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1

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.


2

Assume $T=T^{\star}\in\mathcal{L}(H)$, where $H$ is a complex Hilbert Space. Implication 1: Show $(Tx,x) \ge 0$ for all $x \in H$ implies $\sigma(T)\subseteq [0,\infty)$. To do this, assume that $(Tx,x) \ge 0$ for all $x \in H$, and let $\lambda < 0$. Then $$ 0 \le -\lambda(x,x) \le ((T-\lambda I)x,x) $$ implies $$ |\lambda|\|x\|^{2} \le ...


0

Yes, it is true! This can be seen most easily starting from its diagonalization exploiting the unitarity: $$\quad A^*A=(U^{-1}D^*U)(U^{-1}DU)=U^{-1}D^*DU=U^{-1}DD^*U=(U^{-1}DU)(U^{-1}D^*U)=AA^*$$


1

The sophisticated machinery that is Google has this one for us. For a more general answer see the first generated answer, for your specific question see the second generated answer. http://bit.ly/1oKYXMW


2

Maybe Section 5.6 of my ode book is what you are looking for. There is also a new more comprehensive book Periodic Differential Operators by Brown, Eastham, and Schmidt.


0

This is the positive part of the matrix, commonly denoted $X_+$. More generally, for any bounded self-adjoint operator $X$ on a Hilbert space, there are unique $X_-,X_+\geq 0$ such that $X=X_+-X_-$ and $X_+X_-=X_-X_+$. See also the answer to this question


1

This is just an intuition for the much simpler case of finite dimension. Let $A$ be any matrix and $v$ an associated eigenvector, i.e. $Av = \lambda v$. Then $(A-I)v = Av-v = (\lambda-1)v$, and thus $\sigma(A-I)= \sigma(A)-1$. And now an attempt for infinite dimension, suppose that $(A-i\lambda I)$ is not invertible for every $\lambda \in (-\infty, 0]$ but ...


3

Yes, there are quite a lot easy examples, e.g. consider the momentum operator $-i\frac{d}{dx}$ on $L^2(-a,a)$ where $a$ is some finite number and the Sobolev Space $H^1$ as domain. In this case, $e^{ikx}$ is an eigenfunction for every $k$ and therefore the spectrum is whole $\mathbb{C}$. However, if you choose $H_0^1$ as domain, this operator will be ...


1

Yes, I have seen cases where an operator on $L^1(R^3,d^3x)$ has a much larger spectrum than on $L^2(R^3,d^3x)$. If my memory serves me well this was the case for a Fokker-Planck operator, the $L^2$ spectrum was real but the $L^1$ spectrum contained contributions outside the real axis.


0

I think this problem is Theorem 2.4(iii) of the book written by, L. N. Trefethen, M. Embree, spectra and pseudospectra, 2005.


0

Hint: Since $T$ is finite rank, you can pick an orthonormal basis $g_1,\ldots,g_n$ of the range of $T$, $$ Tv=\sum_{j=1}^n \langle Tv,g_j\rangle g_j. $$ Show that $\mbox{ker } T\supset \mbox{span }\{T^*g_1,\ldots,T^*g_n\}^\perp$. Pick and orthonormal basis of $\mbox{span }\{g_1,\ldots,g_n,T^*g_1,\ldots,T^*g_n\}$, say $e_1,\ldots,e_m$. Show that $m\leq 2n$. ...


1

For $\lambda \in \mathbb{R}$ and $k > 0$, $$ \frac{k^{2}}{2}\left[\frac{1}{\lambda+ik}+\frac{1}{\lambda-ik}\right] =\lambda \frac{k^{2}}{\lambda^{2}+k^{2}} $$ So the above function converges to $\lambda$ as $k\rightarrow\infty$. You have $$ e^{iT_{k}}x = ...


0

Let $\mathcal{D}(L)$ be as stated in the problem, meaning that $f \in \mathcal{D}(L)$ iff $f$ is continuously differentiable on $(-1,1)$, $f'$ is absolutely continuous on $(-1,1)$, $f$ and $Lf =-((1-x^{2})f')'$ are in $L^{2}(-1,1)$. Aymptotics: The two classical solutions of $Lf=0$, both of which are in $L^{2}(-1,1)$: $$ ...


0

An upper bound is $\deg(i) (\deg_\max - 1)^{\ell-1}$. This is best possible in the sense that it is exact in the case of a tree where $i$ is the root, all leaves are at distance $\ell$ from the root, and all nodes other than the root and the leaves have degree $\deg_\max$.


