Questions tagged [self-adjoint-operators]
The self-adjoint-operators tag has no usage guidance.
443
questions
2
votes
1
answer
49
views
When is the image of the functional calculus of a self-adjoint operator also self-adjoint?
Let $A$ be a self-adjoint operator, either bounded or unbounded, and $f$ a Borel function. Using the functional calculus we may define $f(A)$ as a linear operator. Are there any known conditions on $f$...
- 4,323
2
votes
1
answer
33
views
If $T$ is self-adjoint and $\alpha \in \mathbb{R}$, show that $\alpha \langle Tv,v \rangle \geq -|\alpha| \lVert Tv \rVert \lVert v \rVert$
I'm studying Sheldon Axler's "Liner Algebra done right" book, but I'm having some trouble understanding the proof of Lemma 7.11.
Lemma (7.11): Suppose that $T \in \mathcal{L}(V)$ is self-...
- 553
0
votes
0
answers
36
views
About the strong commutativity of unbounded self-adjoint operators?
Let $A$ and $B$ be two unbounded self-adjoint (+ positive if needed) operators with domains $D(A)$ and $D(B)$ respectively. By the spectral theorem, we know that $A=\int_{\mathbb{R}}\lambda dE$ and
$B=...
0
votes
0
answers
21
views
Proof that an operator is self-adjoint if and only if its matrix is self-adjoint
As stated in the wikipedia page on self-adjoint operators, $A$ is a self-adjoint operator on an finite-dimensional inner product space $V$ if and only if, given an orthonormal basis, the matrix of $A$ ...
- 4,708
0
votes
0
answers
17
views
Convergence of self-adjoint operators with converging spectra
Let $A=\int_{\sigma(A)} \lambda dP(\lambda)$ be an unbounded self-adjoint operator on a Hilbert space $H$ with dense domain $D\subset H$.
Let $A_n=\int_{\sigma(A)\cap[-n,n]} \lambda dP(\lambda)$ (see ...
- 11
3
votes
1
answer
85
views
Norm of $Tf(x)=\int_{-\pi}^{\pi} \cos(x-y)f(y)\,dy$
I'm dealing with this operator $T\in \mathcal{L}(H)$ where $H=L^2([-\pi,\pi])$
$$
Tf(x) = \int_{-\pi}^{\pi} \cos(x-y)f(y)\,\mathrm{d}y
$$
One of the questions request to compute the norm of the ...
- 321
0
votes
1
answer
24
views
A formula for a unitary matrix
Proposition: U = (H+iI)(H-iI)^-1 is unitary if H is self-adjoint. I'm having trouble finding a proof. It's straightforward if (H-iI) and (H+iI)^-1 commute. But I don't see it.
0
votes
1
answer
23
views
Operators which are symmetric but not selfadjoint [closed]
Could you please give me some examples of operators defined on inner product space, which are symmetric, but not selfadjoint?
- 1
0
votes
1
answer
41
views
For an essentially self-adjoint operator $T$, why is $(T^{**})^* = T^{**}$?
I am new to unbounded operators and their adjoints. So far I have understood that unlike bounded operators, when $T$ is a densely defined unbounded operator $T^{**} \neq T$ in general.
In Reed & ...
- 4,323
2
votes
0
answers
76
views
A self-adjoint operator related to prime numbers?
I originally posted this question on MO, but since it might not be research related, I deleted it and posted it here again:
Consider the completely additive function $\eta(n) := \sum_{p\mid n} v_p(n)p$...
- 595
2
votes
0
answers
52
views
If $T$ is self-adjoint, then $T^2 + tI$ is bijective for any $t>0$
Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $I:H \to H$ be the indentity map. Let $T:H \to H$ be a bounded linear operator. In preparation for ...
- 16.2k
1
vote
1
answer
47
views
Is a orthogonal projection in a Hilbert space automatically selfadjoint?
Let $P$ be a self adjoint projector on a Hilbert space $H$ i.e. $P: H \rightarrow H$ is linear and continuous, $P^*=P$ and $P^2=P$
Then $P$ is also an orthogonal projection i.e. $\mathrm{ran}P=\mathrm{...
- 622
2
votes
0
answers
33
views
Version of Weyl's criterion for absolutely continuous spectrum
Weyl's criterion says that for a self-adjoint operator $H$, $\lambda\in\sigma(H)$ if and only if there exists a sequence $(\phi_n)_{n=1}^{\infty}$ so that $\vert \vert \phi_n \vert \vert=1$ and
$$\...
