Questions tagged [adjoint-operators]
For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).
2
votes
1answer
20 views
$PQ \; \text{orthogonal projection} \; \Leftrightarrow PQ = QP$
Exercise :
Let $H$ be a Hilbert space and $P,Q \in \mathcal{L}(H)$ are orthogonal projections, then show that :
$$PQ \; \text{orthogonal projection} \; \Leftrightarrow PQ = QP$$
Seeking a formal ...
1
vote
1answer
28 views
$a=cc^*c$ for some $c$. $a \in A$ a $C^*$ algebra.
Let $A$ be a $C^*$ algebra. Let $a \in A$, then there exists $c \in A$ such that $a=cc^*c$.
This fact is used from example (1) of Prop 4.25. How does one show this?
1
vote
0answers
23 views
Continuous adjoint of the one-dimensional Laplace equation
Say I have a problem given by the 1D Laplace equation,
$$
R (T(\alpha), \alpha) = \frac{d^2 T(x)}{dx^2} - \alpha(x) T (x) = 0,
$$
with $x \in [0,1]$, Dirichlet boundary conditions on $x=0$ and $x=1$, ...
0
votes
0answers
25 views
Show that graph of operator with adjoint operator is closed
Let $X,Y$ be inner-product spaces. Let $T\in L\left(X,Y\right)$ be a linear operator with adjoint operator $S\in L\left(Y,X\right)$ such that
$$\langle Tx,y\rangle_Y=\langle x,Sy\rangle_X\quad\forall (...
1
vote
0answers
29 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, ...
0
votes
0answers
20 views
Taking fourier transform of operators with different exponents
I have a coupled differential equation that I wish to take the Fourier transform (FT) of. However, they consists of different operators which also includes an exponent (more will be shown below). This ...
4
votes
2answers
130 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 ...
0
votes
2answers
48 views
Uniqueness of the Adjoint operator
So I was just stuck in the middle of proving the uniqueness of the adjoint operator.
Known theorem(I already know how to prove it): Assume V is a finite dimensional inner product space over a field ...
2
votes
2answers
51 views
If $D$ be the differentiation operator on $V$. Find $D^*$.
Let $V$ be the vector space of the polynomials over $R$ of degree less than or
equal to $3$ with the inner product space $(f|g)=\int_{0} ^{1}f(t)g(t) dt$, and let $D$ be the differentiation operator ...
3
votes
0answers
30 views
Show that $T$ has an adjoint, and describe $T^*$ explicitly.
Let $V$ be an inner product space and $ \beta, \gamma$ fixed vectors in $V$. Show that
$T \alpha = (\alpha\mid\beta) \gamma$ defines a linear operator on $V$. Show that $T$ has an adjoint, and
...
1
vote
0answers
18 views
Find $T^* $, where $T$ is the linear operator defined by $T \epsilon_1 = (1, - 2), \,\,T\epsilon_2 =(i, - 1)$.
Let $V$ be the space $\mathbb C^2$, with the standard inner product. Let $T$ be the linear
operator defined by $T \epsilon_1 = (1, - 2), \,\,\ T\epsilon_2 = (i, - 1)$. If $ \alpha = (x_1, x_2)$, find ...
1
vote
1answer
24 views
Prove that $\ker P(A^\ast) $ is an invariant subspace of $A$.
Let $A$ be a normal linear operator on a finite dimensional unitary space and $P(x)$ a polynomial. Prove that $\ker P(A^{*})$ is an invariant subspace of $A$ (where $A^{*}$ is its adjoint operator).
I ...
0
votes
1answer
44 views
Is this Sturm-Liouville problem self-adjoint?
We are interested in determining whether the problem $\begin{cases}xu''-u'+u = \cos(x)\\u(0) = 0 \\ u(1) = u'(1)\end{cases}$ is self-adjoint.
This is not a Sturm-Liouville problem, the corresponding ...
0
votes
1answer
37 views
Linear operator is compact if and only if its adjoint is compact
Let $H$ be a Hilbert space, and $A:H\rightarrow H$ a linear operator.
Prove that $A$ is compact if and only if $A^*$ is compact.
