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1

The notation $\prod_{i\in I}S_i$ denotes a set of functions. By definition, $f\in\prod_{i\in I}S_i$ if (i) $f$ is a function with domain $I$ and (ii) $f(i)\in S_i$ for every $i\in I$. So $\phi\in\prod_{a\in A}sp(a)$. Because $\phi$ is a function with domain $A$ and $\phi(a)\in sp(a)$ for every $a\in A$. Come to think of it, that raises an obvious ...


1

The product $\prod_{i\in I}A_i$ of an indexed family of sets is, by definition, the set of all functions $f$ whose domain is the index set $I$ and which satisfy, for each index $i\in I$, the requirement that $f(i)\in A_i$. So the product in your question is the set of functions that assign, to each $a$ in your algebra, an element of its spectrum. The ...


1

Yes, eigenvectors corresponding to different eigenvalues are necessarily independent so, in an inner-product space, orthogonal. If an n by n matrix is symmetric then there are n [b]independent[/b] eigenvectors. Therefore, there exist eigenvectors, each of which spans a one dimensional vector space so the eigenvectors form an orthogonal basis for the entire ...


1

Hint: The null space of $A$ is always orthogonal to the range of $A^T$, for any matrix $A$. Now, exploit that $A$ is symmetric. Solution: Let $x$ be in the null space of $A$ and $y=A^T u$ be in the range of $A^T$. Then, we have $$ x^T y = x^T A^T u = (Ax)^T u = 0^T u = 0. $$ Since $x$ and $y$ were arbitrary, it follows that the null space of $A$ and the ...


0

Given the Hilbert space $\ell^2(\mathbb{N}_0)$. Consider the left shift: $$L_0:\mathcal{D}(L_0)\subseteq\ell^2(\mathbb{N}_0)\to\ell^2(\mathbb{N}_0):\quad L_0:=L$$ For nondense domain: $$\mathcal{D}(L_0):=\ell^2(\mathbb{N}):=\{\varphi\in\ell^2(\mathbb{N}_0):x_0=0\}$$ But it has an inverse: $$0\not\in\sigma(L_0):\quad RL_0=1_0\quad L_0R=1$$ Concluding ...


1

To prove what you want, it suffices to prove that if $|\lambda| \neq 1$, then $(A-\lambda)$ is bijective. I was able to prove injectivity - surjectivity seems hard to just brute force, so I will post this and let someone else come up with an elegant answer :) Injectivity: Suppose $Ax = \lambda x$, and $\lambda, x\neq 0$, then $$ \frac{1}{3}x_1 = \lambda x_0 ...


0

As, $R(A(G),x)=\frac{<x,A(G)x>}{<x,x>}$, $R(A^2(G),x)=\frac{<x,A^2(G)x>}{<x,x>}=\frac{<A(G)x,A(G)x>}{<x,x>}=\frac{\|A(G)x\|_2^2}{\|x\|_2^2}=\|A(G)\|_2^2$ for $x=v_{max}$. Also, $\rho(A^2)=\rho(A)^2=\|A\|^2_2$. So, $R(A^2(G),x)=\rho(A^2)$ for $x=v_{max}$. If $\mu$ is the spectral radius of $A^2$, then $\sqrt{\mu}$ is ...


0

Your two expressions for the spectral radius are not identical, but their values on (finite) symmetric matrices are equal. One comment. For an infinite graph of bounded degree, it's perfectly reasonable to view its adjacency matrix as an operator on $\ell^2$, but it's not clear to me that this is even an established convention, let alone a rule.


1

Yes. The functional calculus preserves approximation by polynomials. So, from $$ T^n=\begin{bmatrix}A^n&0\\0&B^n\end{bmatrix}, $$ you get that $$ p(T)=\begin{bmatrix}p(A)&0\\0&p(B)\end{bmatrix} $$ for any polynomial $p$. Now using a sequence of polynomials that converges uniformly to $f$, you get that $$ ...


