Spectral Theory - Science topic

When I did a google search using these keywords, several sources appeared: agriculture hyperspectral download.

Hi professors. I hope you are doing well. This is a mathematical article that proves a new recurrence relation that is fundamental for mathematics. The article proves also that four infinite series are equivalent. Hence, this article opens new opportunities to demonstrate and develop new mathematical findings and observations. This is the link: https://www.researchgate.net/publication/364651911_A_useful_new_equation_of_four_infinite_series_and_sums_by_using_a_new_demonstrated_recurrence_relation

Not sure if it has stainless steel, but a good reference to have saved nonetheless.

Stam Nicolis , you said that

" However it's not necessary for the Hilbert-Polya conjecture to be true, in order for the Riemann Hypothesis to be true. "

Why is that?

First of all it depends to the material studied, and for me i guess that IR is more better then UV cause its wavelength is bigger than UV.

Dear Peter Breuer ,

Gershgorin disks work for columns too.

One can observe that A and A^T have the same spectrum.

Best wishes

The question you ask is a very broad one. Since differential operators are unbounded - the definition of the on which one defines the operator is important. When one goes to infinite dimensions, there might not even be a point spectrum (eigenvalues) - the spectrum may be only the continuous spectrum and residual spectrum. If the operator has a self joint extension or a normal extension then the problem becomes more tractable but for non-normal operators the problem is quite complex.

At one time a class of operators "spectral operators" were defined, see Dunford and Schwartz, "Linear Operators", vol III, to address the issues on non-normal operators. Proving a non-normal operator - even an ordinary differential operator - was not a trivial task. The goal of this effort was to expand the concept of Jordan Canonical form to first bounded linear operators on a Hilbert/Banach space and then to unbounded linear operators, e.g., differential operators. For compact operators or operators with compact resolvents - that has been solved. For bounded normal and unitary operators, that has been solved (spectral theorem). For unbounded operators with a compact resolvent, that has been pretty much solved (by using the spectral theorem the spectral theorem). For unbounded self adjoint operators, there is also a spectral theorem. However, for non-normal bounded operators and non-self adjoint unbounded operators - there is no general result on a canonical form.

In these problems stability questions be quite problem specific.

So far I don't know any Translation. But you find all stuff in H.H. Schaefer: Banach Lattichs and Positive Operators in Chapter 1. 

Another example of a blog that has gone astray completely! Why don't you try to debate the question that has been posed? I think this does not serve any useful purpose!

This would depend on the type of star. In hot stars the hydrogen in the stellar envelope the hydrogen is fully ionized. In cold star it's not. Also, stars like the sun have a hot, low density corona where the hydrogen is ionized.

Dear Francesco,

As the the entries in the array A are complex numbers, I assume A[][].imag = 0 means set the imaginary part of the number to zero.

Steve

Dear  Prof.  Prykarpatski,

According to my previous answer, I think that we can not expect that the operator  T(x)=ax+xa is  a positive operator where a is  a positive element. I realize that there is  a  counterexample in M_{2}(C). There are two positive matrices A and B such  that AB+BA is not positive. Ex: A=\pmatrix(1&0 / /0&2)   and  B=\pmatrix(10.01& 10 \\ 10& 10.01).  In fact the following general statement is true: 

Statement:  If A is  a  C*  algebra such that  ab+ba is  a positive element for all positive elements a and b then A is  a Commutative C* algebra.

This is  a  consequence of the following theorem(which I forget the paper and its Author but it is quoted in the book "K-theory and C* algebra by W. Olsen):

Theorem: Assume that A is  a  C* algebra such that   0<a<b  implies a^{2}<b^{2}, then A is  commutative.

any way it would be interesting to find an example of  a  simple C* algebra with a positive element a, which is  not an scalar element but the operator T(x)=ax+xa is  a positive operator.(Equivalently: the  cones  of  positive element is invariant under T)

 The motivation for   assumption of "simplicity" is that we wish to be most far from commutativity.

Thanks for your original interesting question

Best regards 

Ali Taghavi

thanx dear prod wiwat..

eta_c is an array that you need to compute for each range samples.

eta_c(0) is the first element of this array representing eta_c at the near range of your product

Dear Prof. Peters, thank you. I will take a look.

See the article , I will send you.

Dear Henk, I understand that this letter is addressed to me, and not the one to whom you are writing. I made a mistake in your answer. And it will be a good lesson for me. I'm really not an expert in this area. You can see my profile.

