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

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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 ...
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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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33 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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37 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 ...
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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
46 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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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 ...
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1answer
52 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 ...
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30 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 ...
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1answer
56 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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47 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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28 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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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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23 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 $...
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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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15 views

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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21 views

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
26 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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22 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
24 views

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 ...
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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 ...
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104 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)...
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1answer
20 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
37 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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29 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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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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17 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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35 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(...
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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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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 ...
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38 views

How do I show that if is normal, then if T is a projection then it must be an orthogonal projection?

T is an operator on a finite dimensional inner product space. How do I show that if T is a projection, then it must be an orthogonal projection? I know I have to use the fact that $T^*T=TT^*$, and I ...
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1answer
35 views

Exponential of a self-adjoint operator

Let $\mathcal{H}$ be an Hilbert space. Firstly, I shall define some notions as their definitions may vary: A spectral resolution is a function $E:\mathbb{R}\to\mathcal{L}(\mathcal{H})$ (the space ...
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20 views

Finding two adjoints, and showing boundedness of operators

Let $H = l_2$ and consider the following operators: $T,S:H \to H$ $Tx = (0,x_1,x_2,\ldots)$ and $Sx = (x_2,x_3,x_4,\ldots)$ Show they are bounded, and find the adjoint of both: For $T$, I have $\|Tx\|...
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27 views

adjoint method for computing derivatives

I am curious if anyone has heard of this problem before: Suppose that $u(x,p)$ is a function of $x$ and $p$. These arguments need not be scalars. Let $u(x,p)$ satisfy some differential equation, say:...
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1answer
38 views

The adjoint of a linear isometry on normed space is surjective

I come across with the following question: Let $X$ and $Y$ be normed spaces over the same field and $T:X \rightarrow Y$ be a linear isometry, with $\|Tx\|=\|x\| \, \forall x \in X$, then the adjoint ...
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1answer
74 views

Prove that $R(T^{*})^\perp =N(T)$

Let V be an inner product space, with T being a linear operator on V. How do I prove that $R(T^{*})^\perp =N(T)$? I tried setting $x\in R(T^{*})$ and $Ty\in N(T)$, and set up an inner product = 0 ...
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1answer
50 views

Is there any way to find an explicit formula for the adjoint of a linear transformation?

I know that the definition of the adjoint of a linear transformation is defined to be $\langle T(x), y \rangle = \langle x, T^{*}(y) \rangle$ but is there any way to find an explicit formula for $T^{*}...
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1answer
25 views

Eigen values and self adjoin operator

Can someone give me a clue about how to solve the b part ? All I know is the self adjoint formula $$\langle ku,u\rangle = \lambda\langle u,u\rangle$$
3
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1answer
52 views

Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$

I'm considering the bounded linear operator $T$ on $l^1$ (the space of all absolutely convergent complex sequences) given by (with $e_k=(\delta_{kj})_{j=1,2,...}$) $$T((a_j))=\left( \sum_{j=2}^{\...
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11 views

Let $Q$ be a positive-definite, non-singular, self adjoint linear transformation. Then if $P$ is self adjoint, $Pe_i = \lambda_i Qe_i$

I need to show what's written in the question title, for some set of linear independent vectors $\{e_1,\dots,e_n\}$ and scalars $\lambda_1 \dots \lambda_n$. My hypothesis is that the vectors to use ...
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1answer
62 views

Proving that the eigenvalues of $T^*T$ is non-negative

Suppose $T^*$ is the adjoint of $T.$ I want to prove that $T^*T$ has only non-negative real eigenvalues. I proved that $T^*T$ is self-adjoint (because $\overline{(\overline{A^T}A)^T} = \overline{A^T}...
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2answers
68 views

Relating ${\rm ad}(\phi(X))$ with $\phi \circ\,{\rm ad}(X) \circ \phi^{-1}$.

The exercise is to prove that $\phi: {\frak g}\to {\frak g}$ is an automorphism if and only if ${\rm ad}(\phi(X)) = \phi\circ {\rm ad}(X)\circ \phi^{-1}$. Here $\frak g$ is a Lie algebra, of course. ...
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1answer
47 views

Adjoint action on Lie algebra su(2) ($A \in SU(2), X \in \mathfrak{su(2)} \Rightarrow AXA^{-1}\in \mathfrak{su(2)}$)

I am trying to understand ho $SU(2)/\{\pm I\} \cong SO(3)$ (see: how to show $SU(2)/\mathbb{Z}_2\cong SO(3)$) but i am not sure about the adjoint action. In especially, as I understand, the adjoint ...