2
votes
2answers
65 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
2answers
26 views

help me please about adjoint of operators in L1

A : L₁→L₁ 1) A x=( x₁, x₂,.....xn , 0,0,....) 2) A x= (λ₁ x₁ ,λ₂ x₂,.....) |λ n|≤1 and λ n ∈ R I need to find adjoint of operators A in given space. ...
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 ...
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: $$\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle$$ ...
1
vote
2answers
80 views

Normal Self-Invertible Operator is Self-Adjoint

If $T\in B(H)$ for some Hilbert space $H$, is a normal operator and $T^2=I$, then $T=T^*$. It seemed simple when I first saw the claim, but I'm having trouble showing it. I know that it implies ...
1
vote
0answers
43 views

Adjoint of unbounded Operators: Product and Sum

When precisely does equality hold for sum and product: $$S^*+T^*\subseteq (S+T)^*$$ $$S^*T^*\subseteq (TS)^*$$ So far I checked that for the sum the seemingly weaker condition implies the stronger ...
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 $T$?
0
votes
0answers
84 views

adjoint of a differential equation

Suppose we have the following differential equation $$ \psi''(y)- k^2 \psi(y) - \frac{U''(y)\phi(y)}{V(y)-c}=0$$ where $\psi $ and $\phi$ are complex valued functions, say on the interval $[0,1]$. ...
1
vote
0answers
68 views

Self-Adjoint Operators

If an Operator $L$ is defined as $Lu=u''$ and $a_1u(0)+b_1u'(0)+c_1u(1)+d_1u'(1)=0$ along with $a_2u(0)+b_2u'(0)+c_2u(1)+d_2u'(1)=0$, then for what values of $a_1,b_1,c_1,etc$ is the operator ...
0
votes
4answers
54 views

Help with a 'simple' sum of linear operators and their adjoints acting on an orthonormal basis

Given an orthonormal basis $\{u_1,\cdots, u_n\}$ of a vector space $V$ I am asked to show that $$ \sum_{k=1}^n \|T^*u_k\|^2= \sum_{k=1}^n \|Tu_k\|^2 $$ for all $T\in \mathcal{L}(V)$ where $T^*$ ...
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
133 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
35 views

self adjoint linear operator and integration

is this formula correct ?? $$ \int_{-\infty}^{\infty} Lf(x)\delta (x-1)= \int_{-\infty}^{\infty} f(x)L^{\dagger}\delta(x-1) $$ here $ L $ is a linear operator and $ L^{\dagger}$ is its formal ...
3
votes
2answers
59 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 ...
1
vote
1answer
163 views

What is the adjoint of $x + \frac{d}{dx}?$

I have solved other problems like this using integration by parts. In this case, I can't figure out what to make each part for the integration. The question is true/false. Ultimately to show this you ...
3
votes
1answer
58 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 ...
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 ...
4
votes
2answers
99 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 ...
1
vote
1answer
71 views

Operator inequalities: $0 \leq A \leq B \Rightarrow Tr(A^p) \leq Tr(B^p)$?

It is trivial to show that $0 \leq A \leq B \Rightarrow Tr(A^2) \leq Tr(B^2)$, but does this generally hold for all $p >$ 2 as well?
0
votes
1answer
227 views

Adjoint operator

This is about, a question I answered. Now there is an additional question that I cannot answer and do not want to spend any more time on. I feel like the question will not get any attention, as I ...
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 ...
2
votes
0answers
79 views

Verify a given SVD of an operator

Show that the Singular Value Decomposition of the operator $$ A\colon L^2([0,1])\to L^2([0,1]), x\mapsto\int\limits_0^t x(s)\, ds $$ is given by $$ ...
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 ...
1
vote
1answer
160 views

The convergence of the adjoint operator

If a sequence of operator $A_n$ converges in norm to $A$, i.e. $\lim \lVert A_n-A\rVert=0$)where $A_n$ and $A\in B(H)$ ($H$ is the Hilbert space). Is it true that $A_n^*$ converges in norm to $A^*$?
2
votes
1answer
91 views

Is $\text{rk}L=\text{rk}L^*L $ true for finite rank operators?

Let $L$ be a compact linear operator in an infinitedimensional space that has finite rank. Do the equations $$\text{rk}L=\text{rk}L^*L\ \text{and} \ \text{rk}L^*L=\text{rk}R,$$ where $R$ is the ...
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. ...
1
vote
1answer
148 views

Norm of conjugate Hardy operator

For the classical Hardy operator $T\colon \ell^p\to \ell^p \quad (Tx)_n=\frac{1}{n}\sum_{k=1}^n x_k$ or the integral type $S\colon L^p\rightarrow L^p \quad (Sf)(x)=\frac{1}{x}\int_0^x f(t) dt \ \ $ ...
3
votes
2answers
229 views

eigenfunctions of the adjoint of an operator

If the eigenfunctions of a linear operator are known, is there a way to calculate the eigenfunctions of the corresponding adjoint operator based on the known eigenfunctions? In other words, what's the ...
3
votes
1answer
364 views

Eigenvalues of compact operators and his adjoint.

Let $T: H \to H$ be a compact operator with $H$ a Hilbert space. Let then $\lambda \neq 0$ be an eigenvalue of $T$ with eigenfunction $v$. Is then $\lambda$ an eigenvalue for the adjoint $T^*$ ...