But just to criticize and not to say anything positive, do not explain the person who asked the question, even if he is not right, in ResearchGate, and indeed in the scientific community, in in my opinion it isn't accepted. As I understand, Riyaz Ahemad just studies. And if you so well understand the question, do you really so difficult to explain the incorrectness of its statement?

The same thing I would say and Alexandr N Ryabtsev, who expressed his surprise, too, does not respond directly to the question, and refers to the head. I think that Riyaz Ahemad himself would have understood the need for this. Not difficult to see that you are his answer, actually repeated. Why? And without your speech was clear.

Best Regards, Leonid.

Another interesting thing is that the limit E1->E0 of the single-zone potential corresponds to the single-soliton potential. See gif-file attached as an example

What you are asking for is what people working in C^* algebra's call a functional calculus, i.e. a way to "evaluate" a complex or real function (of some kind), at an element of a C^* algebra (like your algebra of matrices) possibly assuming that the elements are hermitian or positive, or normal or... Now being able to "evaluate" a function is just a way to say that for every element u in the C^* algebra there is a *-algebra homomorphism from the space of functions to the C^* algebra is such a way that f(x) = x gets mapped to u. The canonical example is the functional calculus of the Hermitian operators on a Hilbert space. A Hermitian operator A can be "evaluated" on measurable functions f on the real line which are bounded on the spectrum of A using the spectral decomposition of the Hermitian operator. A general operator A can still be "evaluated" on the complex functions which are holomorphic on its spectrum using a variation of the Cauchy formula. If you don't know this already, see most any book on functional analysis (my favorite one, is the book by Rudin), and it is used extensively in quantum mechanics.

For the multivariate case you immediately run into two problems. For simplicity lets assume k = 2 so that we have to define an "evaluation" of two operators A and B. The easy problem is that we now have to distinguish two functions x and y such that x(A,B) = A and y(A,B) = B. We simply assume that the polynomial algebra R[x,y] is embedded as a sub algebra in our function space. The main problem is non commutativity. Clearly real (or complex) functions being commutative, the functional calculus must end in a commutative sub algebra of the (C^*)-algebra generated by A and B, and now we see that we cannot have our cake and eat it too: if A and B do not commute we cannot have an _algebra_ homomorphism from R[x,y] to the C^* algebra (in quantum mechanics this is known as the Van Hove theorem) so, a fortiori, cannot extend it to a function space. What we can try however is to take a nice sub algebra of R[x,y]. Now, as is well, known, the symmetric polynomials (in two variables) are polynomials in the elementary symmetric polynomials $\sigma_1$, $\sigma_2$ i.e. $R[x,y]^{S_2} = R[\sigma_1, sigma_2]$. Now we _can_ define a homomorphism $\phi_1$

R[\sigma_1, \sigma_2] \to C^*(A,B)^{S_2}

\sigma_1 \mapsto (A+B)

\sigma_2 \mapsto (A+B)^2

Now this is a bit disappointing because one would really like

\sigma_2 \mapsto (AB + BA)/2

but that does not work, because in general (A+B)(AB + BA) \ne (AB + BA)(A+B) !

There is however no problem to extend this homomorphism to continuous or even real measurable functions $f(\sigma_1, \sigma_2)$ of two variables, because for such a function one really has to "evaluate" the real function of one variable f(t, t^2) at $t = A+B$.

Also the homomorphism $\phi_1$ is far from unique if A and B are positive definite, we can define $\phi_2$

R[\sigma_1, \sigma_2] \to C^*(A,B)^{S_2}

\sigma_1 \mapsto A^{1/2}(A^{-1/2}BA^{-1/2})^{-1/2}A^{1/2}

\sigma_2 \mapsto \phi_2(\sigma_1)^2

and this extends to measurable functions for the same (cheap) reason as above.

At the end therefore one really has to be more precise about one wants to preserve.

If the differential operator is not a Riemannian invariant the I donot see a reason why it should relate to the volume. For example the differential operator defined by glueing the laplacian in R^n using partition of unity (i.e.

\sum_{\alpha \in A} \rho_\alpha)(x) \sum_i (-\frac{\partial^2}}{\partial x_{i, \alpha}^2}

where the {\rho_\alpha}} is a partion of unity and the x_i are some arbitrary coordinates defined where \rho_{\alpha} \ne 0 ) will not relate to the volume of a predefined Riemannian metric because it has nothing to do with this metric.

A wild guess could be that you could get non-real eigenvalues by setting some weird boundary conditions, say oblique ones. The operator associated with a variational form of the problem will then be non symmetric, even if the leading term is something standard like the Laplacian.