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0

Since $\rho(A)<1$, there is an induced norm $N(.)$ s.t. $N(A)<1$. Let $\mu=N(A)$; thus $N(A^k)\leq \mu^k$. Since the norms are equivalent, there is a fixed $C$ s.t. $||.||\leq CN(.)$. Finally $||A^k||\leq CN(A^k)\leq C\mu^k$.


1

The Spectral Mapping Theorem allows you to more easily compute the spectrum of some operators. If you know that you can write an operator $A$ as $A=f(a)$ for $f\in hol(a)$ and $a\in\mathcal{A}$, where you already know the spectrum of $a$, you can compute the spectrum of $A$, since $\sigma(A)=f(\sigma(a))$.


1

I accidentally saw this paper today on the internet concerning the case $n=4$: Jeremy Levick, Rajesh Pereira and David W. Kribs (2015), The four-dimensional Perfect-Mirsky Conjecture, Journal: Proc. Amer. Math. Soc., 143: 1951-1956. Abstract: We verify the Perfect-Mirsky Conjecture on the structure of the set of eigenvalues for all $n\times n$ ...


0

This is not true !! to convince you, take $T$ the unilateral shift so : $$ \sigma(T)=\bar{\mathbb{D}}\\ \sigma_p(T)=\mathbb{D}\\ \sigma_c(T)=\mathbb{T} $$ proof page 6-7.


1

Even if it might be to late for answering. The following work provides a characterization of the convergence of the spectrum with respect to the Hausdorff metric. The continuity is satisfied whenever the norms of all polynomials (up to degree 2) of the operator behave continuous. One could do much better: In particular quantitative estimates are provided. ...


3

No. Counterexample: $$ \pmatrix{0&1\\0&0} $$ is not similar to any symmetric matrix. On the other hand, every diagonalizable matrix with real eigenvalues is similar to a symmetric matrix.


3

First, what do you mean by "the eigenvalues of $A$ plus the eigenvalues of $B$"? What is true (under the condition $AB = BA$) is that each eigenvalue of $A+B$ is the sum of an eigenvalue of $A$ and an eigenvalue of $B$. It is not true in general that each sum of an eigenvalue of $A$ and an eigenvalue of $B$ is an eigenvalue of $A+B$. If $A$ and $B$ ...


1

The linear subspace of symmetric matrices is actually of dimension $n(n+1)/2$. The linear subspace of diagoanl matrices is of dimension $n$. The similar transformation (spectral decomposition) maps $n(n+1)/2$ space to $n$ space. You have confused 2 different spaces with a single space.


0

Lemma: Let $A$ be a unital Banach algebra and $\{a_n\} \subset A$ such that $a_n \to a\in A$. Suppose $\lambda_n \in \sigma(a_n)$ are such that $\lambda_n \to \lambda$ in $\mathbb{C}$, then $\lambda \in \sigma(a)$ Proof: Suppose $\lambda \notin \sigma(a)$, then $(a-\lambda 1) \in GL(A)$, which is open. So $\exists \epsilon > 0$ such that $$ \|y - ...


2

You have correctly stated that for any symmetric $A$, there exist rank $1$ matrices $v_iv_i^T$ such that $$ A = \sum \lambda _i v_i v_i^T $$ However, it is impossible to select a fixed set $\{v_1v_1^T,\dots,v_nv_n^T\}$ such that every $A$, there exists a choice of $\lambda_i$ such that $A$ has the above form. That is, there is no basis for the set of ...


0

The dimension space of $n$-symmetric matrices is $\frac{n(n+1)}{2}$. By the spectral theorem every symmetric matrix, with real coefficient, is diagonalizable; in other words every symmetric matrix is similar to a diagonal matrix. But this not means that the dimension of $n$-symmetric matrices in $n$. The set of diagonal matrices is a vector subspace of ...


1

If $\rho$ is a bounded measurable function on $[a,b]$, then $E(t)=\int_{a}^{t}\rho(u)du$ is a function of bounded variation and, for any continuous function $g$, $$ \int_{a}^{b}g(t)dE(t) = \int_{a}^{b}g(t)\frac{dE}{dt}dt = \int_{a}^{b}g(t)\rho(t)dt. $$ In your case, $$ \|E(t)x\|^{2}=\int_{-\infty}^{t}|x(u)|^{2}du $$ The above easily ...


2

Let $E=\{1,1/2,1/3,\cdots\}$ and let $\mu_{a}$ be the finite atomic meausre on $[0,1]$ that is supported on $E$ with $\mu\{1/n\}=1/n^{2}$. Let $\mu = \mu_{a}+m$ where $m$ is Lebesgue measure on $[0,1]$. Let $X=L^{2}_{\mu}[0,1]$. Let $\chi$ be the characteristic function of $E$, and define operators $A, B \in \mathcal{L}(X)$ by $$ Af = ...


1

Okay, so you have $G$ a $srg(n,r,\lambda,\mu)$, $D_{1}$ is the neighborhood graph of a vertex $x_{1}$ (I'm assuming $x_{1}$ is NOT in $D_{1}$); then you take $x_{2}$ in $D_{1}$, and create $D_{2}$ which is the intersection of the two neighborhoods. So $D_{1}$ will contain $r$ vertices, and since each two adjacent vertices share $\lambda$ common neighbors, ...


4

Any normal matrix is diagonalizable (with a unitary matrix, actually). A nilpotent matrix has only $0$ as its eigenvalue, so a normal nilpotent matrix is similar to the zero matrix. The zero matrix is obviously only similar to itself (like any scalar matrix).


2

Not a complete answer, but might be an useful approach for the problem: Use the known spectral decomposition: $$ U = \sum_{a} \lambda_a P_a, \qquad V= \sum_{b}\eta_b \pi_b $$ Where $P_a,\pi_b$ are complete systems of hermitian projectors so that $$ \sum_a P_a = \sum_b \pi_b ={\bf 1}, \qquad P_a P_b = 0, \quad\mbox{and} \quad \pi_a \pi_b =0\quad ...


1

We prove that given $U, V$ unitary matrices, $U^mV^n$ is unitary. For $$ \overline{U^mV^n}^TU^mV^n=\overline{V^n}^T\overline{U^m}^TU^mV^n=I $$ Also given $U_1,\dots, U_k$ unitary matrices, $U_1\cdot U_2\cdots U_k$ is unitary. For $$ \overline{U_1\cdot U_2\cdots U_k}^TU_1\cdot U_2\cdots U_k=\overline{U_k}^T \cdots ...


2

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 ...



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