavin Gonzalez

So, it this actually a limit of a polynomial?

Owen Powell

it is defined in terms of subspaces of H or measurable sets.

Caleb Cooper

Great answer. 0% informative

Gabriel Butler

bump

Colton Long

bump

Jason Hall

Is it defined throught the spectral measure?

Henry Jones

?