Questions tagged [adjoint-operators]

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

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Suppose that $T$ is a normal operator on $V$. Show that $\|T(v+w)\|=10$.

Suppose $T$ is a normal operator on $V$. Suppose also that $v, w \in V$ satisfy the equations $$ \| v \|= \| w \| =2, Tv = 3v, Tw = 4w.$$ Show that $\| T(w+v) \| = 10.$ I thought this problem would ...
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163 views

Determining adjoint operator between spaces with different inner products

I'm given a space of $X \in R^{mxn}$ with inner product $\langle X_1, X_2 \rangle = tr(X_1^T X_2)$ and another space of random vectors $Y \in RV^m$ with inner product $\langle y_1,y_2 \rangle = E[y_1^...
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$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)

I am trying to prove the following statements: Let $X$ and $Y$ be normed spaces (not necessarily complete) Let $T\in L(X,Y)$ (meaning $T:X\to Y$ is a bounded linear map). Let $T^*:Y^*\to X^*$ denote ...
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1answer
931 views

Adjoint representation - A working example

I'm working on Lie Algebra at the moment and while everything I hear and read about it makes sense, I get stuck with some exercises (The question How to determine the matrix of adjoint representation ...
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81 views

Is it true you use adjoint for linear transforms while you use transpose for matrix representation of linear transforms?

In one lectures on linear algebra, my professor wrote on the first day: Let $L$ be in a linear map $L: X \to Y$, then $Im(L) \oplus Ker(L^*) = Y$ On the second day, he wrote: $L: X \to Y$, then $Im(...
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3answers
384 views

Why do we need an orthonormal basis to represent the adjoint of the operator?

For any linear operator on a finite dimensional Inner Product Space, we can get orthonormal basis via Gram Schmidt Process. But what is the necessity of defining the adjoint of the operator using ...
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720 views

Characteristic polynomial of adjoint

I'm trying to show that the adjoint transformation $T^*$ of the endomorphism $T$ on a finite dimensional, real inner product space has the same characteristic polynomial as $T$ in a coordinate free (...
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155 views

Why is $\sqrt{T^*T}$ self-adjoint?

Let $T$ be a bounded linear operator over some Hilbert space $H$. Since $T^*T$ is a positive operator, it has a square root. Let $R=\sqrt{T^*T}$. Prove that $\forall u\in H, ||Ru||=||Tu||$. Here'...
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66 views

An identity regarding linear operators and their adjoint between Hilbert spaces

I came upon a trivial-seeming claim that I can't prove myself: Let $H_1$, $H_2$ and $H_3$ be finite-dimensional Hilbert spaces, let $A:H_1\rightarrow H_2$ and $B,C:H_3\rightarrow H_1$ be linear ...
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128 views

Find the inverse of the $n\times n$ matrix whose entries are given by $a_{ij} = \max (i,j)$

The actual question on the past papers is "Let $n\ge 1$ be an integer and consider the $n\times n$ matrix $A$ whose entries are given by $a_{ij} = \max(i,j)$ for all $1\le i,j\le n$. Show that $A$ is ...
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41 views

If A is included in B, what can I say about their adjoint?

I'm studying the theory of linear operators, but even if I tried to bruteforce my way I can't really continue without understanding this (I believe simple) statement: Let A and B be linear operators ...
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1answer
327 views

Domain of double adjoint

For $T$ a densely defined linear, not necessarily bounded operator on a Hilbert space $\mathscr{H}$, and $T^{**}$ the adjoint of $T$'s adjoint, I read somewhere that $Ran(T)=Ran(T^{**})$. Is that true?...
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178 views

To prove norm on adjoint Banach space?

Case: which reminds me a lot about Bourbaki–Alaoglu theorem i.e. a unit case on a ball because of the last inequality and something with adjoint Hahn-Banach. My proposal by thinking the case on a ...
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65 views

Some claims about spectra

I see the following equalities used sometimes, but couldn't find proofs. How is it done? $$\overline{Ran(L-\lambda\mathbb{1})}^{\perp}=ker(L^*-\overline{\lambda}\mathbb{1})$$ and $$\lambda\in\sigma_p(...
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1answer
376 views

$\operatorname{Im} A = (\operatorname{ker} A^*)^\perp$ [duplicate]

Let $A:\mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation. We know that there is a unique transformation $A^*:\mathbb{R}^n \to \mathbb{R}^m$ such that $$\langle Ax,y\rangle = \langle x,A^*y \...
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Do I have the correct mental map for adjoint operators for inner product spaces?

