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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)... 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\... 1answer 880 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 ... 0answers 3k 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 ... 3answers 1k 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 comapct then AB is compact, so I get at once that if ... 1answer 494 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 ... 0answers 115 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 ... 2answers 687 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 ... 2answers 505 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 ... 3answers 1k 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}. ... 2answers 958 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 ... 2answers 431 views ### Show  \langle Tx,x \rangle \in \mathbb R for all x \in H implies T is self-adjoint Show that a linear operator T: H \rightarrow H is self adjoint if and only if \langle Tx, x \rangle \in \mathbb R for all x \in H. You may use that the equality that for all x,y \in H 4\... 3answers 120 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 \... 0answers 88 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 = \...
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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 ...
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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 ...
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### 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 ...
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### $T$-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 ...
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### 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 ...
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### general representation theorem for bilinear forms

I am interested in representation theorems for bilinear forms, that go beyond treatment of bounded or even coercive bilinear forms. Whilst I am thankful for any references regarding the topic ...
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### 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 ...
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### Restriction of adjoint map on measure space

I am stuck at a certain computation in solving a problem. The problem can be formulated as the following: Let $X, Y$ be locally compact spaces and $C_b(X)$ denotes the space of bounded continuous ...
Let $\varphi$ be an operator on a $k$-vector space $V$ with an inner product $\langle\cdot,\cdot\rangle$. Suppose that $\langle v,\varphi v\rangle = 0$ for every $v\in V$. If we take $k=\mathbb R$, is ...
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 ...