Questions tagged [adjoint-operators]
For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).
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$A^{-1}$ has integer entries if and only if the ${\rm det}\ (A) =\pm 1$
So, $A$ is a $n \times n$ matrix with integer entries. The question is to prove that $A^{-1}$ has all integer entries if and only if ${\rm det}\ (A) =\pm 1$ .
I know that $A^{-1}= {\rm adj}(A)/{\rm ...
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2
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image of adjoint equals orthogonal complement of kernel [duplicate]
Let $T:V\to W$ be a linear map of finite-dimensional spaces. Then
$${\rm im}(T^{\textstyle*})=({\rm ker}\,T)^\perp\ .\tag{$*$}$$
I can prove this as follows:
$${\rm ker}(T^{\textstyle*})=({\rm im}\,T)...
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Motivation for adjoint operators in finite dimensional inner-product-spaces
Given a finite dimensional inner-product-space $(V,\langle\;,\rangle)$ and an endomorphism $A\in\mathrm{End}(V)$ we can define its adjoint $A^*$ as the only endomorphism such that $\langle Ax, y\...
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3
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T compact if and only if $T^*T$ is compact.
I have an operator $T \in B(\mathcal{H})$.
I need to prove that T is comapct if and only if $T^*T$ is compact.
One way is ok, because if $A$ or $B$ is compact then $AB$ is compact, so I get at once ...
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11
votes
1
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Is it possible to define an inner product such that an arbitrary operator is self adjoint?
Given a vector space $V$ (possibly infinite dimensional) with inner product $(.,.)$. We say an operator $A$ is self adjoint if $(Af,g)=(f,Ag)$.
The definition as stated require us to start with an ...
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If a linear operator has an adjoint operator, it is bounded
This is a question I'm struggling with for a while:
Let $H$ be a Hilber space. Let $T,S: H\rightarrow H$ be linear operators (not neccessarily bounded) such that for every $x,y\in H$: $\langle Tx,y\...
user188400
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Intuition of Adjoint Operator
A linear operator $T$ on an inner product space $V$ is said to have an adjoint operator $T^*$ on $V$ if $⟨T(u),v⟩=⟨u,T^*(v)⟩$ for every $u,v\in V$
I know how to proof "why this operator exist$(\text{...
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2
answers
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Self-adjoint operator as difference of two positive operators
The problem comes from my functional analysis homework.
Let $H$ be a complex Hilbert space and $A:H \to H$ be a bounded, self-adjoint linear operator. Prove that there exist positive operators $P$ ...
- 878
9
votes
1
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Derivative of adjoint operator-valued function
Consider an infinite dimensional complex Hilbert space $H$. I think that for a bounded operator-valued function $A: x\mapsto A(x) \in \mathcal B(H)$, where $x\in \mathbb R$, we can define the ...
- 272
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Gradient operator the adjoint of (minus) divergence operator?
Recently I found this statement -- the gradient operator is the adjoint of the minus divergence operator -- in one of my lecture notes. Knowing only a little about functional analysis, I'm looking for ...
- 1,152
8
votes
1
answer
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Geometric intuition for adjoint
Let $V$ be a finite-dimensional inner product space, and let $T$ be a linear operator on $V$. Then $T^*$ ($T$ adjoint) is defined as the unique function such that $\langle T(x), y \rangle = \langle x, ...
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Show $ \langle Tx,x \rangle \in \Bbb R\ \forall x \in H\ \implies T$ is self-adjoint
Show that a linear operator $T: H \to H$ is self adjoint if and only if $\langle Tx, x \rangle \in \Bbb R\ \forall x \in H$. You may use that the equality that for all $x,y \in H$
$4\langle T(x),y \...
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3
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$\operatorname{Adj} (\mathbf I_n x-\mathbf A)$ when $\operatorname{rank}(\mathbf A)\le n-2$
Let $\mathbf B$ denote an $n \times n$ matrix with $r\equiv\operatorname{rank}(\mathbf B)$. I need to prove the following conjecture:
If $r \le n - 2$, then there exists a polynomial matrix $\...
- 1,885
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1
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Intuition behind $\ker(T)=\ker(T^*)$ for $T$ a normal operator
Let $T : V \to V$ be a normal operator and $V$ a finite-dimensional vector space. Show that $\ker(T)= \ker(T^*)$ and $\text{im}(T) = \text{im}(T^*)$.
I know how to rigorously show this, but I'm ...
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0
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Why is it called *adjunction* formula?
Let $X$ be a complex manifold, $Y$ a sub-manifold, and $i \colon Y \to X$ the corresponding embedding. Then one can prove that the corresponding canonical bundles satisfy
$$
\omega_X \big|_Y = \...
- 9,670
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votes
1
answer
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Why is every selfadjoint operator closed?
I've read this theorem multiple times, but never seen a proof:
Every selfadjoint operator is closed.
But it's always been stated without a proof. Is it somehow obvious? I can't see it immediately ...
