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Questions tagged [adjoint-operators]

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

14
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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 ...
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\...
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)...
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\...
8
votes
1answer
880 views

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 ...
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 ...
7
votes
3answers
1k views

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 comapct then AB is compact, so I get at once 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 ...
7
votes
0answers
115 views

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 ...
6
votes
2answers
687 views

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 $P$ ...
6
votes
2answers
505 views

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
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 $( T^{-1} )^{*} = ( T^{*} )^{-1}$. ...
6
votes
2answers
958 views

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 ...
6
votes
2answers
431 views

Show $ \langle Tx,x \rangle \in \mathbb R$ for all $x \in H$ implies $T$ is self-adjoint

Show that a linear operator $T: H \rightarrow H$ is self adjoint if and only if $\langle Tx, x \rangle \in \mathbb R$ for all $x \in H$. You may use that the equality that for all $x,y \in H$ $4\...
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 $\...
6
votes
0answers
88 views

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 = \...
6
votes
0answers
1k views

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 ...
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 ...
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 ...
5
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0answers
39 views

$T$-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 ...
5
votes
1answer
287 views

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 ...
5
votes
1answer
115 views

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}(...
5
votes
0answers
497 views

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 ...
5
votes
0answers
866 views

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. ...
4
votes
2answers
212 views

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 ...
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
129 views

Is this operator $A = \pmatrix{1&1\\0&1}$ self-adjoint?

Is this operator $$A = \pmatrix{1&1\\0&1}$$ self-adjoint? I think not, because $$\pmatrix{1&1\\0&1}^T\neq A$$ where $T$ is the transposition of the matrix. What do you all think?
4
votes
3answers
863 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 ...
4
votes
3answers
246 views

Inner product space over generalized number systems

Apologies for the lengthy setup, but I want to make sure I am clear on how I am using the notation, and what I mean by the phrase "generalized number system". Define a generalized number system $G$ ...
4
votes
2answers
82 views

Can I always extend a selfadjoint Operator in $L^2$?

Assume that we have a self-adjoint operator $T\colon D \to D$ where $D \subset L^2$ is some finite dimensional subspace. Can I conclude that than a self-adjoint operator $S \colon L^2 \to L^2$ exists ...
4
votes
1answer
72 views

Is it true that a functor from a locally small category with a left adjoint is representable?

Consider the functor $F: B \rightarrow Set$ where $B$ is a locally small category. Is it true that if $F$ has a left adjoint then it is representable?
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, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ ...
4
votes
1answer
256 views

Hopf Algebra - Adjoint Representation

I've been asked to prove the following; $$ a \circ (bc) = \sum_{(a)} (a_{(1)} \circ b)(a_{(2)} \circ c)$$ Using the fact that the adjoint representation is as follows; $$ a \circ b = \sum_{(a)} a_{(...
4
votes
2answers
404 views

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 ...
4
votes
1answer
89 views

Minimize f(x, y) = $|x + y -3|^2 + |2x + y + 1|^2 + |3x + y - 2|^2$ using least squares

Problem: minimize f(x, y) = $|x + y -3|^2 + |2x + y + 1|^2 + |3x + y - 2|^2$ using least squares, x and y complex I am getting lost on how to approach this problem. In my book, the least squares ...
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 ...
4
votes
2answers
164 views

System of equations wrt self-adjoint operators

$X = \left( \begin{matrix} 2&s\\ 8&2 \end{matrix} \right)$ and $Y = \left( \begin{matrix} 2&-1\\ 2&2 \end{matrix} \right)$ are two operators wrt the same orthonormal basis $B$ in a 2D ...
4
votes
2answers
1k views

Hilbert Adjoint Operator from Riesz Representation Theorem - $T^{*}y=\frac{\left\langle y,Tx\right\rangle }{\left\langle z_{0},z_{0}\right\rangle}z_0$

Kreyszig's Functional analysis seems to introduce the hilbert adjoint operator by means of an explicit representation. I haven't seen this anywhere else and I would like to confirm this explicit ...
4
votes
2answers
397 views

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 ...
4
votes
1answer
2k views

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 ...
4
votes
1answer
634 views

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 ...
4
votes
1answer
91 views

Block Matrices of Operators

I'm trying to prove the following: Consider the vector space of matrices of size $n\times n$ whose entries in $\mathcal B(H)$. Denote this vector space by $M_{n,n}(\mathcal{B(H)})$. We can define ...
4
votes
2answers
150 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
70 views

Show that $T\neq{T^*}$

Let $V=P_2(\mathbb{R}), T\in \mathcal{L}(P_2(\mathbb{R})),$ where $T(p)=(a_1x)$. Make $V$ an inner product space by defining $$\langle p,q\rangle=\int_0^1{p(x)q(x)\,dx}$$ So I calculate $$\langle T(p)...
4
votes
1answer
282 views

Unitary Operator bounded?

Please read all before giving an answer... Suppose you're given a densely defined not necessarily bounded operator: $\overline{\mathcal{D}(T)}=\mathcal{H}$ Moreover, assume it is injective: $\mathcal{...
4
votes
2answers
552 views

general representation theorem for bilinear forms

I am interested in representation theorems for bilinear forms, that go beyond treatment of bounded or even coercive bilinear forms. Whilst I am thankful for any references regarding the topic ...
4
votes
4answers
50 views

Showing that is a normal operator

Let $H$ is a Hilbert space $I$ is unit operator, $T \in B(H)$ and $\lambda \in \mathbb C$ $T$ is normal operator $\Rightarrow$ $T-\lambda I$ is a normal operator too. I could only write : I must ...
4
votes
1answer
58 views

Restriction of adjoint map on measure space

I am stuck at a certain computation in solving a problem. The problem can be formulated as the following: Let $X, Y$ be locally compact spaces and $C_b(X)$ denotes the space of bounded continuous ...
4
votes
1answer
61 views

Using inner product property to determine if operator is an isomorphism.

Let $\varphi$ be an operator on a $k$-vector space $V$ with an inner product $\langle\cdot,\cdot\rangle$. Suppose that $\langle v,\varphi v\rangle = 0$ for every $v\in V$. If we take $k=\mathbb R$, is ...
3
votes
2answers
420 views

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 ...