# Tag Info

## New answers tagged spectral-theory

0

The discussion at that point in the book was focused on a normal complex matrix $A$ with spectrum (eigenvalues in this case) $\sigma(A)=\{\lambda_{1},\cdots,\lambda_{k}\}$. Normal matrices are unitarily equivalently to diagonal matrices. Equivalently, if $\{ \lambda_{j}\}_{j=1}^{k}$ are the distinct eigenvalues of $A$, then there is an orthonormal basis of ...

0

In order to show that your operator is compact, show that it maps a bounded sequence $\{ f_{n} \}_{n=1}^{\infty}\subset C[0,1]$ to an equicontinuous sequence of functions. So, let $\{ f_{n} \}_{n=1}^{\infty}$ satisfy $\|f_{n}\|_{C[0,1]}\le M$ for all $n$ and some fixed $M$; then, for every $\epsilon > 0$, show that there is a $\delta > 0$ such that $$... 1 I understand your confusion. There's an error in the critical equation of the proof that, once corrected, makes everything obvious. I'll rewrite it for you$$ \left\|A\left(\frac{y_{q}}{\lambda_{q}}\right)-A\left(\frac{y_{p}}{\lambda_{p}}\right)\right\| = ...

1

If $A x_k = \lambda _k x_k$, $x_k \ne 0$, $k \in \{1,...,n \}$, and $\lambda _i \ne \lambda _j$, $i \ne j$, then a vector of the form $$x = \sum _{k=1}^n a_k x_k, \ \ \ a_k \in \mathbb{R}$$ can be an eigenvector of $A$ only if $a_k=0$ for all $k \in \{1,...,n \}$ but one. Indeed, if $A x = \lambda x$, then the linear independence of $x_1,...,x_n$ ...

1

Your operator is already diagonalized, and it is clear that $Te_{n}=y_{n}e_{n}$ for $n \ge 1$. So $\{ y_{n}\}$ are eigenvalues. The eigenvectors of a selfadjoint operator are orthogonal for different eigenvalues. Nothing except the $0$ vector is orthogonal to $e_{n}$ for all $n$. So there can't be any other eigenvalues. Now you know that the spectrum of $T$ ...

2

If $\Im\lambda \ne 0$, and $x \in X$, then $$\Im\lambda \|x\|^{2} = \Im((A-\lambda I)x,x),\\ |\Im\lambda|\|x\|^{2} \le |((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\|,\\ |\Im\lambda|\|x\| \le \|(A-\lambda I)x\|.$$ So $A-\lambda I$ is injective for all $\lambda\notin\mathbb{R}$. The above inequality can be used to show that ...

1

The representation of the elements of $H^*$ is not needed. Let $\lambda$ be in the point spectrum of $A$. Then there is $0\neq x\in H=(H,(\cdot,\cdot))$ such that $Ax=\lambda x$ and by the self-adjointness of $A$: $\lambda(x,x)=(\lambda x,x)=(Ax,x)=(x,Ax)=(x,\lambda x)=\overline{\lambda}(x,x)$. Hence $\lambda=\overline{\lambda}$, thus ...

3

This operator is sometimes called unilateral shift. Suppose $Ax=\lambda x$ where $x=(x_1,\cdots,x_n,\cdots).$ If $(0,x_1,x_2,\cdots)=(\lambda x_1,\lambda x_2,\cdots),\,$you can convince yourself that $\lambda$ must be zero. Also notice that obviously $\|A\|=1.$ Now let $0<|\lambda|<1$ and consider $A-\lambda I$. Then if $(1,0,0,\cdots)=(A-\lambda ... 2 Yes, because if$x$is invertible and$\|y - x \| \le \|x^{-1}\|^{-1}$, $$y = x (1 - x^{-1} (x-y))$$ and$1 - x^{-1} (x - y)$is invertible with $$(1 - x^{-1}(x-y))^{-1} = \sum_{j=0}^\infty (x^{-1} (x-y))^j$$ Moreover $$\|y^{-1} - x^{-1}\| \le \sum_{j=1}^\infty \|x^{-1}\|^{j+1} \|x - y\|^j = \dfrac{\|x^{-1}\|^2 \|x - y\|}{1- \|x-y\| \|x^{-1}\|}$$ -1 Due to the pointwise structure it formally holds: $$\sigma(F)=\bigcup_{x\in X}\sigma(F(x))$$ The difficulties arise as soon as additional requirements are tied upon the functions: 0) The formal inverse exists. 1a) The formal inverse is not necessarily bounded.* 1b) The formal inverse is continuous due to the Neumann series. 2) The formal inverse is again ... 1 It seems fine to me: the only point to be careful of is to note that since you've assumed the normed field$K$to be algebraically closed, its value group must be dense (it cannot be discrete since if it were, there would be a uniformizer and it couldn't have$n$th roots). So for any$\epsilon > 0$, we can find some$a \in K^\times$such that$\rho(A) ...