- 115
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3
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$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)
I am trying to prove the following statements:
Let $X$ and $Y$ be normed spaces (not necessarily complete)
Let $T\in L(X,Y)$ (meaning $T:X\to Y$ is a bounded linear map). Let $T^*:Y^*\to X^*$ denote ...
- 1,312
7
votes
2
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General Cauchy-Schwarz for adjoint positive operators
I'm trying to prove the next inequality, like Cauchy-Schwarz standard inequality:
$$|\langle Tx,y\rangle |\leq\langle Tx,x\rangle ^{1/2}\langle Ty,y\rangle ^{1/2}\space\forall x,y\in\mathcal{H},$$ ...
- 1,287
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2
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Definition of Essentially Self Adjoint Operators
I have two definitions of an essentially self adjoint operator
A symmetric operator with a self adjoint closure
An operator with a unique self adjoint extension.
I can easily show that (1) implies (...
- 7,674
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1
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Proof that every bounded linear operator between hilbert spaces has an adjoint.
As a practice exercises(not an assignment question) for one of the papers I am doing currently at university we are asked to show the following;
I $T:H \rightarrow K$ is a bounded linear operator ...
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0
answers
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$T$-invariance of $U$ is equivalent to $T^{*}$-invariance of $U^{\perp}$
Is the Following argument correct?
Suppose $T\in\mathcal{L}(V)$ and $U$ is a subspace of $V$. Prove that $U$ is invariant under $T$ if and only if $U^{\perp}$ is invariant under $T^*$.
Proof. Given ...
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Find the adjoint under the inner product $\langle f, g \rangle = \int_0^1 f(t)g(t)t \ dt$ of $\mathcal{L}(f)(t) = \frac{d^2 f}{dt^2} + f$.
Find the adjoint under the inner product $\langle f, g \rangle = \int_0^1 f(t)g(t)t \ dt$ of $\mathcal{L}(f)(t) = \dfrac{d^2 f}{dt^2} + f$ with $f(0) = 0$ and $f'(1) = 0$.
Note: The weight function is ...
- 4,437
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2
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views
Is the map sends $T$ to $T^*$ adjoint of $T$ surjective?
Let $B(X)$ denotes the set of all bounded linear operators from $X$ to $X$, where $X$ is a Banach space. Same is defined for the set $B(X^*)$, where $X^*$ denotes the set of all bounded linear ...
- 1,755
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Connection between categorical notion of adjunction and dual space/adjoint in vector spaces
I'm an economist, not a mathematician. I've been trying to make sense of some concepts in functional analysis: dual, bidual, adjoint, natural mapping. The definitions of these notions come out of ...
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If H is Hermitian, show that $e^{iH}$ is unitary
The only approaches I have seen to answering this question have involvd manipulating $e^{iH}$ like an ordinary exponential. i.e.
If $U=e^{iH}$, then $U^\dagger = (e^{iH})^\dagger = e^{-iH^\dagger}=e^{...
- 3,037
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votes
2
answers
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Hermitian Operators and the Spectral Theorem
I understand that in a finite-dimensional vector space $V$, a diagonalizable linear operator $T: V \to V$ decomposes $V$ into a direct sum of its invariant eigenspaces, on each of which it restricts ...
6
votes
2
answers
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When is the restriction of a normal operator not normal?
I was proving the spectral theorem for normal operators on finite-dimensional complex vector spaces today during a test, when I arrived at the point in which
If $T\in\operatorname{End}(V)$ is ...
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1
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Adjoint of an integral operator
I'm reading through a text about integral operators and I've come across the following theorem:
Let $k:\mathbb{R}^2\rightarrow\mathbb{C}$ be a kernel, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ ...
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Adjoint of a nonlinear operator?
For any linear operator $A$, the adjoint $A^*$ is defined as a linear operator that satisfies
$$\langle v, Au\rangle = \langle A^*v, u\rangle$$
Moreover, one has that $(A^{*})^{*} = A$ (proof here). ...
- 1,596
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2
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Help Showing that the Adjoint Operator $T^*$ is Surjective if and only if $T$ is Injective
Let $T\in L(V,W)$,where $L(V,W)$ denotes a linear map from a vector space $V$ to vector space $W$. I want to prove that $T$ is injective iff $T^*$ is surjective, where $T^*$ is the adjoint of $T$. I ...
user227158
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2
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inner product and adjoint operator
This is a problem I found in Schaum's Outlines: Linear Algebra, and I was wondering if someone knew how to solve it. I began using integration by parts, but that approach did not lead to any ...
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1
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Why are self-adjoint operators important?
I am learning about self-adjoint and normal operators.
So far, they have come up in the Spectral theorem, which says self-adjoint operators have an eigenvalue basis and a corresponding diagonal ...
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Schur decomposition of a matrix with distinct eigenvalues is almost unique
Let $M\in \mathbb C^{n,n}$ have $n$ distinct eigenvalues, and let $U_1, U_2$ be two Schur-forms of $M$. Show that if $U_1, U_2$ have equal diagonals, there is a hermitian diagonal matrix $Q$ such that ...