0

Under the assumption that this is a finite dimensional problem and $A(\lambda)$ is holomorphic, it can be characterized with the left and right eigenvectors corresponding to the associated nonlinear eigenvalue problem. Let $T(\lambda):=A(\lambda)-\lambda I$ such that we can formula (14) in http://dx.doi.org/10.1016/j.laa.2011.03.030 with $f(\lambda)=2i\pi$. ...


3

Left/right shift operators are the standard examples, but I personally think that the multiplication operator is the easiest way to see that there may be something else in the spectrum besides eigenvalues. Consider a multiplication operator $A_c$ on $\ell^\infty$ ($c\in\ell^\infty$) $$ (A_c x)_n=c_n x_n. $$ The inverse if exists is clearly a multiplication ...


7

If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen. If it fails to be injective, there's $x \ne y$ such that $(T - \lambda I)(x) = (T - \lambda I)(y)$. So $(T - \lambda I)(x - y) ...


6

For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators $L, R:\ell^2 \to \ell^2$, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The map $L$ ...


3

$T-\lambda I$ being non-invertible does not imply there is a non-zero $x$ with $(T-\lambda I)x=0$. That is true when $H$ is finite-dimensional, but not necessarily when $H$ is infinite-dimensional. The classic counterexample is the right-shift operator $R:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$. Take a look at the Wikipedia article on the notion of ...


2

This is too large for a comment, but I thought it might help to shed some light on the underlying reason that the matrix should be expected to be upper-triangular, and indeed nilpotent (so that you would sooner suspect an error in formatting, or perhaps in the choice of basis order). Your transformation (call it $L[f]$) can be broken down into $L[f] = ...


4

Your matrix is a strictly upper triangular matrix. Upper triangular matrices $n\times n$ are all nilpotent since their characteristic polynomials, $\det (A-XI_n)$, is equal to $(-1)^nX^n$. According to Cayley-Hamilton's Theorem, $(-1)^nA^n=0\Rightarrow A^n=0$.


0

Hint: $E\{z:|z|>\epsilon\} = N f(N)$ where $f(z) = \ldots$


0

Firstly I just wanted to clarify what you meant by "twice absolutely continuous", my assumption is that means "twice absolutely continuously differentiable". If that is the case, then the fact that the question is posed on $\Bbb R$ is important, of course $f\in L^2(\Bbb R)$ is a strong assumption. Assume $Hf=-f''+x^2f\in L^2(\Bbb R)$, this tells us there ...


3

I'm assuming that by $\sigma(\Sigma P)\leq1$ you mean that $\|\Sigma P\|\leq1$. Suppose that $\|P+Q\|\leq1$. So $0\leq P+Q\leq 1$. Then $(P+Q)^2\leq P+Q$ (just conjugate with $(P+Q)^{1/2}$). That is, $$ P+Q+QP+PQ\leq P+Q, $$ or $QP+PQ\leq0$. If we conjugate this inequality with $Q$, we get $QPQ+QPQ\leq0$. But $QPQ\geq0$, so $QPQ=0$. Then $$ ...


-1

Let's enumerate the conditions: (1) $P\perp P'$; (2) $0=PP'=P'P$; (3) $\Sigma P$ is a projection. We can assume that $A\subset B(H)$ for some Hilbert space $H$ (this basically follows from the GNS construction); also, remember that for positive operators $T\in B(H)$ we have $\langle T\xi,\xi\rangle\geq 0$ for all $\xi\in H$. For (1)$\Rightarrow$(2), assume ...


2

(note that you are using $m$ for two different things; I will just ignore the Haar measure as I don't think it is needed) Since the characters $\chi_m$ evaluate at single points $m\in\mathbb Z^d\cap\mathbb T^d$ (which are the $d$-tuples consisting of $1$ and $-1$) , you have $$ \int_{\mathbb T^d}\chi_m\,d\mu_v=\mu_v(\{m\}). $$ And you want this to equal ...


2

This is linked with the spectral theorem. There are multiple formulations of this powerful theorem, one is the following : Spectral theorem (multiplication operator form) : Let $\mathcal{H}$ be a separable Hilbert space and $A\in\mathcal{L}\left(\mathcal{H}\right)$, that is, $A$ is a linear bounded mapping $\mathcal{H}\to\mathcal{H}$. Then there is ...