- 4,123
1
vote
2
answers
85
views
How is a spectral subspace of a bounded linear operator defined?
Let $A$ be a bounded linear operator on some Hilbert space. In a previous question (How to interpret spectral projections?) I learned that the spectral projectors, which are defined using the Borel ...
- 4,323
3
votes
0
answers
44
views
Matrices of self-adjoint operators
Fix a finite-dimensional real inner product space. Does a self-adjoint operator have a symmetric matrix with respect to all bases, or just with respect to orthonormal bases? Also, does this change at ...
- 1,169
0
votes
0
answers
50
views
How is the functional calculus explicitly computed for functions that are not analytic?
Let $A$ be a bounded self-adjoint linear operator on a Hilbert space $H$. For any analytic function $f$ whose radius of convergence contains the spectral radius of $A$ we may compute the functional ...
- 4,323
5
votes
1
answer
206
views
Isolated points of the spectrum are always eigenvalues?
Let $A=A^*:D(A)\subset H \rightarrow H$ be a linear operator, where $H$ is a seperable Hilbert space.
The discrete spectrum of $A$ is defined to be
$\sigma_{disc}(A):=\left\lbrace \lambda \in \sigma(A)...
- 622
0
votes
1
answer
31
views
$T$ and $U$ self-adjoint on $V$, $T$ is positive definite. Prove $TU$ & $UT$ are diagonalizable linear operators that have only real eigenvalues.
This is a problem from Freidberg linear algebra (4th edition) chapter 6.4.21
The whole problem and the hints are
Let $V$ be a finite-dimensional inner product space, and let $T$ and $U$ be
self-...
- 181
1
vote
1
answer
31
views
Given the expression of the matrix find the adjoint and the square of the matrix
Let the matrix $U=\mathbb{I}-ww^* $ with $w\in\mathbb{C}^3$ being a column vector, find the expression for $U^*$ and for $U^2$.
So for the adjoint $U^* $ I gave it a try and got $U^* =\mathbb{I}-(ww^*...
- 131
3
votes
1
answer
82
views
Strong limit of $A^n$ and $A^{1/n}$ for a positive operator
I want to prove the following fact: Let $A$ be a positive operator (thus self-adjoint) over a Hilbert space $H$. (Notation: $s-\lim$: strong limit of operators; $P_K$: orthogonal projection onto a ...
- 1,323
1
vote
2
answers
49
views
Commuting self adjoint operator exercise.
Let $V$ be a complex vector space with positive definite inner product $(, )$, and $T: V \to V$ a linear map. Recall that the adjoint $T^*$ of $T$ is defined by:
$$(T(x), y) = (x, T^*(y))$$
for all $x ...
- 1,123
0
votes
1
answer
37
views
Prove that matrix is Hermitian iff it's equal to its adjoint with general hermitian inner products
The Wikipedia article for Hermitian matrices gives an alternative characterization for Hermitian matrices link. The statement I want to prove is:
$$ A^* = A \iff \langle w,Av \rangle_H = \langle Aw,v \...
0
votes
0
answers
32
views
Does $A^*$ being self-adjoint imply that $A$ is self adjoint?
The question is in the title. This arose when I read that for some linear densely defined symmetric operator A that is not self-adjoint, and B a self-adjoint extension of $A$, we had $A \subsetneq B \...
- 414
0
votes
1
answer
25
views
Spectral decomposition of a normal operator: non-distinct eigenvalues
A normal operator $T$ in some inner product space usually has spectral decomposition $T=\sum_j\lambda_j P_j$ where $P_j$ are orthogonal projections and $\lambda_j$ are usually distinct values often ...
- 143
1
vote
0
answers
31
views
self-adjoint projection
for the following exercise i have some questions, i appreciate some help or hints for this exercise.
Let be V an euclidean or unitary vectorspace and $ p:V \to V$ a self-adjoint Projection.
$(a)$ Show ...
- 13
0
votes
0
answers
23
views
is this (unbounded) operator normal?
I have some question regarding normal operators.
I have been given the following definition for the case of bounded linear operators $A\in\mathcal{A}(H)$: $A^*A=AA^*$.
My question is: what happens if ...
- 101
0
votes
1
answer
78
views
endomorphism and self adjoint operators and isometry
Let $F=\mathbb{R}$ or $F=\mathbb{C}$, let $(V,\langle.,.\rangle)$ be a finitely generated $K$-vector space with scalar product, let $f \in \operatorname{End}_{F}(V)$. Show that:
a) If $f$ is self-...
