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

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Determine the adjoint of $\tilde Q(x)$ for $\tilde Q(x)u:=(Qu)(x)$ where $Q:U→L^2(Ω,ℝ^d$ is a Hilbert-Schmidt operator and $U$ is a Hilbert space

Let $d\in\mathbb N$ $\lambda$ denote the Lebesgue measure on $\mathbb R^d$ $\Omega\subseteq\mathbb R^d$ be open $H:=L^2(\Omega,\mathbb R^d)$ $U$ be a separable $\mathbb R$-Hilbert space $Q:U\to H$ ...
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1answer
37 views

Hermitian adjoint

I'm trying to solve this task, but I'm not sure, if my solution for a) is correct. For b), i dont find a starting point. Did someone have an idea how to solve this? Thanks in advance. Be $V$ the set ...
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25 views

Calculate the adjoint map

i'm trying to solve this task, but I don't find a starting point. Did someone have an idea how to solve this? Be V the set $\{f \in \mathbb{R}[X]| grad\,f \leq 2 \}$. This becomes to an euclidic ...
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19 views

$ (k\otimes h^\ast)^\ast=h\otimes k^\ast$?

Let $H,K$ be Hilbert spaces with $h\in H,k\in K$. Let $k\otimes h^\ast(g)= \left\langle g,h \right\rangle k$. I'm supposed to prove $ (k\otimes h^\ast)^\ast=h\otimes k^\ast$, but I don't see how this ...
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32 views

Exercise on the operator $g(t)\mapsto \sqrt \lambda g(\lambda t)$

I have the linear operator $A_\lambda :g(t)\mapsto \sqrt \lambda g(\lambda t)$ on $C[0,1]$ with $\lambda\in (0,1)$ (which extends to $L^2[0,1]$). I need to find its adjoint and then prove that while $...
0
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1answer
40 views

Calculating $\|A\|_2$ in terms of eigenvalues of $A^\ast A$

Let $A$ be a real matrix. I'm supposed to calculate $\|A\|_2$ in terms of the eigenvalues of $A^t A$. I thought to just diagonalize $A^t A$ as $UD^2U^t$ but then I have $\|Ax\|_2 =x^tUD^2U^tx$ instead ...
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1answer
28 views

Can we talk about the adjoint of a linear operator defined on a distribution space?

Let $d\in\mathbb N$ $\Omega\subseteq\mathbb R^d$ be open $\mathcal D:=C_c^\infty(\Omega,\mathbb R^d)$ $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the inner product on $L^2(\Omega,\mathbb R^d)$ and $$...
2
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30 views

Seemingly strange exercise on unitary Hermitian operator

Let $V$ be a finite dimensional inner product space. Let $T$ be a Hermitian unitary operator. Prove there's a subspace $W$ such that for each $v\in V$ we have $Tv=w-w^\prime$ where $w\in W,w^\...
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49 views

Finding the adjoint of a compact operator?

Let $H = L^2([0,1])$ and $A:H \to H$ be defined as $$ Af(t) = \int_0^t f(s) ds, \quad f \in H $$ By the Ascoli-Arzela theorem, $A$ is compact. Now the adjoint of $A$ is given as $$ A^\ast g(s) = \...
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31 views

The adjoint of matrix operator?

I have a 2 by 2 matrix as follows: $$ A:= \begin{bmatrix} d &0\\ -c& 1 \end{bmatrix} $$ where $d$ $c$ are two constant. Then the paper I am reading, page 21, after equation $(5.3)$ claiming ...
4
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2answers
51 views

Prove or disprove: $\operatorname{Adj} (A)$ is diagonlizable $\implies A$ is diagonalizable

For $2X2$: $$ A:\\ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ $$ \operatorname{Adj}(A):\\ \begin{bmatrix} d & -c \\ -b & a \end{bmatrix} $$ So the statement is true. The ...
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10 views

Property of the orthogonal projection $\tilde{\operatorname P}$ in the definition of the Stokes operator

Let $d\in\mathbb N$ $\Omega\subseteq\mathbb R^d$ $\langle\;\cdot\;,\;\cdot\;\rangle_2:=L^2(\Omega,\mathbb R^d)$ and $$\langle u,v\rangle_{1,\:2}:=\langle u,v\rangle_2+\sum_{i=1}^d\langle\nabla u_i,\...
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1answer
24 views

