Let $ A, B\in M_n(\mathbb{C})$ s.t. $ AB=BA$ . Show that the eigenvalues of $ A+B$ is the sum of eigenvalues of $ A$ and $ B$ .

First I tried to find a counterexample and I found out this property without proof.

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# Tag: eigenvalues

## Show that the eigenvalues of $A+B$ is the sum of eigenvalues of $A$ and $B$.

## Eigenvalues decrease with power

## Rough conjecture about eigenvalues of Maass forms

## Eigenvalues of the Laplace-Beltrami operator on a compact Riemannnian manifold

## Wrong Eigenvalues result in Version 12.0

## If A = u*u^T where 0 ≠ u ∈ R^n , then find the eigenvalues of A and show that A is diagonalizable.

## General solution and eigenvalues and eigenector

## Eigenvalues of the Laplacian of the directed De Bruijn graph

## Eigenvalues and eigenfunctions of an integral operator

## Eigenvalues of a block matrix from the eigenvalues of the blocks

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Let $ A, B\in M_n(\mathbb{C})$ s.t. $ AB=BA$ . Show that the eigenvalues of $ A+B$ is the sum of eigenvalues of $ A$ and $ B$ .

First I tried to find a counterexample and I found out this property without proof.

Take $ n \in \mathbb{N}$ , and consider a square matrix $ A$ of size $ n \times n$ , with real and positive entries, and such that $ \|A\|_2 \leq 1$ .

I think the following statement holds from simulation, but fail to show that it is true.

$ p \mapsto \lambda_i(A^p(A^T)^p)$ decreases when $ p$ increases, where $ \lambda_i(\cdot)$ is the $ i$ -th largest eigenvalue (in complex norm).

Basically, as I know, we know almost nothing about Maass forms. For example, Cohen constructed first (maybe not) example of a Maass cusp form by using one of Ramanujan’s $ q$ -series, as a non-definite theta series. (This is from his Inventiones paper, *$ q$ -identities for Maass waveforms*.) Also, this article from Buzzard gives some explicit examples, both “algebraic” (corresponds to even 2-dimensional Galois representations) and “non-algebraic” (something that mysterious?) Algebraic ones has an eigenvalue $ 1/4$ , while non-algebraic ones have (conjecturally) transcendental eigenvalues (such as $ $ \frac{1}{4} + \frac{\pi^{2}}{4\log^{2}(\sqrt{2} – 1)} $ $ Also, this paper provides an algorithm to compute eigenvalues effectively, and the first eigenvalue for $ \mathrm{SL}_{2}(\mathbb{Z})$ is $ \lambda = \frac{1}{4} + r^{2}$ , where $ r = 9.5337…$ .

I think the eigenvalues should be some special numbers, which can be transcendental, but maybe a little *algebraic*. More precisely, I hope that these numbers are related to *periods*. Periods are defined in Zagier-Kontsevich’s article (this) as numbers that can be obtained by integral of rational functions over a domain defined by inequalities of rational coefficient polynomials. For example, any algebraic numbers, $ \pi, \log 2, \zeta(3)$ are periods. Especially, $ $ \log(\alpha) = \int_{1}^{\alpha} \frac{1}{x} dx $ $ is a period for any $ \alpha\in \overline{\mathbb{Q}}$ . Periods form a proper subring of $ \mathbb{C}$ that contains $ \overline{\mathbb{Q}}$ , denoted by $ \mathcal{P}$ , and $ \mathcal{K} = \mathrm{Frac}(\mathcal{P})$ is a field between $ \overline{\mathbb{Q}}$ and $ \mathbb{C}$ . So here is my conjecture:

If $ \lambda$ is an eigenvalue of a Maass waveform (cusp form) on $ \Gamma_{0}(N)$ , then $ \lambda \in \mathcal{K}$ .

