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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).

This tag is intended for questions about adjoints of operators in inner product spaces (see the corresponding tag ), or more generally in vector spaces equipped with a nondegenerate bilinear form. If $T : X \to Y$ is a bounded linear operator between two inner product spaces, the adjoint of $T$ is the operator $T^* \colon Y \to X$ such that: $$(\forall x\in X)(\forall y\in Y):\langle Tx, y \rangle = \langle x, T^*y \rangle.$$

If $X$ and $Y$ are complex Hilbert spaces, then $T^*$ is called the Hermitian adjoint of $T$. This notion is conceptually similar to the notion of of a matrix.

This tag is not intended to be used for adjoint functors from category theory; use the tag instead.

1153 questions
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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 from the graph.
Belen
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How to Find the Adjoint of a Differential Equation?

I am a bit confused about the entire topic of the adjoint and am not totally sure how to apply the definition of it to a given differential equation in order to obtain the adjoint equation. I have perused several examples but have not found one…
tsteimle
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What is the adjoint operator between two inner product space with different scalar fields

Is the adjoint operator defined for any two inner product spaces with possibly different scalar fields? Here is a confusing example. Assume that $H_{1}$ is an inner product space defined on the set of hermitian matrices over the real vector space.…
patrick
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Why does the adjoint exist for densely defined operators?

It is my understanding that when an operator is densely defined, then the adjoint operator exists. But, if $X,\, Y$ Banach spaces and $A:D(A)\subset X\to Y$ linear operator. Why is it necessary for $D(A)$ to be dense in $X$ to have an existence of…
eraldcoil
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Adjoint of a linear application in infinte dimensional spaces

Let $E$ be a vector space and $u \in L(E)$. For the following case, determine whether $u$ has an adjoint with respect to the given bilinear form $\phi (x,y)$. $E = \{ f: \mathbb{R} \to \mathbb{R}: f $ is a polynomial $\}$, $\phi(f,g) = \int_0^1…
karnan
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properties about adjoint operators

Let $T:V \rightarrow V$ a linear map such that $Tv = w$ then the adjoint linear map $T^*$ is $T^*v = u. \forall u,v,w \in V $. My professor defined the linear map $T^*$ as follow: For any linear operator $T:V \rightarrow V$ on a…
Joãonani
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how to find adjoint of this operator on the space of C[0,1]?

We are given $T$ is an operator on $C[0,1]$ as follows $T(g(x))=\sum\limits_{k=1}^{m}p_kg(f_k(x)), p_k\in [0,1], f_k\in C[0,1]$, could anyone tell me how to show adjoint of this operator is as follows?
Myshkin
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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 $\mathcal{A}^*(X_{11})= \begin{bmatrix} X_{11}&0\\ 0&0 \end{bmatrix}$ is the adjoint. To me I understand it by…
Dan
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How do I find adjoint operator in different bases?

I know the Gram matrix $G$ of euclidean space my operator is in and the matrix $A$ of it. How do I use them to calculate adjoined operator? I know I can use $A^*=\overline{G^{-1}A^TG}$ So for example if $G= \begin{pmatrix} 1 & 1 & 1 & 1\\\ 1 & 2 &…
Alexander Kraynov
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Proof $Ly = (ky')'$ is self-adjoint

I'm looking for a proof that $Ly = (ky')'$ is self-adjoint. My attempt: \begin{align} \langle Ly(x), w(x)\rangle &= \int_a^b (k(x)y(x)')'w(x) \,\mathrm{d}x \\ &= \int_a^b (k'(x)y(x)' + k(x)y''(x))w(x) \,\mathrm{d}x \\ &=\quad\color{red}{??} \\ &=…
Frank Vel
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difference between adjoint and inverse

I am not sure if i'm understanding this correctly but if I have a map from $T:X \rightarrow Y$ then the adjoint $T^*: Y \rightarrow X$ Is the inverse not defined in the same way? I figure I am missing some assumption or restriction. I would really…
knk
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Basic operations on adjoint operators

How do you prove : $a^{\dagger}b^{\dagger}ba=a^{\dagger}ab^{\dagger}b$ knowing that $[b,b^{\dagger}]=[a,a^{\dagger}]=1$ and where $^{\dagger}$ denotes the adjoint. do you need extra properties on $a$ and $b$ or is it straight forward?
Ronan Tarik Drevon
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Proving the adjoint nature of operators using Hermiticity

How can the fact that $\hat x$ and $\hat p$ are Hermitian be used to prove that $\hat x - \frac{i}{m \omega} \hat p$ and $\hat x + \frac{i}{m \omega} \hat p$ are adjoints of each other?
nightmarish
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Applying the definition of adjoint operator

How do I apply the definition of adjoint operator in this problem? U and V are two arbitrary operators, not necessarily Hermitian. Show that (UV )† = V †U†.
RK Ali
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