Verify my proof: For an invertible $A$ , $V\in F^n$ is an invariant subspace $V$ of $A$ $\iff$ $V$ is an invariant subspace $V$ of $Adj(A)$

$$ v_0,v_1\in V $$ $$ V \text{ is an }A\text{-invariant subspace} $$ $$ Av_0=v_1 $$ $$\iff$$ $$ A^{-1}Av_0=v_0=A^{-1}v_1 $$ $\text{Using }A^{-1}=\frac{1}{\operatorname {det}(A)}\operatorname{Adj}(...
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19 views

Adjoint of the gauge covariant derivative

Suppose $A=A_1dx_1+A_2dx_2$ is a 1-form connection in $\mathbb{R}^2$ and $D_A \phi=d\phi-iA\phi$ is the gauge covariant derivative with $\phi=\phi_1+i\phi_2$ is a complex scalar field. May I ask what ...
2
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1answer
19 views

Show that $q(T)(x)=\sum_{n=1}^\infty q(\lambda_n) \langle x,e_n\rangle e_n$ coincide with $q(T)=\sum_{k=0}^n a_kT^k$

Let $Tx=\sum_{n=1}^\infty \lambda_n \langle x,e_n\rangle e_n$ be bounded where $\{\lambda_n\}_n$ are the complex eigenvalues and $\{e_n\}_n$ are an orthonormal basis of the separable space $H$. For ...
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34 views

Sufficient conditions for $f(T)$ to be compact and self adjoint whenever $T$ is compact and self adjoint

Let $Tx=\sum_{n=1}^\infty \lambda_n \langle x,e_n\rangle e_n$ be bounded where $\{\lambda_n\}_n$ are the complex eigenvalues and $\{e_n\}_n$ are an orthonormal basis of the separable space $H$. For ...
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38 views

explicit self adjoint operator which has no diagonalization

Let a linear operator $T : H \to H$ be diagonalizable if $H$ has an orthonormal basis composed of eigenvectors of $H$ Give an example of an explicit self adjoint operator which has no diagonalization ...
2
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1answer
37 views

What is the relation between the matrix of an operator and that of its adjoint?

Let $X$ and $Y$ be finite-dimensional normed spaces, either both real or both complex, and let $T \colon X \to Y$ be a linear operator. (Then $T$ is bounded since its domain is finite-dimensional). ...
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Prob. 5, Sec. 4.5 in Kreyszig's functional book: The adjoint of the composite of two bounded linear operators

Let $X$, $Y$, and $Z$ be normed spaces, either all real or all complex. Let $T \colon X \to Y$ and $S \colon Y \to Z$ be bounded linear operators. Let $X^\prime$, $Y^\prime$, and $Z^\prime$ denote the ...
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1answer
47 views

Prove that $\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$. [closed]

Let $T$ be a self adjoint bounded linear operator in a Hilbert space $H$. Prove that $$\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$$
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19 views

Proving that $-\Delta+V$ on some domain is self-adjoint

This question may look as a "proof-reading" question, but what I ask is if I correctly understand the way these concepts work, by showing how I think about them. Suppose I have the following three ...
0
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1answer
54 views

Prove there is a compact self adjoint $S:H\to H$ such that $S^3=T$.

Let $T:H\to H$ be compact and self adjoint. Prove there is a compact self adjoint $S:H\to H$ such that $S^3=T$. Is the $S^3$ means power of 3 or applying the operator 3 times? What is there to prove ...
3
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1answer
32 views

For positive self adjoint $T$, show $|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$

As in title, $T$ is a positive self adjoint, bounded linear operator on a Hilbert Space $X$ and I'd like to show $$|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$$ Self adjoint ...
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1answer
22 views

How does the usual properties of Hilbert adjoint operator follow from this definition?

Given two hilbert spaces $X,Y$, and a bounded linear $T:X\to Y$, define $S:Y\to X$ by $$ S=J_{X}^{-1} \circ T' \circ J_Y $$ Where $T':Y'\to X'$ is given by $T'(y')=y'\circ T$ for $y'\in Y'$ and $J_X ...
0
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1answer
58 views

How can we compute the adjoint of the inclusion between two Hilbert spaces?