Obviously, there’s no reason that this conjecture is true. The reason I belive this is because the eigenvalues shouldn’t be just random transcendental numbers. To *prove* this conjecture, the first thing we have to figure out is the exact value of the smallest eigenvalue for $ \mathrm{SL}_{2}(\mathbb{Z})$ , which is approximately 91.14. Thanks in advance.

p.s. According to this question, if we assume some conjectures about motivic stuff, then $ \mathcal{P}$ is not a field.

Let $ (M,g)$ be a compact Riemannian manifold, and let $ \Delta_g$ be its Laplace–Beltrami operator. A “well-known fact” is that the eigenvalues of $ \Delta_g$ have finite multiplicity and tend to infinity. What is the easiest/simplest way to estblish this fact? By simplest, I mean an abstract, formal, approach to the problem, not too heavily based on technical calculation. In the same line of thought, is there a philosophical/intuitive reason to see why the spectrum should behave like this?

The following code calculates the eigenvalues of a certain complex matrix, which should come in pairs of opposite complex numbers. The code plots the real part of adding each pair. So the correct plot should be just zeros everywhere.

This is indeed the case in Version 10.1 & 11.3 as far as I tested. However, Version 12.0 gives something seriously wrong as shown below.

`NN = 200; R = 0.05; xlist = Table[x, {x, -0.2 \[Pi], 0.2 \[Pi], 0.01}]; modl[n_] := 2*^-3 (Quotient[n, 2] - NN/2); t1 = -1 + Cos[x] - I Sin[x] + I R; t1p = -1 + Cos[x] + I Sin[x] + I R; t2a[n_] := -1 - modl[n]; t2b[n_] := -1 + modl[n]; mat[x_] = DiagonalMatrix[ Table[If[EvenQ[n], t1, t2a[n]], {n, 0, 2 NN - 1 - 1}], 1] + DiagonalMatrix[ Table[If[EvenQ[n], t1p, t2a[n]], {n, 0, 2 NN - 1 - 1}], -1] + DiagonalMatrix[ Table[If[EvenQ[n], t2b[n], 0], {n, 0, 2 NN - 1 - 3}], 3] + DiagonalMatrix[ Table[If[EvenQ[n], t2b[n], 0], {n, 0, 2 NN - 1 - 3}], -3]; list0 = Sort@Re@Eigenvalues[mat[xlist[[3]]]]; list0p = Table[list0[[i]] + list0[[2 NN - i + 1]], {i, NN}]; ListPlot[Tooltip@list0p, PlotRange -> All] `

If A = u*u^T where 0 ≠ u ∈ R^n , then find the eigenvalues of A and show that A is diagonalizable.I mean how to find eigenvalues of A?

3 x 3 matrix is given below 1 -1 -2 1 -2 -1 2 -1 -1

a. find the eigenvalues and eigenvectors b. write a particular solution

We will denote by $ DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of $ \{0,1,\dots,k-1\}^n$ , and $ \sigma_1\cdots \sigma_n$ is connected to $ \tau_1\cdots \tau_{n-1}$ if and only if $ \sigma_i=\tau_{i+1}$ for every $ i=1,\dots,n-1$ .

We would like to calculate the spectrum of the Laplacian matrix of $ DB(n,k)$ . Note that since $ DB(n,k)$ is $ k$ -regular, then $ \lambda \in Spec(L_{DB(n,k)})$ if and only if $ k-\lambda \in Spec(A_{DB(n,k)})$ , so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.

We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:

https://core.ac.uk/download/pdf/82810454.pdf

Thanks!

Let $ T$ be an integral operator with kernel $ K(x,y)=|x-y|$ on $ L^2(-1,1)$ . How can we find the eigenfunctions and eigenvalues of $ T$ ?

I am trying to find the eigenvalues of the following complex matrix \begin{align} M=\left(\begin{matrix} A & B \ B^\dagger & A^\dagger \end{matrix}\right) \end{align} where the symbol $ \dagger$ stands for the conjugate transpose of the matrix.

Is there an expression of the eigenvalues of $ M$ as a function of the eigenvalues of $ A$ and $ B$ ?

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