1

If $A$ is selfadjoint with $\sigma(A)=\{\lambda\}$, then we may assume that $\lambda = 0$ by replacing $A$ with $A-\lambda I$ if necessary. However, the norm and spectral radius for a selfadjoint operator are the same because $\|A\|^{2}=\|AA^{\star}\|=\|A^{2}\|$. Therefore, $\|A\|=0$. The counter-example is any niplotent operator $N$ of order $n > 1$.


3

The matrices having at least one eigenvalue with geometric multiplicity smaller than its algebraic multiplicity are exactly the non-diagonalizable ones. The comments and answer illustrate this: non-zero nilpotent matrices are examples of non-diagonalizable matrices (but there are many other examples !).


1

As a counterexample consider $$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$over $\mathbb C^2$. It's spectrum is $\{0\}$.


4

Consider $$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ As $\chi_A(t) = t^2$, $0$ has algebraic multiplicity $2$, but geometric multiplicity $1$.


1

Since $T$ is self adjoint, the residual spectrum is empty, so $\lambda\notin \sigma_r(T)$. This means either $\lambda \in \sigma_p(T)$ (point spectrum) or $\lambda \in \sigma_c(T)$ (continuous spectrum). Note: Since $T$ is self adjoint, $<Tu_n,v>=<u_n,Tv>$ and $\lambda$ will be real, so $<\lambda u_n,v>=< u_n,\overline \lambda ...


3

Giuseppe Negro already mentioned it, but I wanted to go into a bit more detail with his first point, the Sturm-Liouville methods. A strategy which is used very often in practice is: Calculate the Greens function or respectively the fundamental solution for $L-\lambda \mathbf{1}$ (you can actually allow $G$ to be a distribution), this induces a resolvent ...


3

There is a whole theory dedicated to this, so the short answer is: there are a lot. I can think of three: Solving analytically the resolvent differential equation (i.e. the equation $Lu - \lambda u = v$). This tends to work when the geometrical domain is one-dimensional (Sturm-Liouville's theory) or when it is very symmetrical (separation of variables, ...


3

Lemma: Let $B : \mathcal{D}(B)\subseteq X\rightarrow X$ be a symmetric linear operator on a Hilbert space $X$ for which $(Bx,x) \ge 0$. If $I+B$ is surjective, then $B$ is closed, densely-defined and selfadjoint. Proof: Let $B$ as stated. Suppose that $B+I$ is surjective. To show that $B$ must be densely-defined, suppose that $y \perp \mathcal{D}(B)$ ...


3

If you use unbounded operators, then $$ H = \int_{0}^{\infty}\lambda dE(\lambda). $$ The spectrum theorem for unbounded selfadjoint operators has the excellent provision that $$ \mathcal{D}(H) = \left\{ x \in X : \int_{0}^{\infty}\lambda^{2}d\|E(\lambda)x\|^{2} < \infty\right\}. $$ You have $H=A^{2}$, where ...


1

The domain of $A^{\star}$ is identical to the domain of $A$ in this case. However, when you compose the two, then you get $\mathcal{D}(A^{\star}A)$ as $$ \mathcal{D}(A^{\star}A)= \{ f \in L^{2} : f\in \mathcal{D}(A) \mbox{ and } Af \in \mathcal{D}(A^{\star}) \}. $$ In particular, $f$ is twice absolutely continuous with $f(0)=f(1)$ and $f'(0)=f'(1)$ . ...


1

If $H:X \to X$ is a bounded operator, there is indeed always such a decomposition available. We may construct one such decomposition as follows: By the spectral theorem, we have $H = UTU^*$ Where the operator $T$ is given by $$ [T(\phi)](x) = f(x)\phi(x) $$ For some $f:\Bbb R \to \sigma(H) \subset [0,\infty)$. We can simply define $\sqrt T$ by $$ [\sqrt ...


3

This is a continuation of what I posted earlier. What follows are two examples of how the theory is applied to the trigonometric functions where $V=0$. The equation is in the limit point case on $[0,\infty)$ because $e^{i\sqrt{\lambda}x}\in L^{2}[0,\infty)$ while $e^{-i\sqrt{\lambda}x}$ is not, where $\sqrt{\lambda}$ is the branch whose branch cut is along ...



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