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For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).

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How to find the inverse of a 4x4 order matrix using adjoints for example $$A=\begin{pmatrix} 2 & -6 & -2 & -3 \\ 5 &-13 &-4 &-7 \\ -1 & 4& 1& 2 \\ 0 & 1 &... 2answers 292 views Give an example of a non-self-adjoint operator on a Hilbert space H whose range is H and which is not invertible. Give an example of a non-self-adjoint operator on a Hilbert space H whose range is H and which is not invertible. I cannot think of an example to save my life. Any solutions/hints are greatly ... 1answer 101 views How to prove this is a self-adjoint operator? I have this operator from H^1_0 to H^1_0 defined by:$$Au(t)=\int_0^1 G(t,s) f(s,u(s))\mathsf ds$$where$$G(t,s)=\begin{cases} t(1-s), &t\leq s\\s(1-t), &s\leq t.\end{cases}$$I want to ... 1answer 108 views On the space l_2 we define an operator T by Tx=(x_1, {x_2\over2}, {x_3\over3}, . . . ). Show that T is bounded, and find its adjoint. [duplicate] On the space l_2 we define an operator T by Tx=(x_1, {x_2\over2}, {x_3\over3}, . . . ). Show that T is bounded I know that ||T||\leq 1, but I don't know how to show this. Any solutions or ... 3answers 133 views Is this operator A = \pmatrix{1&1\\0&1} self-adjoint? Is this operator$$A = \pmatrix{1&1\\0&1}$$self-adjoint? I think not, because$$\pmatrix{1&1\\0&1}^T\neq A$$where T is the transposition of the matrix. What do you all think? 1answer 48 views Proving the adjoint nature of operators using Hermiticity How can the fact that \hat x and \hat p are Hermitian be used to prove that \hat x - \frac{i}{m \omega} \hat p and \hat x + \frac{i}{m \omega} \hat p are adjoints of each other? 2answers 441 views When is the restriction of a normal operator not normal? I was proving the spectral theorem for normal operators on finite-dimensional complex vector spaces today during a test, when I arrived at the point in which If T\in\operatorname{End}(V) is ... 1answer 40 views Work out the adjoint of T(x,y) = (y,-x) this seems like a simple question but I don't understand it. We define a transformation T(x,y) = (y,-x). We want to work out what the adjoint is. I know the answer: T^*(x,y) = (-y,x) but how? ... 1answer 126 views Confirm my understanding of adjoints adjoints seem REALLY important and useful so I don't want to move onto the next topic without really understanding them; I have too many a times moved on and been lost because I don't have the ... 3answers 698 views Find the adjoint of this non-standard inner product space I'm really blanking out (a lot of late nights these past 10 weeks). The point of the exercise I'm about to type up is to show that the adjoint structure may possibly change when the inner product ... 1answer 63 views Adjoint of T_A = Ax Is it true that if T_A(x) = Ax then T^*_A(x) = A^*x? I tried to prove this for the standard inner product$$ \newcommand{\innp}{\left\langle #1,#2 \right\rangle} \innp{Ax}{x} = x^tA^t\overline{...
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Let $F$ denote $\mathbb R$ or $\mathbb C$. Let $T : V → W$, $S : V → W$ and let $R: U → V$ be linear transformations between inner product spaces $U$, $V$, $W$ over $F$. Verify the following facts: (...
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Prob. 8, Sec. 3.10 in Kreyszig's functional analysis book: An isometric linear operator has its adjoint as its left inverse

Let $H$ be a Hilbert space, and let $T \colon H \to H$ satisfy $$\langle Tx, Tx \rangle = \langle x, x \rangle \ \mbox{ for all } \ x \in H.$$ Then $T$ is bounded and norm $\Vert T \Vert = 1$ (unless ...
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Prob. 6, Sec. 3.10 in Kreyszig's functional analysis book: Powers of self-adjoint operators

Let $H$ be a Hilbert space. If $T \colon H \to H$ is a bounded self-adjoint linear operator and $T \neq 0$, then $T^n \neq 0$ for all $n \in \mathbb{N}$. How to show this? I've managed to show ...
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What is the adjoint of an inverse matrix? [duplicate]

What is the adjoint of an inverse matrix? Is $(T^{-1})^{*} = (T^{*})^{-1}$?
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$T \in B(X,Y)$ is an isometry if and only if $T^*$ is an isometry

I would like to prove that $T \in \mathscr{B}(X,Y)$ is an isometry of $X$ onto $Y$ if and only if $T^*$ is an isometry of $Y^*$ onto $X^*$. I am not really sure what to do. I started the argument as ...
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Product of compact, bounded and self adjoint operator.

$T \in B(H)$, and $T = S^2$ for some self adjoint operator $S \in B(H)$. I need to prove that T is compact if and only if S is compact. If S is compact, it is easy to show that T is compact since S ...
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Adjoint operator on Banach space

Suppose $X$ and $Y$ are Banach spaces and $T:X\to Y$ is a bounded linear operator. Show that $T$ is an isometric isomorphism if and only if its adjoint $T^*$ is also an isometric isomorphism. Given an ...