Tagged Questions

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

14 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 ...
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 ...
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 ...
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 ...
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). ...
19 views

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 ...
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 $$\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$$
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 ...
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 ...
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 ...
22 views

73 views

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 ...
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 ...
21 views

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 ...
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 ...
104 views

23 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 ...