Questions tagged [adjoint-operators]

For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).

9
votes
3answers
2k views

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\...
14
votes
2answers
9k views

$A^{-1}$ 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 ...
5
votes
1answer
3k views

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 ...
0
votes
3answers
308 views

Prove that if $A$ is regular then $\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{n-2} A$

$\newcommand{\adj}{\operatorname{adj}}$Let $A\in \mathbb{M}_n$ ($n \geq\ 2$) be a regular matrix and $\adj(A)$ its adjoint. Prove that if A is regular then $\adj(\adj(A)) = (\det A)^{n-2} A$ (where $...
1
vote
1answer
521 views

Prove that every self-adjoint unitary linear operator can be expressed in the form $U\alpha = \beta - \gamma$

This problem is from Kunze Hoffman book. I think I go in the right direction to solve this but I miss some point to finish it. Can anyone help me? Suppose $U$ is a self-adjoint unitary linear ...
10
votes
2answers
4k views

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)...
1
vote
1answer
114 views

Compute $ad_X$, $ad_Y$, and $ad_Z$ relative to a basis

For a lie algebra $\mathbb{g} $ we can define the adjoint representation as: $ ad: \mathbb{g} \rightarrow End(\mathbb{g}) $ as the map such that $ad_x(y)=[x, y] $ for all $\in \mathbb{g} $ I am ...
6
votes
3answers
120 views

$\operatorname{Adj} (\mathbf I_n x-\mathbf A)$ 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 $\...
2
votes
2answers
199 views

$T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)$

$T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)$ I need to know whether it is self adjoint and unitary operator given that $x_i\in\mathbb C$ I am not able to do it please tell me how ...
0
votes
2answers
295 views

Is intersection of a dense subspace and a closed subspace of a Hilbert space also Dense?

I have a Hilbert space $H$ and a closed operator $T$ defined on its domain $D(T)$ which is dense in H. Also $M = \text{range} \ T^n$, for some $n$, is given to be closed. Consider the restriction of $...
1
vote
1answer
313 views

Adjoint of a matrix and inverse of a matrix

As everyone know that we can use a matrix $A$ to represent an operator $T$. The adjoint of a matrix $A$ is denoted as $A^*$, which takes complex conjugate of $A$ and then transpose. My problem ...
0
votes
3answers
147 views

Selfadjoint Operator: Basic Criterion

For symmetric operators one has: $$A\text{ symmetric}:\quad\mathcal{R}(A\pm\imath)=\mathcal{H}\implies A^*=A$$ How to prove this in an unveiling way?
-1
votes
2answers
272 views

Isometric <=> Left Inverse Adjoint

Is it true that: $$T\text{ isometric}\iff T^*\text{ left inverse}$$ Obviously: $$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle$$ $$...
12
votes
2answers
2k views

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\...
8
votes
0answers
3k views

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 ...
4
votes
3answers
2k views

$T$ 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 ...
4
votes
3answers
867 views

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 ...
5
votes
2answers
527 views

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 ...
4
votes
2answers
257 views

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 ...
3
votes
2answers
217 views

General Cauchy-Schwarz for adjoint positive operators

I'm trying to prove the next inequality, like Cauchy-Schwarz standard inequality: $$|\langle Tx,y\rangle |\leq\langle Tx,x\rangle ^{1/2}\langle Ty,y\rangle ^{1/2}\space\forall x,y\in\mathcal{H},$$ ...
3
votes
1answer
350 views

Gelfand Triples / Rigged Hilbert Spaces - Reflexivity necessary?

There have been several questions asked on various aspects of Gelfand triples. However, I have not yet found an answer to the following question: Let $V$ be a Banach space, $H$ be a Hilbert space ...
2
votes
1answer
45 views

Understanding $\overline{Im\: }A=\bigcap_{A'f=0}\ker f$ proof

Theorem: Let $X,Y$ be Banach spaces and $A$ an linear bounded operator. The closure of the image is $\overline{Im\: }A=\{y\in Y:f(y)=0,\forall f\in Y'$ such that $A'f=0$}. $(A'f)(x)=f(A(x))$ is the ...
2
votes
1answer
954 views

self-adjoint operator without eigenvalues?

