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The connection with the spectral measure $P$ is $$P(E) = \chi_{E}(A).$$ So, for example, $$P[a,b] = \chi_{[a,b]}(A), \;\; P(a,b) = \chi_{(a,b)}(A) \\ P[a,b] = P(a,b) + P\{a\}+P\{b\}$$ The spectral measure $P$ is regular in the strong topology, which gives you \begin{align} P[a,b]x & = \lim_{\...

0

Note that the norm of an isometry is 1, and thus the spectral radius is at most one, so the spectrum is contained in the closed unit disk $\overline{\mathbb{D}}$. We will show the set $A = \{ z \in \mathbb{D} \mid z \notin \sigma(T) \}$ is relatively open and closed in $\mathbb{D}$, and so by connectedness, is either $\varnothing$ or $\mathbb{D}$. If $\... 1 Another way of writting Jeb's solution is the following. Let$v_1,v_2,v_3$be an orthonormal basis of$\mathbb{R}^3$of eigenvectors of$A$associated to the eigenvalues$\lambda_1\leq\lambda_2\leq\lambda_3$. Let$V=\text{span}\{v_2,v_3\}$. If$X$is a 2-dimensional subspace of$\mathbb{R}^3$then$\dim(X\cap V)\geq 1$. Let$u\in X\cap V$with norm 1. ... 2 Seems like you're jumping the gun with this question... I'd like to take a few steps back then work up to whats going on here. If$A$is a hermitian matrix ( or symmetric for real matrices) then all the eigenvalues of$A$are real (why?) and the eigenvectors are orthogonal (also why?). This allows us to rewrite our basis in terms of these eigenvectors. i.e. ... 2 You seem to be getting twisted in a knot. The actual problem is, as I'm sure you realize, pretty simple: you are trying to minimize$x^\top A x$subject to the constraint$x^\top x = 1$. For that you can use any of several techniques, such as Lagrange multipliers. You seem to be doing fine for a while: you've written$A = P^\top D P$where the diagonal ... 1 Notice that$\|Lx\|^2 = \langle x, L^2 x\rangle$, and by assumption, this is$\le \|Ax\|^2=\langle x, A^2 x\rangle$. In other words,$L^2\le A^2$, and since the square root function is operator monotone, this implies that$L\le A$and thus also$L+t\le A+t. Now we obtain in the same way that $$\|(L+t)^{1/2}x\|^2 = \langle x, (L+t) x\rangle\le \|(A+t)^{1/2}... 1 Yes. The necessary and sufficient condition is that A is bounded. Indeed, the domain of AE_I is the whole Hilbert space, and the domain of E_IA is the domain \mathcal D(A) of A. 1 First, note that X_2 is an A-invariant subspace, so that A_2:X_2\to X_2. We can show formally that the adjoint of A_2 should be the restriction of A^* to X_2, which is again A^*. It then suffices to note that the image of the closed unit ball under A_2 is a closed subset of the image of the closed unit ball under A and is therefore ... 1 It follows from the definition of \sup. If$$ M=\sup\left( f(x)\ : x \in B\right), $$then there exists a sequence x_n\in B such that$$ M=\lim_{n\to \infty} f(x_n).$$Apply this observation with B=\text{unit sphere}, f(x)=|(Ax_n, x_n)| and M=\|A\|. 2 A simple example Let us look at a specific example in two dimensions (the main point would probably come forward better in three dimensions, but that is harder to draw, and is left as an exercise to you in the end). We study, as an example, the symmetric matrix$$ A= \begin{bmatrix} 4 & -2\\ -2 & 7 \end{bmatrix}. $$A standard calculation shows ... 0 There are multiple interpretations. The usual scenario is that you have a single heavy particle (like a pollen grain) in a sea of light particles (like water molecules). The heavy particle experiences collisions with the light particles. These light particles endow the heavy particle with force, and the heavy particle is also under the influence of a spatial ... 