2
votes
1answer
22 views

Role of metric in the matrix representation of Hermitian adjoint

I'm working through Jeevanjee's "An Introduction to Tensors and Group Theory for Physicists", and while trying to prove that the matrix representation $M(A^\dagger)$ of a Hermitian adjoint $A^\dagger$ ...
0
votes
2answers
27 views

Proof: adjoint map of projection is a projection and …

Let $V$ be a pre hilbert space and $\pi \in \mathrm{End}(V)$. Show: the adjoint map $\pi^+$ of a projection (meaning: $\pi^2 = \pi$) is a projection itself. Show then: a projection $\pi$ is ...
0
votes
1answer
29 views

help,example about disjoint operators

$T\colon L^2[0,1]→L^2[0,1]$ is given by $$ Tx(t)=∫_0^1 tx(s)\,ds $$ How can we find adjoint operator of $T$ in this space? $\langle Tx,y\rangle= \langle x,T^*y\rangle$ should be okay.But what ...
0
votes
1answer
20 views

Finding the Hilbert Adjoint in this case

If we let $H$ be a Hilbert space with inner product $\langle.,.\rangle$. And we fix $y, z \in H$. Then let $T:H\rightarrow H$ be the bounded linear operator $Tx = \langle x,y\rangle z$. Then what is ...
1
vote
2answers
76 views

Isometry <=> Adjoint left inverse [duplicate]

Is it true that: $$T\text{ isometric}\iff T^*\text{ left inverse}$$ Obviously: $$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle$$ ...
-1
votes
2answers
65 views

Isometric <=> Left Inverse Adjoint

Is it true that: $$T\text{ isometric}\iff T^*\text{ left inverse}$$ Obviously: $$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle$$ ...
0
votes
2answers
73 views

Checking whether an operator is self-adjoint. Problem with domain of an operator.

I want to check whether the position operator $A$, where $Af(x)=xf(x)$ , is self-adjoint. For this to be true it has to be Hermitian and also the domains of it and its adjoint must be equal. The ...
2
votes
1answer
83 views

How to find adjoint operator?

Let $(X,\langle\cdot,\cdot\rangle)$ be a Hilbert Space over $K$ with orthonormal basis $(x_n)$, and let $(\lambda_n)\in K$ be a bounded sequence. The mapping $T:X\to X$ is defined by ...
5
votes
1answer
83 views

Adjoint of multiplication by $z$ in the 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 ...
2
votes
2answers
71 views

Clarifying the definition of essential self-adjointness

If a Hilbert space operator $T$ has a unique self-adjoint extension, must the extension be the closure of $T$?
1
vote
0answers
27 views

Proof of Rayleigh trace

I found the following statement without proof: Let us given a self-adjoint Operator $T\colon L^2 \to L^2$ which has n eigenvalues $ \lambda_n \leq \dots \leq\lambda_{n-1} < \lambda_1 =1$ counted ...
3
votes
0answers
68 views

Proving that a certain differential operator is self-adjoint

Consider the differential operator $T:u\mapsto -iu'$ for any $u\in D(T):=\{f\in AC[-\pi,\pi]~|~f'\in L^2(-\pi,\pi),f(-\pi)=f(\pi)\}$; we consider $T$ as a densely-defined operator on $L^2(-\pi,\pi)$. ...
1
vote
1answer
232 views

Dual and adjoint operator

Let $X$ be a Hilbert space with associated canonical isomorphism $I:X\rightarrow X^\ast$ (by the Riesz representation theorem). If $A:X\rightarrow X$ is a linear operator on $X$, then its dual ...
0
votes
2answers
64 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 ...
3
votes
2answers
135 views

Is it a unitary, self adjoint and normal operator?

Let $A\colon H\to H$ be a bounded linear operator on a complex Hilbert space such that $\|Ax\|=\|A^*x\|\forall x$, given that there is a nonzero $x$ for which $A^*x=(2+3i)x$. Then I need to know ...
0
votes
0answers
95 views

Showing an operator is self adjont

I am trying to show that the operator: $$Tf(s)=5s^2\int_0^1t^2f(t)dt+2\int_0^1f(t)dt$$ is self adjoint where $H=L(0,1)$ with real scalars and $t\in \mathcal{L}(H)$. So I can re-write this operator ...
2
votes
1answer
197 views

Invertible operator norm bound

Let $H$ be a Hilbert space and that $X$ are bounded. Suppose $X$ is self-adjoint. Show that $Y=X+iI$ is invertible and the inverse $Y^{-1}$ has the norm $\lVert Y^{-1} \rVert \le 1$. I can prove $Y$ ...
2
votes
0answers
126 views

Find the adjoint operator

I would like to find the adjoint operator in the Hilbertspace $L^2(0,\infty)$ of the operator $$ (Ax)(t)=x(at), x\in L^2(0,\infty), a>0. $$ My calculation is the following; I use the ...
3
votes
2answers
231 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 ...
2
votes
2answers
412 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 ...
5
votes
0answers
337 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. ...