I have a self-adjoint operator $d$ which acts on vector fields defined on $\mathbb{R}^n$. I am interested on its eigenvalues. That is, I study the equation $d(X)-\lambda X=0$. I have found that if $\...
7
votes
1answer
494 views

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
votes
2answers
121 views

Prove that a regular $\phi : E \to E$ can be uniquely decomposed as a composition of self-adjoint map and a rotation(Unitary trick of Weyl)

In the book(archive.org) of Linear Algebra by Greub, at page 226 it is asked that: Note: $E$ is a $n$-dimensional real vector space. Prove that a regular linear transformation $\phi$ of a ...
2
votes
2answers
145 views

Show $\alpha$ is selfadjoint. [duplicate]

Question: Let $V$ be an inner product space finitely generated over $\Bbb C$ and let $\alpha$ be an endomorphism of $V$ satisfying $\alpha \alpha^* = \alpha^2$. Show that $\alpha$ is selfadjoint. **...
2
votes
2answers
651 views

If $\|Tv\|=\|T^*v\|$ for all $v\in V$, then $T$ is a normal operator

I have solved a question but I am not sure the last step of the question. If someone can verify it that would be great. Let $V$ be a finite dimensional vector space with complex inner product. Let $...
1
vote
1answer
375 views

$\operatorname{Im} A = (\operatorname{ker} A^*)^\perp$ [duplicate]

Let $A:\mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation. We know that there is a unique transformation $A^*:\mathbb{R}^n \to \mathbb{R}^m$ such that $$\langle Ax,y\rangle = \langle x,A^*y \...
1
vote
0answers
963 views

An idempotent operator $P$ is an orthogonal projection iff it is self adjoint

$V$ is a finitte dimension vector space. If for some $P\in End(V)$ we have $P=P^2$ then $P$ is an orthogonal projection $\iff$ $P$ is self-adjoint. I can show that since $P$ is idempotent then $V=ker(...
0
votes
1answer
202 views

How can we compute the adjoint of the inclusion between two Hilbert spaces?

Let $\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$ $(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)$ and $(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$ be $\mathbb K$-Hilbert spaces such that $U\subseteq V$ and ...
0
votes
0answers
114 views

Adjoint representation, standard basis - $\mathfrak{sl}(2)$ and Killing form [duplicate]

Let $L$ be the Lie algebra $\mathfrak{sl}_{2}$ (char ${F}$ $\neq$ 2). Take as standard basis for $L$ the three matrices: $x=\begin{pmatrix} 0&1\\0&0\end{pmatrix}, y= \begin{pmatrix} 0&...
0
votes
2answers
399 views

Isometry <=> Adjoint left inverse [duplicate]

Is it true that: $$T\text{ isometric}\iff T^*\text{ left inverse}$$ Obviously: $$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle$$ $$...
6
votes
3answers
1k views

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 $( T^{-1} )^{*} = ( T^{*} )^{-1}$. ...
4
votes
2answers
151 views

$T=AU \iff T $ is a normal operator on Hilbert space

This is Exercise 16.(c) from Conway's Functional Analysis book. Suppose $H$ is a Hilbert space and $T$ is a compact operator on $H$. Assuming the result that $\exists A$ positive operator and $U$ a ...
4
votes
1answer
1k views

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, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ ...
3
votes
1answer
137 views

Prove that $\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$. [closed]

Let $T$ be a self adjoint bounded linear operator in a Hilbert space $H$. Prove that $$\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$$
3
votes
2answers
491 views

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 ...
2
votes
1answer
44 views

Are all adjoints lattice homomorphisms?

