Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Questions about the properties of vector spaces and linear transformations, including linear systems in general.
1
vote
0
answers
96
views
Determinant formula related to solutions of a second-order recurrence
Let $A$ be the linear map on the space of complex sequences acting as
$$(Au)_{n}=u_{n-1}+a_{n}u_{n}+u_{n+1}, \quad n\in\mathbb{Z},$$
where $\{a_{n}\}$ is a fixed sequence. Let $f=f(z)$ and $g=g(z)$ be …
2
votes
1
answer
84
views
Information on special matrices similar to Jacobi matrices
Jacobi matrices are well known and deeply investigated mathematical objects from various point of view. One can arrive at these operators while studying discrete systems of particles interacting with …
2
votes
Accepted
The norm of a Finite Hilbert matrix
I think the answer is no, indeed. Even in the particular case $p=2$ the formula is not known (to my best knowledge). Concerning $\|H\|_{2}$, is known $\|H\|_{2}\leq\pi$. More precisely, we have the fo …
2
votes
1
answer
153
views
A problem from linear algebra and difference equations
Let $A$ be a linear second-order difference operator acting on the space of complex sequences as
$$(Af)_{n}=f_{n-1}+a_{n}f_{n}+f_{n+1}, \quad n\in\mathbb{Z},$$
where $a_{n}\in\mathbb{C}$. Further, let …
1
vote
Simple Spectrum of Jacobi matrices
In the case of finite or semi-infinite Jacobi matrix, the first entry of an eigenvector uniquely determine other entries since they are related by the tree-term recurrence. This is the reason why the …
0
votes
Eigenvalues of symmetric tridiagonal matrices
Withnout loss of generality, one can put $a=0$. For sure, there is no closed-form (or explicit) formula for the eigenvalues in general. However, at least the characteristic polynomial of $A_n$ can be …
16
votes
1
answer
874
views
Hankel determinants of binomial coefficients
For $\{h_{n}\}_{n=0}^{\infty}$ a real sequence, denote by $H_{n}$ the $n\times n$ Hankel matrix of the form
$$
H_{n}:=\begin{pmatrix}
h_{0} & h_{1} & \dots & h_{n-1}\\
h_{1} & h_{2} & \dots & …