1

You essentially have proven the result. I will restate and embellish upon your calculations to show that $Z$ satisfies the definition of a Vector Measure. Let $\mathcal{F}$ be the Borel field for the interval $(-\pi, \pi]$. Let $X$ be the set of complex valued $L^2$, mean-zero, random variables on the probability space $(\Omega, \mathcal{U}, P)$. We ...


0

Let $B = C^*(x) \subseteq A$ be the $C^*$-subalgebra generated by $x$. Note that $B$ is commutative, since $x$ is normal. Furthermore, recall that $\text{spect}_B(x) \cup \{0\} = \text{spect}_A(x) \cup \{0\}$ holds whenever $B$ is a $C^*$-subalgebra of $A$, so we have $\text{spect}_B(x) \subseteq \mathbb{R}$ as well. The Gelfand representation now gives an ...


1

Eigenvalue problems were first seriously considered in separation of variables problems arising out of Partial Differential Equations. The separation parameter was the eignevalue, and general eigenfunction analysis came out of Fourier's method. The matrix methods came out of this method, which is the opposite direction of abstraction that one would naturally ...


0

Let me leave out technicalities.. Closed Operators Given a Banach space $E$. Consider a closed operator: $$T:\mathcal{D}(T)\subseteq E\to E:\quad T=\overline{T}$$ Denote its resolvent: $$R(\lambda):=(\lambda-T)^{-1}\in\mathcal{B}(E)$$ Then one can construct: $$\eta\in\mathcal{H}(\mathbb{C}):\quad\eta(T):=\oint\eta(\lambda)R(\lambda)\mathrm{d}\lambda$$ ...


1

Without loss of generality we can assume that $z=0$ (otherwise apply the proof to the operator $A-z$ instead). The sequence $\{f_n\}$ is Weyl, i.e. $\|f_n\|=1$ and $\|Af_n\|\to 0$ when $n\to+\infty$. The vector $(f_n,Af_n)$ belongs to the graph $\Gamma(A)$. Since $A=\bar A_0$ we know that $\Gamma(A)=\bar\Gamma(A_0)$, so we can approximate $(f_n,Af_n)$ ...


1

This is immediate from the definitions, plus the ordinary scalar-valued Cauchy's Integral Formula. Suppose $\Lambda\in X^*$, and let $g=\Lambda\circ f$. So $g$ is analytic (by definition or not, depending on which definition of "analytic" you took). CIF shows that $$\Lambda(n(\gamma;\lambda)f(\lambda))=n(\gamma;\lambda)g(\lambda)=\frac1{2\pi ...


1

The full quote is: If $\partial\mathbb{D}=\{z\in\mathbb{C}:|z|=1\}$, let $B=$ the uniform closure of the polynomials in $C(\partial\mathbb{D})$. This means: consider the set of all continuous functions on $\partial\mathbb{D}$, equipped with the uniform norm $\|f\|=\sup_{\partial \mathbb{D}}|f|$. This space is denoted by $C(\partial\mathbb{D})$. ...


1

The definition of core is given in the same Kato's book that you cited at page 317 for a closed sectorial form and at page 166 for a closed operator. I refer to this last (but the first is a particular case). If $T: X \rightarrow Y$ is a closed operator and $\mathbf{D}(T)$ its domain, than a subspace $\mathbf{D}$ of $\mathbf{D}(T)$ is a core of $T$ ...


2

Assume $T$ has empty spectrum. Then $T$ is invertible, $T^{-1}$ is a bounded selfadjoint operator and, for $\lambda \ne 0$, $$ (T^{-1}-\lambda I) =(I-\lambda T)T^{-1}=\lambda(\frac{1}{\lambda}I-T)T^{-1} $$ has bounded inverse $$ \frac{1}{\lambda}T\left(\frac{1}{\lambda}I-T\right)^{-1} $$ So $\sigma(T^{-1})=\{0\}$ because only $\lambda=0$ ...