I saw the following proof in my book -
What I don't understand ...
1
vote
2answers
32 views
computing the adjoint of operator $T$ on the space $P_2(\mathbb{R})$
Suppose that the inner product on $P_2(\mathbb{R})$ is defined by $$\langle f,g \rangle:= f(-1)g(-1)+f(0)g(0)+f(1)g(1).$$
Consider the operator $T \in B(P_2(\mathbb{R}))$ which is defined as $Tf=f'$, ...
2
votes
1answer
35 views
Does the concept of “adjoint map” determine the metric up to scaling?
Let $V$ be a real finite-dimensional vector space, and $g$ and inner product on $V$. $g$ induces a concept of "adjoint map" , i.e. a linear map $\text{Hom}(V,V) \to \text{Hom}(V,V)$ given by $S \to S^...
0
votes
1answer
29 views
Inner product of a vector field and gradient - Adjoint of the gradient
On page 9 in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.639.5952&rep=rep1&type=pdf
it is being shown why the negative divergence is the adjoint of the gradient. $V: \mathbb R^n \...
0
votes
1answer
27 views
How can the adjoint be defined if $f$ is not one to one
Let $f \in \mathcal{L}(V, W)$. Moreover let's suppose $(e_1, ..., e_n)$ is a basis of $V$ and $f(e_i) = v_i$ where the $v_i$ aren't distinct (so there is at least $i \ne j$ such that $v_i = v_j$) so ...
1
vote
1answer
33 views
How to use operators on tensor products
This problem is from physics, but I have trouble understanding the math.
I have some problem understanding how to use tensors. Let's say in Quantum Optics if I have the state in mode $b$ (where I can ...
1
vote
1answer
31 views
Computing the Adjoint using the Definition $\langle Tv,w\rangle = \langle v, T^*w\rangle$
Let $T$ be the linear operator on $\mathbb{C}^2$ defined by $$T(a,b)=(2ia+3b,a-b).$$ I am trying to compute the adjoint. The answer is $$T^*(c,d)=(-2ic+d,3c-d)\tag{$1$},$$ which can be seen by (i) ...
0
votes
1answer
20 views
Defining Hermitian Adjoints Non-degenerate Hermitian Forms that are NOT positive definite.
I was looking around different textbooks and websites for the definition of a Hermitian adjoint. All the resources that I have checked including the one I am studying at the moment (Jeevanjee's Intro. ...
1
vote
1answer
34 views
$\underset{_{a\rightarrow 0}}{\lim }L_{a}^{\ast }\left( L_{a}L_{a}^{\ast }\right) ^{-1}L_{a}=?$
Let $\varepsilon >0$ and let $L$ be a bounded operator acting on a Hibert space such that $L_{\lambda
}=L-\lambda I$ is surjective of every $\lambda $ such that $\varepsilon
>\left\vert \lambda \...
0
votes
0answers
13 views
Existence of solution of some functional equation involving integral
I want to prove that there exists always solutions to this equation
$$g(x) = \int\limits_a^b {f(s,x)ds} $$
where $g \in {L^2}(0,1)$ is given and $f \in {L^2}((a,b) \times (0,1))$ is the uknown, $0<...
3
votes
0answers
32 views
Adjoint of a polynomial in a closed linear operator.
Let $ H $ be a Hilbert space and let $ T $ be a closed densely defined linear operator in $ H $ with domain $ D(T) $ and with nonempty resolvent set. We define the following polynomial in T:
$ P(T) :...
0
votes
1answer
33 views
Understanding the defintion of dual operators
I'm reading a book about Functional Analysis, and now I've reached to the part about dual operators.
I'm having some difficulties understanding the following definition -
Why $A^*$ is $Y^*\...
0
votes
0answers
21 views
Adjoint operator on $\mathbb{L} ^2 _\mathbb{R}$ of $\frac{d}{dx}$
If A is a linear operator defined on $\mathbb{L} ^2 _\mathbb{R}$ of $\frac{d}{dx}$ then its adjoint must satisfy the property:
$$(A^*f,g) = (f,Ag) \\ f,g \in \mathbb{L} ^2 _\mathbb{R}$$$
Now if $A = \...