- 159
1
vote
0
answers
25
views
endomorphism and square root self-adjoint operator
Let $F=\mathbb{R}$ or $F=\mathbb{C}$, let $(V,\langle.,.\rangle)$ be a finitely generated $K$-vector space with scalar product, let $f \in \operatorname{End}_{F}(V)$. Show that:
If $f$ is self-adjoint ...
- 159
1
vote
0
answers
25
views
Difference between time reversible and stationary distribution. "watching a movie backwards"
A time homogeneous Markov process on Ω
with semi-group $P_t$
is said to stationary w.r.t a distribution $π$
if
$$∫_{Ω}P_tf(x)dπ(x)=∫_{Ω}f(x)dπ(x), \text{for $f$ bounded measurable}.$$
and reversible ...
- 21
2
votes
1
answer
63
views
Assume $|u_n| =1$ for all $n$ and $\langle Tu_n, u_n\rangle \xrightarrow{n \to \infty} \alpha$. Then $\alpha u_n - Tu_n \xrightarrow{n \to \infty} 0$
Let $(H, \langle \cdot, \cdot\rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $T:H\to H$ be a self-adjoint bounded linear operator. Let
$$
\alpha := \sup \{\langle Tu, u\rangle : ...
- 5,128
2
votes
1
answer
66
views
If $f$ is a eigenfunction of $-\Delta$ in $L^2[0,1]$, is it necessarily $C^\infty$?
I am a little bit confused about the properties of the Laplacian $-\Delta$ on $L^2[0,1]$ with the periodic boundary conditions.
At least I know that $-\Delta$ is an unbounded self-adjoint operator on $...
- 6,880
0
votes
1
answer
65
views
If an element $b$ of a $C^*$-algebra $B$ is self-adjoint with $\delta=\|b-b^2\|<\frac14$ then there is a projection $p\in B$ with $\|b-p\|\leq2\delta$
If an element $b$ of a $C^*$-algebra $B$ is self-adjoint with $\delta=\|b-b^2\|<\frac14$, then there is a projection $p\in B$ with $\|b-p\|\leq2\delta$
I believe the proof as following however I am ...
0
votes
0
answers
27
views
Restriction of self-adjoint operator to the vectors orthogonal to a finite-dimensional subspace
Let $A:\mathcal{D}(A)\subset\mathcal{H}\rightarrow\mathcal{H}$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$, with $A\geq I$, and let $\mathcal{N}\subset\mathcal{H}$ be a finite-...
- 349
2
votes
1
answer
58
views
$T : H \to H$ bounded self-adjoint, when is $\sup \langle Tx, x \rangle$ achieved?
If $H$ is a Hilbert space and $T : H \to H$ is a self-adjoint bounded operator, under what conditions can we guarantee that $\sup\limits_{||x|| = 1} \langle Tx, x \rangle \le ||T|| < \infty$ is ...
- 4,924
1
vote
1
answer
52
views
Show that operator is self-adjoint in $\mu L^2(\mu dv)$
I am referring to this paper, page 9.
The background of my question is the equation
$$
\partial_tF-\partial_v(\partial_v+v)F=0,\quad F_{|t=0}=F^0,
$$
where $F=F(t,v)$ is unknown for $t>0$.
An ...
- 495
2
votes
1
answer
47
views
Deficiency indices of the restriction of a self-adjoint operator to the kernel of a rank-one functional
Let $\mathcal{H}$ be a complex Hilbert space, $A$ a self-adjoint unbounded operator on $\mathcal{H}$ with domain $\mathcal{D}(A)$, and $g:\mathcal{D}(A)\rightarrow\mathbb{C}$ be an unbounded linear ...
- 349
0
votes
0
answers
85
views
Why does $\int_{-\infty}^\infty e^{ar}\nabla^2\nabla^2 e^{br} = \int_{-\infty}^\infty e^{br}\nabla^2\nabla^2 e^{ar}$ for a self-adjoint operator?
If the Laplacian operator in spherical coordinates is:
$$ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)$$
then for $a<0$, $\int_{r=0}^\infty e^{ar}\...
- 700
1
vote
1
answer
54
views
If $T$ is symmetric and has at least one real number in its resolvent set, then it is self-adjoint?
If a closed symmetric operator $T$ has at least one real number in its resolvent set, then does it follow that $T$ is self adjoint? Since $T$ is closed and symmetric, we know that it is extended by ...
