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Schrodinger operators, operators on manifolds, general differential operators, numerical studies, integral operators, discrete models, resonances, non-self-adjoint operators, random operators/matrices
0
votes
Accepted
Spectrum of the operator PAP, with A self-adjoint and P strictly positive
The only reasonable condition I can think of is that $P = g(A)$ where $g \ge 1$ is a bounded continuous function on $\mathbb R$.
It is instructive to consider the case where $A$ is the multiplicatio …
5
votes
What does multiplying a matrix by its transpose have to do with spectral theorem?
The spectral theorem is for normal operators (in particular, hermitian, or real symmetric, matrices). If $A$ is a real matrix, $A A^T$ is a real symmetric matrix.
2
votes
Convergence of eigenvectors
For any compact set K of complex numbers disjoint from the spectrum of T, there is $\epsilon > 0$ such that for every operator S with $\|S-T\| < \epsilon$, K is disjoint from the spectrum of $S$. Nam …
2
votes
Accepted
Spectrum of compact operator between different Banach spaces
By "$I$ is well-defined", I presume you mean you have a continuous injection $\iota$ of $X$ into $Y$. If $X$ and $Y$ are not isomorphic it will not be a bijection, because there is no continuous line …
4
votes
Accepted
Examples of potentials for which Schrödinger equation lacks discrete points in continuous sp...
The spectrum of an operator is always a closed set. But perhaps they are defining the "continuous spectrum" to be all points of the spectrum that are not in the point spectrum (i.e. not eigenvalues). …
2
votes
Holomorphic functional calculus and idempotents
Let $f$ be your function that is $0$ in a disk $D_0$ around $0$ and $1$ in a disk $D_1$ around $1$. There is (by Runge) a sequence of polynomials $g_n$ such that $g_n \to f$ uniformly on $D_0 \cup D_ …
3
votes
Why do we distinguish the continuous spectrum and the residual spectrum?
The most important thing about residual spectrum, I think, is that some classes of operators, e.g. normal operators on Hilbert space, don't have any.
1
vote
A is a nonnegative matrix; the only principal submatrix having spectral radius above 1 is A ...
Let's assume the matrix $A$ has all its row sums equal to $\lambda$, the largest eigenvalue.
We can rescale the rows and columns of any other nonnegative irreducible matrix by a similarity transformat …
2
votes
Spectral decomposition function
My guess is that this is referring to the resolution of the identity corresponding to the operator, which is sometimes called the "spectral decomposition" (e.g. in Rudin, Functional Analysis) or "spec …
2
votes
Analogue of PSD matrices for permanents?
The set of $n \times n$ matrices with positive permanent is connected. Given such a matrix
$A$, consider the expansion of $P(A)$ in minors along the first row: say
$P(A) = \sum_{j=1}^n a_{1j} P(A_{1 …
5
votes
Accepted
Estimating spectral radius with a Gaussian vector
No: given $\eta > \epsilon > 0$, there are $n \times n$ symmetric matrices $A_n$
with spectral radius $> \eta$, such that $Pr\left[|v^TA_nv|/\|v\|^2 > \epsilon\right] < e^{-cn}$ for some $c > 0$.
I a …
3
votes
Accepted
How does $E$ closed follow from the upper semicontinuity of the spectrum?
Upper semicontinuity of the spectrum is the following statement: if $U$ is a neighbourhood of $\text{Sp}(x)$, then there is a neighbourhood $V$ of $x$ such that $\text{Sp}(y) \subset U$ for all $y \in …
5
votes
Simplicity of eigenvalues of an elliptic operator under a symmetry assumption
For example, consider the Laplacian on $[-1,1] \times [-1,1]$ with Dirichlet boundary conditions. If $m$ and $n$ are distinct positive integers, $\sin(m \pi x) \sin(n \pi y)$ and $\sin(n \pi x) \sin( …
2
votes
Significance of the length of the Perron eigenvector
For example, if $$ \eqalign{A &= \pmatrix{1 & t\cr 1 & 1\cr},\
v = \pmatrix{\sqrt{t}\cr 1},\ u =\pmatrix{1\cr \sqrt{t}},\cr
\frac{(u^T v)^2}{(u^T u)(v^T v)} &= \frac{4t}{(1+t)^2} \to 0 \ \text{as}\ t …
3
votes
Accepted
Spectral radius's relation with row sum
No. For example, the spectral radii of
$$ A = \pmatrix{0 & 1 & 1\cr 1 & 0 & 1\cr 0 & 0 & 0\cr},\ A' = \pmatrix{0 & 0 & 3\cr 1 & 0 & 1\cr 0 & 0 & 0\cr}$$
are $1$ and $0$ respectively.