Questions tagged [adjoint-operators]
For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).
1,231 questions
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The multiplication operator and sigma-finiteness
I'm now learning basics of functional analysis. My question is about the following statement on the multiplication operator:
Theorem (?). Let $(X,\mathfrak{A},\mu)$ be a measure space not necessarily ...
2
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1
answer
60
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Question on controlling norm of output of a semidefinite operator by inner product
I'm asked to prove the following proposition:
Prop. If $H$ is a complex Hilbert space, $A\in L(H)$ satisfies that
$$
(Ax|x)_H\ge0,\ \forall x\in H,
$$
then
$$
\|Ax\|_H^2\le\|A\|_{H\rightarrow H}(Ax|x)...
1
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1
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173
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Find adjoint operator of $L(f) = f' + 3f$
Problem Setup: Let $V = \mathcal{P}_{1}$ with $\langle f,g \rangle = \int_{-1}^{1} f(x)\,g(x)\,\mathrm dx$ be the inner product space. Here $\mathcal{P}_{1}$ is the space of all linear polynomials. ...
1
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0
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The domain of the adjoint of an unbounded closed operator is dense
Let $ \mathcal H $ be a Hilbert space. Let $ A $ be an unbounded operator on $ \mathcal H $, and call $ \mathcal D(A) $ its domain (we assume that $ \mathcal D(A) $ is dense). Let $ A^* $ denote the ...
4
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3
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Proving equivalence of $T$-invariance of $W$ and $T^{*}$-invariance of $W^{\perp}$
I'm working in my homework of linear algebra and there is this problem I'm strugling with
Let $V$ be a finite-dimensional vector space with an inner product, and let $T$ be a linear operator on $V$. I'...
6
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0
answers
111
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Norm convergence in Lie-Trotter formula
I am studying the Lie-Trotter formula for operator exponentials. Let $H$ be a Hilbert space and $A$, $B$ be self-adjoint operators on $H$. The classical Lie-Trotter formula (see M. Reed, B. Simon. ...
1
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1
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38
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The adjoint of $\partial^{a + ib}$ on $H^k([0, \infty))$ for functions that do not vanish at $x = 0$.
Let's say I have a pseudodifferential operator $\partial^{a + ib}$ for $a + ib \in \mathbb C$ that is defined on the Sobolev space $H^k([0, \infty))$ for a sufficiently large $k$. (In fact, I really ...
7
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1
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102
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Existence of at least one injective skew-Hermitian operator
Let $V$ be any infinite-dimensional real inner product space. Is it always possible to find a linear operator $T$ which is injective, Hilbert adjoint $T^*$ exists and $T^* = - T$ i.e. $T$ is skew-...
5
votes
0
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145
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Verifying surjectivity of the function $(\sigma_0,v_0)\mapsto-\Delta v_0 -\lambda_*f^{\prime}(u_*)v_0-\sigma_0f(u_*)$
I'm currently working on a bifurcation problem involving a nonlinear elliptic PDE, and I need to verify the surjectivity of the linearized operator at a critical point.
$\textbf{Setting:}$
Let $\Omega ...
1
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1
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114
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${\rm ker}(T^*\mp i) = {\rm range}(T \pm i)^{\perp}$
Proposition: Let $T: {\rm dom}(T) \rightarrow H$ be a densely defined operator.
(i) ${\rm ker}(T^*\mp i) = {\rm range}(T \pm i)^{\perp}$. In particular ${\rm ker}(T^*\mp i)=\{0\} \Leftrightarrow {\rm ...
1
vote
1
answer
66
views
Computing truncated singular value decomposition (SVD) in alternative inner products
Let $\mathbf{N}\in \mathbb{R}^{n \times n}$ and $\mathbf{M} \in \mathbb{R}^{m \times m}$ be symmetric positive definite matrices. How can we efficiently compute the truncated singular value ...
3
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2
answers
149
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Showing the adjoint operator is closed.
I want to solve the following exercise:
Let $T:\text{dom}(T) \rightarrow H$ be a densely defined linear
operator. Show that the Hilbert-adjoint $T^*$ is a closed operator.
Definition of Hilbert-...
2
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1
answer
84
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Do we really need $\phi$ to be normal in order for $\phi \circ \phi^*$ to have real, nonnegative eigenvalues?
I'm working through a problem from the book Lineare Algebra by Bosch (in the chapter on self-adjoint endomorphisms), which states:
Let $V$ be a finite-dimensional real Euclidean or complex unitary ...
4
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2
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485
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Understanding Hilbert-adjoint for (possibly unbounded) operators
I want to understand how one defines the Hilbert-adjoint of a (possibly) unbounded operator.
Here is what I know for the bounded case:
For a bounded linear operator $T$, the Hilbert-adjoint is defined ...
4
votes
1
answer
127
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Closure of the adjoint operator
Let $\mathcal{H}$ be a Hilbert space, $T:\mathfrak{D}_T\subseteq\mathcal{H}\rightarrow\mathcal{H}$ a densely defined linear operator that is closable. Consider its closure $\overline{T}$, and its ...