- 283
0
votes
0
answers
32
views
Representation of semigroup of trace-class operators in terms of projectors
Let $H$ be a Hilbert space. For all $t\in\left]0,\infty\right[$ the operator $A_t\in L(H,H)$ is assumed to be trace-class and symmetric. Furthermore $A$ is a semigroup, i.e.
$$A_{t+h}=A_tA_h$$
for all ...
- 3,093
1
vote
0
answers
32
views
Can an eigenbasis be divided into bases of the eigenspaces?
I am considering an operator $A$ defined on a subspace of some Hilbert space $H$ and I am given a Hilbert-basis $(v_n)_{n\in\mathbb N}$ such that all $v_n$ are eigenvectors of $A$:
$$Av_n=\lambda v_n$$...
- 3,093
0
votes
2
answers
61
views
Is the kernel finite-dimensional if only finitely many basis-vectors of a Hilbert basis are mapped to zero?
Let $H$ be a Hilbert space, $A$ a self-adjoint operator on a subspace of $H$ and $(v_n)_{n\in\mathbb N}$ a Hilbert basis (all $v_n$ are assumed to be in the domain of $A$). If $Av_n=0$ for at most ...
- 3,093
0
votes
1
answer
89
views
Hermitian matrices and inner products
Let $V:=\mathbb{C}^n$ and $\langle \cdot, \cdot \rangle: V\times V\to \mathbb{C}$ be an inner product, which by definition is a positive-definite sesquilinear form. The adjoint of a linear operator $A:...
- 892
0
votes
0
answers
47
views
Does the existence of an eigenbasis imply boundedness?
The following question arose while studying the proof of proposition 2.36 in "Heat Kernels and Dirac Operators":
Let $H$ be a Hilbert space and $A$ a self-adjoint operator defined on a ...
- 3,093
0
votes
0
answers
26
views
For self-adjoint automorphism $T$, to what extent is $U \circ T \circ S$ equivalent to $U \circ S$?
I would like to know to what extent can composition with self-adjoint isomorphism affect composition of linear operators. More specifically, for linear operators of Hilbert spaces as $H \xrightarrow{S}...
- 75
2
votes
1
answer
62
views
Inequality for $\langle T^2x, x\rangle$ if $T$ is positive operator
Let $T \in \mathcal{L} (\mathcal{H})$ is positive operator (i.e. $\langle T x, x \rangle \ge 0$). Prove that $$\langle T^2 x, x \rangle \le \langle Tx, x \rangle^{\frac{1}{2} + \frac{1}{4} + \ldots + \...
- 129
1
vote
1
answer
98
views
Proof verification: Characterising the domain of a self-adjoint operator
Consider a separable, complex Hilbert space $H$ and a self-adjoint operator $A: D(A) \longrightarrow H$, where $D(A)$ is a dense subspace. Assume that $A$ admits an orthonormal basis $\{\varphi_n\}_{n\...
- 374
1
vote
1
answer
38
views
$T = A+i B$ is normal operator, if $\Vert T x \Vert^2 = \Vert Ax \Vert^2 + \Vert Bx \Vert^2$.
Let $T = A+iB$, where $A$ and $B$ are self-adjoint and bounded. Prove that if $\Vert T x\Vert^2 = \Vert Ax\Vert^2 + \Vert Bx\Vert^2$, then $T$ is normal operator.
So it's obvious that $\Vert T^* x \...
- 129
1
vote
2
answers
64
views
Spectral resolution of discrete Laplacian on $\ell^2(\mathbb{Z})$
Let $\Delta$ be the discrete Laplacian on $\ell^2(\mathbb{Z})$, that is, $$\Delta = 1- \frac{S+S^*}{2},$$
where $S$ is the right shift operator. I know that $\Delta$ is a bounded self-adjoint operator....
- 1,038
0
votes
0
answers
37
views
Self-adjoint operator in $L^2(\mathbb Q_p)$.
Let $g:\mathbb Q_p^n\longrightarrow{\mathbb Z}$ increasing and radial. We define the following operator
$$Hf(x)=\frac{1}{p^{n\left|{g(x)}\right|}}\displaystyle\int_{B_{g(x)}^n}\left |{f(y)}\right |dy$$...
- 21
2
votes
0
answers
21
views
Analitycity of band functions
Let $H=-\dfrac{d^2}{dx^2}+V(x)$, where $V(x)$ is a $2\pi$-periodic and $V\in L^{\infty}(\mathbb R)$.
If $\mathcal H':=L^{2}(0,2\pi)$, we have the decomposition
$\mathcal H=\int_{\mathbb [0,2\pi)}^{\...
- 2,665