2

The separability of $H$ has little influence, the only difference between the separable and the non-separable case is that in the non-separable case, $V^\perp$ has an uncountable Hilbert basis (of the same cardinality as any Hilbert basis of $H$, naturally). Some authors prefer to only treat countable Hilbert bases and therefore restrict to the separable ...


3

Note that $\|B^n\| \le 1/n!$. From this you can show that the spectral radius of $B$ is $0$.


-1

For a symmetric matrix you want to maximise $||Ax||$ for $||x||=1$, which is the same as maximising $||Ax||^2 = xA^TAx$ Since we know that if $A$ is symmetric, it is diagnolaisable, so we write $$A = Q^T\Lambda Q$$, with $Q^{-1}=Q^T$then $$xA^TAx= (Qx)^T\Lambda^2(Qx)$$ where $Q$ is orthonormal so $||Qx||=1$ It is straightforward to see what you have ...


3

For any square $A$, $\rho(A)\leq\|A\|_2$ with the equality (not necessarily) if $A$ is normal. Besides the general inequality, $\rho(A)$ and $\|A\|_2$ can be completely unrelated. Consider, e.g., $$ A_\alpha:=\pmatrix{0&\alpha\\0&0} $$ with $\rho(A_\alpha)=0$ but $\|A_\alpha\|_2=|\alpha|$. All the eigenvalues are zero but the 2-norm can be an ...


0

Whether you use $L^{2}(\mathbb{T})$ or $L^{2}[0,2\pi]$ is a matter of preference. I prefer the classical Fourier setting $$ U : L^{2}[0,2\pi]\rightarrow \ell^{2}(\mathbb{Z}) $$ defined by $$ Uf = \left\{ \int_{0}^{2\pi}f(t)\frac{e^{-int}}{\sqrt{2\pi}}dt\right\}_{n=--\infty}^{\infty} $$ The inverse of $U$ is $$ ...


2

The answer to this question is yes. If $A$ is real and orthogonally diagonalizable, then $A = UDU^T$ for some orthogonal matrix $U$ and real diagonal matrix $D$. We find that $$ A^T = (UDU^T)^T = UDU^T = A $$ so that $A$ is symmetric. Similarly, if $A$ is complex and unitarily diagonalizable, then $A = UDU^*$ for some unitary matrix $U$ and (complex) ...


0

A very good reference is Schmüdgen: "Unbounded Self-adjoint Operators on Hilbert Space", if you want to focus on Hilbert spaces. http://www.springer.com/en/book/9789400747524


0

References Operator theoretic treatments: William Arveson (Spetral Measures, Pettis Integral)* M. S. Birman, M. Z. Solomjak (Spectral Measures, Pettis Integral) J. Blank, P. Exner, M. Havlíček (Spectral Measures, Pettis Integral) John B. Conway (Spectral Measures, Pettis Integral) Alexander Frei (Spectral Measures, Pettis Integral) Emmanuel Kowalski ...


5

There are two scales at play here. Let $M$ be a closed Riemannian manifold and let $\Delta$ be the Laplace-Beltrami operator on it. There is an orthonormal eigenbasis $\{\phi_k\}_{k=1}^\infty$ of $L^2(M)$ and an increasing sequence of eigenvalues $\lambda_k\geq0$ so that $\Delta\phi_k=-\lambda_k\phi_k$. The piece of an article you linked to discusses the ...


1

Note that all of the work is in Exercise 8, which says that if $0\leq x\leq y$ in a C*-algebra and $x$ is invertible, then $y^{-1}\leq x^{-1}$. You will apply this with $x=1+\alpha a$ and $y=1+\alpha b$, and the rest follows easily. For a solution to Exercise 8, see Inversion in a unital C* algebra or positive invertible operators or Inverse of ...


2

Your reasoning is correct, for a continuous (bounded) everywhere defined operator $A$ on a Banach (in particular on a Hilbert) space, the denseness of the range of $\lambda\mathbb{I} - A$ together with the boundedness of the inverse already implies the surjectivity of $\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting ...



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