Obviously something must be wrong in the following reasoning proving that any linear operator $T:X\to Y$ between Banach lattices has a lattice homomorphic adjoint: $\forall a,b\in E':$ $$T'(a\wedge b)=...
2
votes
1answer
2k views

Kernel of adjoint and orthogonal complement images

Alright, suppose we are given $V$, a finite dimensional inner product space, and a linear map, $T:V \rightarrow V$, with its corresponding adjoint, $T^\star :V \rightarrow V$. I want to show: $[im(T)]...
2
votes
1answer
460 views

Adjoint differential equations

Consider the vector differential equations \begin{equation} \mathbf{x}^{\prime}=\mathbf{A}(t)\cdot\mathbf{x}\tag{1} \end{equation} and \begin{equation} \mathbf{y}^{\prime}=-\mathbf{A}^{\ast}(t)\cdot\...
1
vote
1answer
169 views

Finding dimension of range of T*T, where T* is the adjoint

Consider T ∈ L(U, V ). U, V are finite dimensional. Show that dim range (T*T) = dim range T. Then, show that if dim range T = dim U then T*T is invertible. T* is the adjoint of T. So it is obvious ...
1
vote
0answers
31 views

Does an adjoint of an internal Hom functor of a prounital closed category define a tensor product?

A closed category is a category equipped with internal Hom functors along with a unit object. Now this answer shows that if $C$ is a closed category whose internal Hom functor has a left adjoint, ...
1
vote
1answer
59 views

Can we find a concrete representation of $\iota\iota^\ast y$, if $\iota$ is a Hilbert-Schmidt embedding between Hilbert spaces?

Let $U$ and $H$ be real Hilbert spaces $\iota:U\to H$ be a Hilbert-Schmidt embedding $Q:=\iota\iota^\ast$ Can we find a concrete representation of $Qy$ for some $y\in H$? By Riesz' representation ...
1
vote
1answer
130 views

Prob. 6, Sec. 3.10 in Kreyszig's functional analysis book: Powers of self-adjoint operators

Let $H$ be a Hilbert space. If $T \colon H \to H$ is a bounded self-adjoint linear operator and $T \neq 0$, then $T^n \neq 0$ for all $n \in \mathbb{N}$. How to show this? I've managed to show ...
1
vote
4answers
257 views

Let $A$ be a $3\times3$ matrix. Given $\mathrm{adj}(A)$, find $\det(A)$.

Let $A$ be a $3\times3$ matrix such that $$\mathrm{adj}(A) = \begin{pmatrix}3 & -12 & -1 \\ 0 & 3 & 0 \\ -3 & -12 & 2\end{pmatrix}.$$Find the value of $\det(A)$. I know that ...
1
vote
2answers
142 views

Adjoint of projection onto direct sum of Hilbert spaces

Let $K_n$ be Hilbert spaces and define \begin{equation*} K := \bigoplus_{\ell_2} K_n = \left\{ (x_1,x_2,\ldots) \in \bigoplus_{n=1}^\infty K_n : \sum_{n=1}^\infty \|x_n\|^2 < \infty \right\} \end{...
1
vote
1answer
102 views

Two “adjunct” (quasi-inverse) functions

Let $A$, $B$ be fixed sets. What "means" the formula $Y \cap \alpha X \neq \emptyset \Leftrightarrow X \cap \beta Y \neq \emptyset$ for functions $\alpha:\mathscr{P}A\rightarrow\mathscr{P}B$ and $\...
0
votes
1answer
334 views

Inverse of a positive definite automorphism over infinitely generated inner product space

Let $\alpha$ be a positive definite automorphism of an inner product space $V$. Is $\alpha^{-1}$ necessarily positive definite? I know the answer is true for invertible positive definite matrices, ...
0
votes
1answer
89 views

Show $\sigma(T)=\sigma{(\overline{T^{*}})}$

Let $T \in B(H)$ be a bounded operator. Is $\sigma(T)=\sigma{(\overline{T^{*}})}$ true for $T$? $\textbf{TRY-}$ I have proved it is true for normal operator but could not do it for bounded operator....