1

I think very likely the question you might wish to be asking includes more structure than the question you literally asked... based on your example of a Laplacian. That is, your Hilbert space $H$ is really a Sobolev space $H^1$ on some compact Riemannian manifold. Then, yes, the Laplacian maps $H^1$ to $H^{-1}$ continuously, and $H^{-1}$ is the Hilbert space ...

1

For simplicity, assume $f(x,y)=f(y,x)$ is a real function. Because $f$ is in $L^{2}([0,1]\times[0,1])$, then the integral operator $K$ given by $$Kg = \int_{0}^{1}f(x,y)g(y)\,dy$$ is a selfadjoint Hilbert-Schmidt integral operator on $L^{2}[0,1]$. So there is an orthonormal basis $\{\varphi_{n}\}_{n=1}^{\infty}$ consisting of real eigenfunctions ...

0

Okay, I think I have it. The strategy is to generalize the proof of "No eigenvalues => product ergodic" which is usually given in standard texts when introducing the notion of weak mixing. The following is a proof I have written just before in a document: Let $E \subset L^2(X,\mu)$ be the set of eigenfunctions for $U_T$ that have constant absolute value ...

2

The functional calculus works as stated. However, the spectral mapping result that you want to prove is not true in general. For example, consider the operator $M$ of multiplication by $x$ on $L^{2}[0,1]$. The spectrum of $M$ is $[0,1]$. If you let $g(x)=x$ for $x \in [0,1/2)\cup(1,2,1]$ and $g(1/2)=50$, then $g(M)$ does not have $50$ in its spectrum. On the ...

6

here's my solution: The function $\min \{ x, y \}$ can be written as follows: $$\min\{x, y \}= \begin{cases} & y, \mbox{ if } 0 \le y \le x \\ & x, \mbox{ if } x \le y \le 1 \end{cases}$$ so we found the form for $T$: $$Tf(x) = \int_0^x yf(y) \, dy + x \int_x^1 f(y) \, dy.$$ Now let $Tf = \lambda f, \lambda \ne 0$. So we ...

1

One possible description of this set is the following. Let $\Im(C)$ denote the image of the matrix $C$. Thus, $S(A)=\{B,\ \Im(AB)\subset\Im(B)\}$. Before proving it, notice that we can provide many examples of matrices in $S(A)$ with this description. For example, any matrix $B$ such that $\Im(A)\subset\Im(B)$ belongs to $S(A)$, because ...

2

If you have a bounded operator $A$, then the holomorphic functional calculus is always an option, and it is based on Cauchy's integral representation: $$f(A) = \frac{1}{2\pi i} \oint_{C} f(\lambda)\frac{1}{\lambda I-A}"\,d\lambda = \frac{1}{2\pi i} \oint_{C} f(\lambda)(\lambda I-A)^{-1}\,d\lambda.$$ The contour $C$ is any simple ...

2

Physically, Hamiltonian operators in Quantum Mechanics should be semibounded, meaning that $(Ax,x) \ge M(x,x)$ for all $x\in\mathcal{D}(A)$ and for some fixed $M$. This has to be done with energy considerations. Second order ODES and PDES, in order to be symmetric, are quadratic in nature, and usually end up being semibounded--again, this is related to ...

0

Suppose $\overline{G}$ is the complement of the graph $G$. Then $A(G)+A(\overline{G}) = J-I$ (where $J$ is the matrix with all entries equal to one). If $n=|V(G)|$, this implies that $\lambda(G)+\lambda(\overline{G}) \ge n-1.$ We get equality here if $G$ is a regular self-complementary graph. In particular if we take $G$ to be the Paley graph on ...