Let $\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$ $(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)$ and $(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$ be $\mathbb K$-Hilbert spaces such that $U\subseteq V$ ...
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1answer
33 views

Can we find a concrete representation of $\iota\iota^\ast y$, if $\iota$ is a Hilbert-Schmidt embedding between Hilbert spaces?

Let $U$ and $H$ be real Hilbert spaces $\iota:U\to H$ be a Hilbert-Schmidt embedding $Q:=\iota\iota^\ast$ Can we find a concrete representation of $Qy$ for some $y\in H$? By Riesz' ...
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48 views

When is $adj(A)$ nilpotent?

Are there any conditions regarding the $adj(A)$ being nilpotent for some square matrix A?
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1answer
30 views

Clever way to prove $\langle A,X\rangle=x^TAx$ with $X=xx^T$, $A\in S^n$?

How to prove $\langle A,X\rangle=x^TAx$ with $X=xx^T$, $A\in S^n$? (inner product of matrices) $xx^T$ is rank one. The following is one way to prove it: $$\langle A,X\rangle=\text {tr}(AX)$$ ...
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66 views

Is every invertible matrix $A$ an adjugate matrix of some other matrix $B$? If so, is $B$ unique?

Is it true in general, true for a specific field ($\mathbb R$/$\mathbb C$) or false? could it be that $A=adj(B),A=adj(C)$ but $B\not=C$?
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16 views

Rank of adjoint of a matrix [duplicate]

I need to prove these 3 statements, and I don't know how to start... A is an nxn matrix: 1) if rank(A) = n then rank(adj(A)) = n 2) if rank(A) = n-1 then rank(adj(A)) = 1 2) if rank(A) < n-1 ...
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29 views

Finding the adjoint of the left translations semigroup on $L^p (\Bbb R)$

If $t \mapsto T_l (t)$ is the left translation operator by $t$ on $L^p (\Bbb R)$ given by $\Big( T_l (t) (f) \Big) (s) = f (t + s)$, find the adjoint of the left translations semigroup. Note that on $...
3
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1answer
73 views

Operator $f \mapsto u(f)$ solution of non-homogeneous Laplace equation is compact and self-adjoint

Let $u : L^2_0(D) \to L^2_0(D): = \lbrace f \in L^2 : \int_D f = 0 \rbrace $ be the linear operator which associates $f$ to $u(f)$ the solution of $$ \begin{cases} \Delta u = f & \text{in } D \\ \...
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1answer
30 views

determinants of matrix with adjoints of order 2

Let $A$ be a square matrix of order $2$ with $\lvert A \rvert\ne 0$ such that $\big\lvert A+\lvert A \rvert \operatorname{adj} (A)\big\rvert=0$, then the value of $$\big\lvert A-\lvert A \rvert \...
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59 views

Does every invertible matrix A has a matrix B such that A=Adj(B)?

I'm trying to understand if it's always true, always true over $\mathbb C$ or never true. I know that if $A$ is invertible, than there exists $A^{-1}$. $$A=\frac{1}{det (A^{-1})}Adj(A^{-1})$$ So I ...
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Singular values of the differentation operator

I was trying to solve this exercise but I can't get to answer given in book: Find the singular values of the operator D over $R_2[x]$ (That is polynomials with degree equal or less than 2) define as ...
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Showing that $Im(L^*)=(Ker\: L)^\perp \space \:\mathrm and \:\:Ker(L^*)=(Im\: L)^\perp$

Let $V,W$ be finite-dimensional euclidean or unitarian Spaces and $L: V \to W$ a linear map. I have to show the following: $$Im(L^*)=(Ker\: L)^\perp \space \:\mathrm {and} \:\:Ker(L^*)=(Im\: L)^\perp $...
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1answer
28 views

If $(U,〈\;⋅\;,\;⋅\;〉),H$ are Hilbert spaces, $W\in U$, $Y\in H$, $Z\in L(U, H)$ and $f\in L(H,L(H,\mathbb R))$, then $〈Y,fZW〉=〈ZW,fY〉$