Let $X$, $Y$ be finite dimensional inner product spaces, let $A: X \to Y$ be a linear operator, let $A^*: Y \to X$ be the adjoint operator to the linear operator, defined using $<y, Ax>_Y = <...
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397 views

Linear map is diagonalizable iff its adjoint is diagonalizable

Problem Let $V$ be a finite inner product space and let $T:V \to V$ be a linear transformation. Prove that $T$ is diagonalizable if and only if the adjoint transformation $T^{*}$ is diagonalizable. ...
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141 views

The existence of adjoint operation on Banach space

I have a question about adjoint operator. I have known that bounded linear operator on Hilbert space has a unique adjoint operator, but I am wondering whether there is similar existence result about ...
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2answers
65 views

How to find T* if the product inner space is not explicitly

I have a little question. I was studying for my Linear Algebra test and I tryed the following problem: If $T\in\mathcal{L}(K^n)$ is such that $T(z_1,\ldots,z_n)=(0,z_1,\ldots,z_{n-1})$. Determine a ...
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Adjoint of $\lambda I - T$

Given a selfadjoint (maybe unbounded) operator $T$ on a Hilbert space $H$, I want to calculate the adjoint of $\lambda I - T$ for a $\lambda \in \Bbb C$. I am tempted to argue as follows: According ...
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1answer
33 views

In what condition we have $(K^{-1})^\ast = (K^\ast)^{-1}$?

Suppose $X$ $Y$ are two finite dimensional Hilbert space. Assume $K$: $X\to Y$ is linear. My question is, in what condition of $K$ that $$(K^{-1})^\ast = (K^\ast)^{-1}?$$
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1answer
102 views

Two “adjunct” (quasi-inverse) functions

Let $A$, $B$ be fixed sets. What "means" the formula $Y \cap \alpha X \neq \emptyset \Leftrightarrow X \cap \beta Y \neq \emptyset$ for functions $\alpha:\mathscr{P}A\rightarrow\mathscr{P}B$ and $\...
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1answer
70 views

Coadjoint action $\operatorname{Ad}^*_\phi(h)$ respects coproduct $\Delta$?

In Majid's quantum group primer at the beginning of Chapter 3, page 18, he's proving that if $H'$ and $H$ are dually paired bialgebras or Hopf algebras, the coadjoint action $$ \operatorname{Ad}^*_\...
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60 views

how to find inverse of a matrix

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 &...
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282 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 ...
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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 ...
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1answer
107 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 ...
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129 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?
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47 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?
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2answers
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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 ...
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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? ...
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1answer
125 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 ...
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3answers
665 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 ...
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1answer
61 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}[2]{\left\langle #1,#2 \right\rangle} \innp{Ax}{x} = x^tA^t\overline{...
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1answer
78 views

Proving facts about adjoints

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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2answers
147 views

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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1answer
132 views

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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1answer
315 views

Prob. 10, Sec. 3.9 in Kreyszig's functional analysis book: The null space and adjoint of the right-shift operator

Let $(e_n)$ be a total orthonormal sequence in a separable Hilbert space $H$, let $T \colon H \to H$ be defined as follows: Since span of $(e_n)$ is dense in $H$, for every $x \in H$, we have $$x = ...
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Prob. 2, Sec. 3.9 in Erwine Kreyszig's INTRODUCTORY FUNCTIONAL ANALYSIS WITH APPLICATIONS: Inversion and adjointness

Let $H$ be a Hilbert space, and let $T \colon H \to H$ be a bijective bounded linear operator whose inverse is bounded. Then how to show that $(T^*)^{-1}$ exists and $$(T^*)^{-1} = (T^{-1})^*?$$ My ...
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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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1answer
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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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452 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\...
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3answers
731 views

Proof that All Entries in an Inverse Matrix are Integers

"Prove that if $det(A)=1$ and all the entries in $A$ are integers, then all the entries in $A^{-1}$ are integers." I began by setting up the adjoint method for finding the inverse. $A^{-1} = \cfrac {...
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1answer
455 views

$T \in B(X,Y)$ 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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424 views

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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1answer
2k views

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 ...
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2answers
9k views

$A^{-1}$ has integer entries if and only if the ${\rm det}\ (A) =\pm 1$

So, $A$ is a $n \times n$ matrix with integer entries. The question is to prove that $A^{-1}$ has all integer entries if and only if ${\rm det}\ (A) =\pm 1$ . I know that $A^{-1}= {\rm adj}(A)/{\rm ...
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1answer
36 views

About the self-adjoint extension of an operator.

Let $B$ be a selfadjoint extension of an operator $A$ on a Hilbert space $H$. Let $\varphi \in \ker(A^\ast-z_0)$. Then i want to show that $\varphi + (z- z_0)(B-z)^{-1} \varphi \in \ker(A^\ast-z)$. I ...
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1answer
29 views

Relation between a function and its norm

While reading up on Sturm-Liouville system theory, I came across something I didn't fully understand. At one point, in the midst of proving the existence of solutions to the Sturm-Liouvill problem, ...
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334 views

Adjoint of Exponential Map

If $\exp: T_p(G) \rightarrow G$ is the expoenential map of a lie group, then what does the adjoint operator (as in $\langle Ax,y\rangle=\langle x,A^*,y\rangle$) of the derivative of exp look like? ...