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Adjoint of multiplication by $z$ in a Hilbert Space (Bergman space)
I am learning Hilbert space theory from Halmos' "Introduction to Hilbert space and the theory of spectral multiplicity".
While talking about understanding adjoints (p. 39), he calls special ...
- 3,287
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2
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Exponential Law for based spaces
I realize most people work in "convenient categories" where this is not an issue.
In most topology books there is a proof of the fact that there is a natural homeomorphism of function spaces (with ...
- 1,486
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Inverse vs. adjoint operators
I hope this is enough of a question (and less of a start of a discussion) to be allowed here. I am trying to get my head around the ideas of inverse and adjoint operators. To keep it simple, let's ...
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2
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Self adjoint operators on a Hilbert space
Let $H$ be a Hilbert space and let $T\in \mathcal{B}(H)$ such that $T$ is self-adjoint. I want to show that if $T$ is non-zero, then $T^n\neq 0$ for all $n\in \mathbb{N}$.
Suppose $n$ be the least ...
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2
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For positive self adjoint $T$, show $|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$
As in title, $T$ is a positive self adjoint, bounded linear operator on a Hilbert Space $X$ and I'd like to show
$$|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$$
Self adjoint ...
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Invertibility in a finite-dimensional inner product space
Let $T$ be an invertible linear operator on a finite-dimensional inner product space. I just want a hint as to how I should prove that $T^{*}$ is also invertible and $( T^{-1} )^{*} = ( T^{*} )^{-1}$.
...
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1
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Spectral gap and Poincaré inequality
Consider the PDE
$$\partial_t u = L u$$
where $L = \Delta + \nabla V \cdot \nabla $ is a self-adjoint operator.
I read that if $L$ has a spectral gap $\lambda > 0$ then "[convergence of the ...
- 1,716
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votes
2
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Derivative of inner product via adjoint operator vs. complex derivatives
Dear math enthusiasts,
I need to take the derivative of an inner product involving an operator on one side and I'd like to do this via the adjoint operator. However, it seems I'm doing something ...
- 2,530
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2
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661
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Find adjoint operator of an operator T
I would like to find the adjoint operator of
$$
T\colon L^2([0,1])\to H^1([0,1]),\quad x\mapsto\int\limits_0^t x(s)\, ds.
$$
Here $H^1([0,1])$ is the Sobolev space $W^{1,2}([0,1])$.
I tried to find ...
user34632
5
votes
1
answer
204
views
Why is the Compactness of an Operator so important? What is the use of compact operators in Mathematics?
Compact Operators have been the major topic in our Operator Theory course for the past few weeks.
All the theorems which tell us whether a operator is compact or not are clear to me, but I still don't ...
- 513
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votes
1
answer
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Schrodinger with real potential is self-adjoint?
Suppose I define the operator
$$
-\frac{d^2}{dx^2}+V(x)
$$
on the space of Schwartz class functions $\mathcal{S}(\mathbb{R})$ and take its closure to form the operator $H$ acting on a domain in $L^2(\...
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1
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Adjoint operator on Banach space
Suppose $X$ and $Y$ are Banach spaces and $T:X\to Y$ is a bounded linear operator. Show that $T$ is an isometric isomorphism if and only if its adjoint $T^*$ is also an isometric isomorphism. Given an ...
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prove that if $\lambda$ is an eigenvalue of T then $\bar\lambda$ is eigenvalue of $T^*$
I have to prove that if $\lambda$ is an eigenvalue of T then $\bar\lambda$ is eigenvalue of $T^*$ (adjoint)
I know that $<Tv,u> = <\lambda v,u> = \bar\lambda<v,u>=<v,\bar\...
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1
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How to find formal adjoint operators for operators $\Gamma(E) \to \Gamma(T^*M \otimes E)$
Let $(M,g)$ be a Riemannian manifold and let $E \to M$ be a real vector bundle over $M$. Let $d_A = d+A$ be a covariant derivative on $E$. It is an $\mathbb R$-linear map $d_A \colon \Gamma(E) \to \...
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Show properties of the linear operator $L((a_n)_n)=\left(\frac{1}{n}*a_n\right)_n$
Let $L\colon \ell^p \rightarrow \ell^p$ such that $L((a_n)_n)= \left(\frac{1}{n}*a_n\right)_n$.
1) Determine $L'$(adjoint operator), $\ker(L)$, $\ker(L')$, $\operatorname{rg}(L)$, $\operatorname{rg}(...
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Sum of operator and adjoint is self-adjoint
In abstract Hodge theory there is the following lemma:
Let $H$ be a Hilbert space and $A \in \mathcal{C}(H)$ a densely defined, closed operator (so possibly unbounded) and $A^*$ its adjoint operator. ...
- 1,134
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2
answers
682
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Prove or disprove: $\operatorname{Adj} (A)$ is diagonlizable $\implies A$ is diagonalizable
For $2X2$:
$$
A:\\
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$$
$$
\operatorname{Adj}(A):\\
\begin{bmatrix}
d & -c \\
-b & a
\end{bmatrix}
$$
So the statement is true.
The ...
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