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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 $L$ ...


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 ...


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, $(-1)^nA^n=0\Rightarrow A^n=0$.


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 ...


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 ...


3

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


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 ...


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 $$ ...


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 ...


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 ...


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

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) ...


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] = ...


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 $\lambda=0$ ...


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 ...


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

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

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 ...


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 $T$ ...


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 $n$ independent 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 ...


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 $$ ...


1

Let's fix some notations. Denote the Haar measure on $\Bbb{T}^d$ by $\lambda$ instead of $m$. The measures suggested should probably be normalized by $$ d\mu_{v,N}(y)=\frac{1}{(2N+1)^d}\left\|\sum_{m\in\mathbb{Z}^d, \lvert m_i \rvert \leq N}\chi_{-m}(y) U(m)v\right\|^2 d\lambda(y).$$ The characters of $\Bbb{T}^d$ are the functions $$\chi_m: \Bbb{T}^d \to ...


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 ...


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

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 $(f_n,Af_n)$ ...



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