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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 ...