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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19
votes
2answers
17k 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 ...
16
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2answers
10k 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)...
15
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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\...
11
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3answers
2k 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 compact then $AB$ is compact, so I get at once ...
10
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1answer
2k 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 ...
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\...
9
votes
2answers
1k views

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 \...
8
votes
1answer
249 views

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, ...
8
votes
3answers
202 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 $\...
8
votes
1answer
976 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 ...
8
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0answers
4k 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
2answers
2k views

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
votes
1answer
701 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
167 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 ...
7
votes
1answer
952 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 ...
7
votes
0answers
170 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 function is ...
7
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0answers
117 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
1answer
4k 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 ...
6
votes
2answers
125 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 ...
6
votes
2answers
2k 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
738 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
2answers
700 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 ...
6
votes
1answer
4k 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, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ ...
6
votes
2answers
1k 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
1answer
1k 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 ...
6
votes
2answers
636 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 ...
6
votes
1answer
414 views

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 ...
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
3answers
3k 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 ...
5
votes
2answers
648 views

General Cauchy-Schwarz for adjoint positive operators

I'm trying to prove the next inequality, like Cauchy-Schwarz standard inequality: $$|\langle Tx,y\rangle |\leq\langle Tx,x\rangle ^{1/2}\langle Ty,y\rangle ^{1/2}\space\forall x,y\in\mathcal{H},$$ ...
5
votes
3answers
2k 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 $( T^{-1} )^{*} = ( T^{*} )^{-1}$. ...
5
votes
1answer
118 views

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 ...
5
votes
2answers
761 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 ...
5
votes
2answers
625 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
votes
1answer
76 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 ...
5
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 ...
5
votes
1answer
3k views

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\...
5
votes
1answer
130 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
1k 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
622 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
2answers
473 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 ...
4
votes
2answers
5k views

Derivation of Adjoint for SO(3)

I am in the process of learning about Lie Algebras and Lie groups, specifically for $SO(3)$ and $SE(3)$. I've been reading a tutorial here: http://www.ethaneade.org/lie.pdf, but I'm getting stuck at ...
4
votes
2answers
157 views

Eigenvalue of self-adjoint

I am solving an exercise: Let $T: V \rightarrow W$ be a linear transformation. $V$ and $W$ are finite-dimensional inner product spaces. Prove $T^*T$ and $TT^*$ are semidefinite. This is a solution ...
4
votes
2answers
182 views

a linear map on $W$

Define $W = \{(a_1, a_2,\cdots) : a_i \in \mathbb{F}, \exists N\in\mathbb{N}, \forall n \geq N, a_n = 0\},$ where $\mathbb{F} = \mathbb{R} $ or $\mathbb{C}$ and $W$ has the standard inner product, ...
4
votes
3answers
144 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
1k 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
1answer
2k 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 $\...
4
votes
3answers
477 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
86 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
254 views

What is the motivation behind the definition of adjoint of a linear operator?

Given $T:V\to V$ linear and $V$ being an inner product space, we define $T^*$ by a linear operator on $V$ such that $\langle Tx,y\rangle=\langle x,T^*y\rangle$ for each $x,y\in V$. We later see that, ...

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