2
votes
1answer
30 views
Prove that there is no self-adjoint extension using deficiency indices
Consider an operator $P =-i\frac{d}{dx} : dom(P) \to L^2(\mathbb{R}^+)$ where
$$ dom(P) = \{ f \in \mathcal{D}(\mathbb{R}^+) : f(0)=0\}$$
where $\mathcal{D}(\mathbb{R}^+)$ - smooth compactly ...
0
votes
1answer
58 views
if $AA^*=BB^*$ what are the relations between A and B [closed]
I'm wondering if we have two linear operators $A, B \in \ell(V)$. and we know that $AA^*=BB^*$. then what informations can this give to us about relationships between $A$ and $B$?
I think they have ...
1
vote
1answer
39 views
Prove that $(Au)(t)=\frac{d^{2}u(t)}{dt^{2}}$ is self-adjoint
Let $D(A)=\{ u \in L_2(0,T)| u, \frac{du}{dt}$ are absolutly continuous with $\frac{du}{dt} \in L_2(0,T)$, $u(0)=u(T)=0\}$
and, $(Au)(t)=\frac{d^{2}u}{dt^{2}}$ prove that $A$ is self-adjoint.
Trial
...
1
vote
2answers
48 views
prove $\dim(\operatorname{range}(T)) = \dim(\operatorname{range}(\sqrt{T^*T}))$
I'm a student and I'm studying linear algebra. in Polar Decomposition we have:
for a linear operator $T$, there exist a linear isometry $S$ that:
$$ T =S\sqrt{T^*T}$$
so if $S$ is a linear ...
0
votes
1answer
36 views
Prove $Tx=x$, for $x\in H$, if and only if $(Tx,x)=\|x\|^2$ and $\ker(I-T)=\ker(I-T^*)$
Let $H$ be a complex Hilbert space and $T:H\rightarrow H$ an operation such that $\|T\|\leq 1$. Show that
$Tx=x$ if and only if $(Tx,x)=\|x\|^2$
$\ker(I-T)=\ker(I-T^*)$.
My attempt
1. ...
1
vote
1answer
39 views
Existence and uniqueness of adjoints with respect to pairings
Let $V,W,L$ be $R$-modules over a commutative ring $R$. A pairing is an $R$-linear map $V\otimes W\to L$. An adjoint of an endomorphism $f:V\to V$ w.r.t a pairing $V\otimes W\overset{g}{\to}L$ is an ...
4
votes
4answers
46 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 ...
3
votes
1answer
26 views
OC of Adjoint Operator’s Image is subset of Kernel
Let $T \in B(H,K)$ when $H,K$ are Hilbert spaces and $T^{\star}$ is adjoint of $T$
Show that
$(ImT^{\star})^{\perp} \subseteq KerT$
( $ImT$ means Image of $T$ and $KerT$ means kernel of $T$)
My ...
1
vote
1answer
27 views
Adjoint of operator composition in Hilbert spaces
Let $H,K,L$ are Hilbert spaces, $R \in B(H,K)$ and $T \in B(K,L)$.
Show that $(TR)^\star = R^\star T^\star$ (where $\star$ denotes the adjoint operator).
My attempt:
Let $x\in H$, $y\in K$, $z\in L$...
1
vote
1answer
35 views
Existence of adjoint in normed space
In Luenbeger's book Optimization by Vector Space Methods, chapter 6, the adjoint of a linear operator is defined in the following way:
Let $X$ and $Y$ be normed spaces and let $A: X \mapsto Y$ be a ...
2
votes
0answers
31 views
adjoint operators in the vector space of real polynomials
This problem is about the space $V$ of real polynomials in the variables $x$ and $y$. If $f$ is
a polynomial, $d_f$ will denote the operator $f(d/dx,d/dy)$ , and $d_f(g)$ will denote the result of
...
0
votes
1answer
52 views
Proving adjoint operator identity $(I-\lambda T)^*=I^*- \bar{\lambda}T^* $
Let $T\in \cal{B}$$(H,H)$, where $H$ is a Hilbert space and $\lambda \in \mathbb{C}$. I need to show that $(I-\lambda T)^*=I^*- \bar{\lambda}T^* $ .