0

I'm no expert on this, so I might say stupid things, but lets have a go. Let me assume that $A$ is unital (with unit $\mathbb{1}$). The answer to your first question is positive, since $$\sigma(F)=\bigcup_{\omega\in\Omega} \sigma(F(\omega)).$$ Indeed, the inverse $h$ (if it exists) of the function $\lambda I-F$ satisfies $h(\lambda I-F)=(\lambda I-F)h\equiv ... 0 I decided to make a separate answer for the Numerical Range question. They're different. Yes, the closure of the numerical range is the same as the closed convex hull of the spectrum. Numerical Range: Suppose$\lambda_{1},\lambda_{2} \in \sigma(A)$with$\lambda_{1}\ne \lambda_{2}$. Using the spectral theorem, you can find sequences$\{ ...

3

Every bounded linear operator on a complex Banach space has non-empty spectrum $\sigma(B)$. A simple way to argue this is to assume the contrary, and conclude that $(B-\lambda I)^{-1}$ is an entire function of $\lambda$ which vanishes at $\infty$; then Liouville's theorem gives the contradiction that $(B-\lambda I)^{-1}\equiv 0$ for all ...

0

Note that the spectrum of any operator can't be empty. I think what you mean is that the operator might not have an eigenvalue- an example is given here.

1

The fact that the range of $P$ is invariant under $A$ can be written as $PaP=aP$ for all $a\in A$. Now fix $a\in A$; since $a^*\in A$, $Pa^*P=a^*P$. Take adjoints and you get $PaP=Pa$. So we have shown that $Pa=aP$ for all $a\in A$.

2

Let's first check that $P^\perp$ has invariant range. Let $a \in A$ and let $\eta \in [A\xi_1]^\perp = (A\xi_1)^\perp$. Then, for any $b \in A$, $$\langle a\eta, b\xi_1 \rangle = \langle \eta, a^\ast b \xi_1 \rangle = 0,$$ where $a^\ast b \in A$ and hence $a^\ast b \xi_1 \in A\xi_1$ precisely because $A$ is a self-adjoint algebra of operators in $B(H)$, ...

1

You can find hint at Neumann series wikipedia article. Furthermore the question is maybe a duplicate, I think you can find the answer at math.se.

3

This is trivial from the definition of spectrum. If the spectral radius is less than one, then in particular $1$ is not in the spectrum, which means $I-A$ is invertible. Cameron Buie has answered a more interesting question.

1

$||AB-BA||=||AB-A_{\alpha}B+A_{\alpha}B-BA|| \leq ||AB-A_{\alpha}B||+||A_{\alpha}B-BA||$. But $A_{\alpha}B=BA_{\alpha}$, and $||AB-A_{\alpha}B|| \leq ||B||||A-A_{\alpha}||$. So $S'$ is closed in the norm topology as well. A "high-level" explanation can be given: $S'$ is a von-neumann algebra, and every von-neumann algebra is a $C^*$ algebra- and so $S'$ ...

2

Let $C$ be a nonnegative matrix and $M(t)=M+tC$ (a primitive matrix). Then $\rho(M(T))$ is the maximal eigenvalue of $M(t)$. Let $spectrum(M(t))=(\lambda_i(t))$ with $\lambda_1(t)>|\lambda_2(t)|\geq |\lambda_3(t)|\geq\cdots$. Since $\lim_{t↓0}M(t)=M(0)$, $\lim_{t↓0}\rho(M(t))=\rho(M)$. Moreover , when $t$ decreases, $\rho(M(t))$ decreases too. EDIT:(with ...

1

Injectivity of the Gelfand transform is equivalent to the assertion that characters separate points. This can be verified by using that in the commutative case characters are precisely the pure states and so they have to separate points since the states do.

2

Hint as rhetorical question: If it's isometric, and $\hat{x} = 0$, what does that tell you about $\lVert x\rVert$?

1

Any matrix is similar to its Jordan Canonical form. The matrices implementing the "similarity" need not be unitary.

2

It seems that you are looking for interwining operators.

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