ITT I will try to summon Emma-Stone-fag. We recently had a discussion, allegedly with you, on computable/constructive spectral theorem.
Namely, if [math]A[/math] is an [math]n \times n[/math] complex-valued matrix, then there NEEDS TO EXIST an approximate spectral decomposition that is effectively computable/constructive. This means that there need to exist approximate eigenvalues [math]\alpha_1, \ldots, \alpha_n[/math], approximate eigenvectors [math]v_1, \ldots, v_n[/math] such that [math] \| A v_i - \alpha_i v_i \| \leq \varepsilon [/math], approximate projections [math]P_1, \ldots, P_n[/math] such that [math] \| A - \sum_{i=1}^{n} \alpha_i P_i \| \leq \varepsilon [/math] projecting onto invariant subspaces whose dimension and bases can be effectively computed and the whole space decomposes into a direct sum of those.
You recommended to contact some bigshot in this field. Apparently, they don't even notice messages among shittones of spam. If you have contacts to them by any chance, please ask. This problem has a number of serious consequences.
nigga no need for no big shot 1/fix an epsilon 2/start with the classical algorithm to find the biggest eigenvalue and an associated eigenvector (power method or a variation of the power method) once you have that, you see that you only need to scale the eigenvector v1 you found so that ||Av1 - a1 v1||
Colton Fisher
>to find the biggest eigenvalue You immediately pooped your pants here.
Samuel Adams
Seems hard to believe. A can be any matrix? Not just hermetian or normal? Are the v_i supposed to be nearly orthogonal?
Seems like you are saying all matrices are nearly diagonalizable.
Michael Murphy
ADDENDUM: MATRIX IS HERMITIAN
Cooper Hughes
Sorry for that. It's hermitian.
Austin White
how? Because you're a mathfag and don't know you can perturbate matrices?
The post was just the gist of it. I won't attach a fucking functional code that works in all cases.
OP can adapt this to whatever he needs to do.
Levi Young
If it's Hermetian then it can be diagonalized exactly.
Which part is bullshit? Hermetian matrices are diagonalizable by a unitary matrix. The question is either has an obvious answer or poorly stated.
Lucas Ramirez
try to do the finite case first, like the finite matrices acting on the complex vector space C^n
Juan Lopez
That's exactly the question
Brandon Morales
bump
Camden Gray
BUMP FOR WHAT
Gabriel Ross
What do with degenerate eigenvalues?
Jack Lopez
Is this from Rudin's Functional Analysis? Looks like the print.
Leo Watson
>"degenerate"
Jacob Morgan
do you guys have some kind of autism? what the fuck are you talking about? what's a "computable/constructive" spectral theorem?
Xavier Gutierrez
It's right there in Wikipedia...
"The finite-dimensional spectral theorem says that any Hermitian matrix can be diagonalized by a unitary matrix, and that the resulting diagonal matrix has only real entries. This implies that all eigenvalues of a Hermitian matrix A with dimension n are real, and that A has n linearly independent eigenvectors. Moreover, Hermitian matrix has orthogonal eigenvectors for distinct eigenvalues. Even if there are degenerate eigenvalues, it is always possible to find an orthogonal basis of C^n consisting of n eigenvectors of A."
It always works.
Hunter Lewis
Just fuck you, piece of an idiot.
Nathan Hill
bump
Hudson Clark
I think OP is not right in the head.
You think Wikipedia is wrong, or what?
Josiah Gray
Leave the thread
Juan Green
bump
Hudson Powell
If I leave you will be alone. You are asking for the solution to a problem that is well known and solved.
The book Harold M. Edwards-Essays in constructive mathematics-Springer (2004) has a section on the spectral theorem and a detailed proof of existence. He claims that the explicit formulations of the spectral matrices is through another algorithm:
>Therefore, the spectral decomposition of S can be given once the Moore-Penrose generalized inverses of the matrices I — S + pil are found. Note that the computation of the Moore-Penrose generalized inverse of a matrix requires exact computations with the entries, so it becomes possible only after a splitting field for the minimum polynomial is constructed; the interpretation of the p's and P's as real numbers requires an identification of the splitting field with a subfield of the field of real numbers.
Does it answer your question?
Aaron Mitchell
No. Look at the statement. The matrix entries are integers. Furthermore, I don't need an exact decomposition.
So again, if they do not care, you must do the work yourself, at leas tin finite dimension. Real well Chapter 6 and chapter 8 I guess.
Kevin Gonzalez
I saw that Paul Taylor himself answered you on about the reals. Nice.
Anyway, I just come here to recall, to anybody, that Henri lombardi published his book about constructive algebra (on libgen) which is not about linear algebra. hlombardi.free.fr/publis/LivresBrochures.html
Alexander Harris
Well, it seems the conclusion can be made. This turns out to be an open problem as for now. But addressing it might be not a cakewalk. Perhaps, I should work on it.
John Russell
From I understood, the reals numbers used in ASD have bad behavior in that there is no computable bound on iterations that allows you to compute a given number up to a predefined precision.
I personally feel much better with the usual Cauchy real numbers.