Let $(U,\langle\;\cdot\;,\;\cdot\;\rangle)$ and $H$ be Hilbert spaces $W\in U$, $Y\in H$ and $Z\in\mathfrak L(U, H)$$^1$ $f\in\mathfrak L\left(H,\mathfrak L\left(H,\mathbb R\right)\right)$ How ...
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1answer
24 views

Prove that $A = G^{4}$ for some self-adjoint matrix $G$

Let $A\in M_{n}(\mathbb{C})$, $n\geq 2$. Let $P_A(x) = (x-\lambda_1)....(x-\lambda_n)$ be characteristic polynomial of $A$ and all $\lambda_i$ are positive reals. Show that if $A$ is normal, then $A ...
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1answer
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When is this matrix unitary

If we have the matrix $$U=\begin{bmatrix} a & b & \frac{1}{\sqrt{2}} \\ c & 0 & 0 \\ d & e &\frac{-1}{\sqrt{2}} \end{bmatrix}$$ what are the ...
2
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2answers
48 views

Self adjoint and symmetric operator

I am wondering whether for an operator defined on a real Hilbert space to be positive we need to show that it is self-adjoint at first. It seems to me that they are two different property and can be ...
4
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122 views

image of adjoint equals orthogonal complement of kernel

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)...
2
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1answer
21 views

Does this inner product manipulation make sense?

Suppose A is a normal matrix over $M_n(\mathbb{C})$, with diagonalization $A = PDP^*$. Consider the inner product $<A\mathbf{v}, \mathbf{v}>$ $<A\mathbf{v}, \mathbf{v}> = <P(DP^*\...
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1answer
38 views

Compute $ad_X$, $ad_Y$, and $ad_Z$ relative to a basis

For a lie algebra $\mathbb{g} $ we can define the adjoint representation as: $ ad: \mathbb{g} \rightarrow End(\mathbb{g}) $ as the map such that $ad_x(y)=[x, y] $ for all $\in \mathbb{g} $ I am ...
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1answer
30 views

Is it true that when $X$ is a positive matrix and $Y$ is a self adjoint matrix then $XY$ is positive??

Let $X_{n \times n}$ be a positive matrix i.e $<Xy,y> \ge 0$ for all $y \in \mathbb{C^n}$ and $Y_{n \times n}$ be a self adjoint matrix. Show that $XY$ is positive or find a counterexample. So ...
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2answers
41 views

Proving an complex operator is self-adjoint

Let V be a finite dim complex inner product space. Let T be a linear operator on V. Prove that T is self-adjoint if and only if $\langle T\alpha,\alpha\rangle$ is real $\forall \alpha \in V$. The ...
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2answers
18 views

Proving Kernel Equality of a Linear Transformation and Adjoint

Let $T$ be a linear transformation in an inner product space $V$. Determine if the following it true or false: $$Ker (T)= Ker (T^*T)$$ Where $*$ donates the adjoint operator. Would it help proving ...
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37 views

Is the adjoint of an isometry $S$ the inverse of $S$?

Let $S$ be an isometry with adjoint $S^\ast$. Prove that $S^\ast S=SS^\ast=I$. Here is what I have so far: Since $S$ is isometry, there is an orthonormal basis of $e_j$'s such that $\|Se_j\|=\|e_j\|...
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1answer
48 views

Hodge decomposition thm. Why $\Delta(E^p)=d\delta(E^p)\oplus\delta d(E^p)=d(E^{p-1})\oplus\delta(E^{p+1})?$

$\DeclareMathOperator{Img}{Im}$ In Warner's "Foundations of differentiable manifolds and Lie groups" we read that $$E^p\stackrel{(1)}{=}\Delta(E^p)\oplus H^p\stackrel{(2)}{=}d\delta(E^p)\oplus\delta d(...
2
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1answer
23 views

Are all adjoints lattice homomorphisms?

Obviously something must be wrong in the following reasoning proving that any linear operator $T:X\to Y$ between Banach lattices has a lattice homomorphic adjoint: $\forall a,b\in E':$ $$T'(a\wedge b)=...
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2answers
54 views

self-adjoint and orthonormal basis

Suppose $F=\mathbb{R}$. Let $A: V\to V$ (where $V$ is a finite dimensional inner product space over $F$) so that $A=A^*$ ("self-adjoint"), then there exists an orthonormal basis of eigenvectors and ...