My most likely wrong attempt:
$\langle y,(I-\...
1
vote
1answer
26 views
Finding the polar decomposition of a $2\times2$ matrix.
Specifically, the question is as follows:
Let $B$ be the standard basis of $\mathbb{R}^2$. Let $T\in\mathcal{L}(\mathbb{R}^2)$ be such that
$$\mathcal{M}(T,B)=\begin{bmatrix}2&3\\0&2\end{...
1
vote
1answer
31 views
Eigenbases in the infinite dimensional case
Let $H$ be a Hilbert space and $T \in \mathcal{L}(H)$ be self-adjoint. I know that this is a basic question, but I do not understand the infinite dimensional case of linear algebra very well. My ...
0
votes
2answers
32 views
Normal operator in Hilbert Complex space share an eigenvalue.
Everyone, I get stuck in an exercise of Functional Analysis.
Let $T \in B(H)$ (H a complex Hilbert space) and $T^*$ adjoint of $T$. Supose $T$ is a normal operator.
1) Prove that $Ker(T)= Ker(T^*) = ...
1
vote
1answer
49 views
Relation between range of adjoint of T and kernal of T.
I was given two problems in homework
Let $V$ be a finite dimensional inner product space and $T$ be a linear operator on it then,
Prove the following
(1) $ \operatorname{range}$$ ({T^{\dagger}})^{...
0
votes
1answer
110 views
Orthogonal complement of a vector space
I am working on a problem where I have to find the orthogonal complement of a specific vector space. Let $V = \{f: \mathbb{R} \rightarrow \mathbb{C}$, such that f is continuous and periodic with ...
0
votes
1answer
42 views
Proof regarding self-adjoint linear operators.
Let $H$ be a Hilbert space, and let $T : H \rightarrow H$ be a bounded self-adjoint linear operator, with $T \neq 0.$
I need to show that $T^{2^k} \neq 0$ $\forall k \in \mathbb{N}$.
Here's what I'...
0
votes
1answer
25 views
Linear Map Adjoint or Inverse?
In my lecture notes, we have a linear map $\mathcal{A}(X)=X_{11}$ that maps $X\in S^2$ to its first element. It then claims that $\mathcal{A}^*(X_{11})=
\begin{bmatrix}
X_{11}&0\\
0&0
\end{...
2
votes
0answers
35 views
Find $T^*(x,y)$ with the given inner product
Let $T:\mathbb{C}^2 \to \mathbb{C}^2$ be a inner product space, $T(x,y)=(2ix+y, -x-iy)$ and the inner product $<(x_1, x_2), (y_1, y_2)>=4x_1\overline{y_1}+9x_2\overline{y_2}$.
Find $T^*(x,y)$
...
3
votes
2answers
60 views
$Av=\lambda v \Rightarrow A^*v=\bar \lambda v$ (general case)
Suppose $V$ is a finite-dimensional complex inner product space and $v_1,v_2,...,v_3$ is an orthonormal basis of $V$.
Define $A:V \to V$ by $Av_i=\lambda_i v_i$ for some $\lambda_i \in \mathbb{C}.$
...
0
votes
1answer
60 views
If U and T are normal operator which commute with each other then U+T is normal
we had question in our end-semester exam.
State true or false with explanation.
If U and T are normal operator which commute with each other then U+T is normal.
On inner product space.
It is not ...
3
votes
2answers
62 views
Is this differential operator Hermitian?
The operator is
$$\hat{A} = -i \left(x \frac{d}{dx} + \frac{1}{2} \right).$$
Is it true that
$$\langle \hat{A} \psi_1(x)|\psi_2(x)\rangle = \langle \psi_1(x)|\hat{A}\psi_2(x)\rangle\ ?$$
Here, $\...
1
vote
0answers
30 views
$<L(v), v> = 0 $ for all $v$ if and only if $L + L* = 0$.
How do I prove that <$L(v), v$>$ = 0$ for all $v$ if and only if $L + L$*$ = 0$. Here, $L$ * is the adjoint.
I did something like <$L(v), v$> = <$v, L*(v)$> then added the two to get the $L +...