0 you can write it in a similar way as stated in wikipedia as M \ddot{X} = - \nabla U(X) - \gamma \dot{X} + \sqrt{2\gamma k_{B}T}R(t), where M are the masses of N particles and coordinates X=X(t). U(X) is the particle interaction potential, so - \nabla U(X) is the force calculated from the particle interaction potentials. \gamma is a small ... 2 As mickep observes in the comments, the spectra are not the same! The Dirichlet spectrum is$$ \lambda_{kl} = \bigg(\frac{\pi k}{a}\bigg)^2 + \bigg(\frac{\pi l}{b}\bigg)^2,\ k,l = 1,2,3,\ldots $$while the Neumann spectrum is$$ \nu_{kl} = \bigg(\frac{\pi k}{a}\bigg)^2 + \bigg(\frac{\pi l}{b}\bigg)^2,\ k,l = 0,1,2,3,\ldots The intuition is that the ... 4 Definition [C_0 Semigroup]: Let X be a Banach space, and let T : [0,\infty)\rightarrow\mathcal{B}(X) be a function into the bounded linear operators on X. Then T is a semigroup if T(0)=I and T(t)T(t')=T(t+t') for all t,t' \ge 0. T is a C_0 semigroup if \lim_{t\downarrow 0}T(t)x=x for all x\in X. Suppose X is a Banach space, and ... 1 If \lambda^n \ne 1, then you can directly verify that T-\lambda I is invertible by showing \begin{align} I&=(T-\lambda I)\left[\frac{1}{1-\lambda^n}(T^{n-1}+\lambda T^{n-2}+\cdots+\lambda^{n-2}T+\lambda^{n-1}I)\right] \\ &=\left[\frac{1}{1-\lambda^n}(T^{n-1}+\lambda T^{n-2}+\cdots+\lambda^{n-2}T+\lambda^{n-1}I)\right](T-\lambda I).... 0 The contour \Gamma doesn't matter, as long as all \lambda_j are inside. You have, with D=\text{Diag}\,(\lambda_1,\ldots,\lambda_n), \lambda I-A=\lambda V^* V-V^*DV=V^*(\lambda I-D)V=V^*\,\begin{bmatrix}\lambda-\lambda_1&0&\cdots&0\\ 0&\lambda-\lambda_2&\cdots&0\\ \vdots& &\ddots&\vdots \\0&0&\cdots&\... 0 Case 1. You know approximations of the eigenvalues ofA$; then you know also (the calculation is in$O(n^3)$) a convenient matrix$V$and finally$f(A)=V^*diag(f(\lambda_1),\cdots,f(\lambda_n))V$. Of course, you can also use the mickep's method. Yet, the calculation of$(\lambda I_n-A)^{-1}$becomes complicated when$n$grows. Case 2. You know only bounds ... 1 Your first inequality is usually known as the Kadison-Schwarz inequality. It only requires$2$-positivity. Claim.$A=\begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\geq0$if and only if$a^*a\leq b$. Proof. If$A\geq0$, then for any$\xi\in H, $$\langle (b-a^*a)\xi,\xi\rangle=\left\langle \begin{bmatrix}I &a\\ a^*&b\end{bmatrix}\,\begin{... 0 Let \lambda\notin \sigma(A) then A-\lambda 1 is invertible, denote it's inverse by C\in B(H_2). We have C(A-\lambda1)=(A-\lambda1)C=1. Therefore, D=U^{-1}CU\in B(H_1) is the inverse of (B-\lambda1) (why? write B=U^{-1}AU). Thus, \lambda \notin \sigma(B). Now, if \lambda\in \sigma_p(A) that means \exists v\ne 0 such that Av=\lambda v. ... 1 You're not going to find an explicit function for N. (For example, N is not even continuous.) To prove the asymptotic expansion$$N(\lambda) = c_n|\Omega|\lambda^{n/2} + o(\lambda^{n/2})for the Dirichlet counting function on a domain \Omega, the simplest way in bounded Euclidean domains --- and one that I assume you're to use, since you appear to ... 2 You can write unravel this definition by writing \begin{align} Af(x) & = \int_{0}^{x}K(x,y)f(y)dy+\int_{x}^{1}K(x,y)f(y)dy \\ & =\sinh(1-x)\int_{0}^{x}\sinh(y)f(y)dy+\sinh(x)\int_{x}^{1}\sinh(1-y)f(y)dy. \end{align} This is a typical kind of Green function expression. First note that (Af)(0) = 0,\;\;\; (Af)(1)=0. $$Then, \begin{... 0 If f(x,y,z)\in P^k(x,y,z), then it must be of the form$$f(x,y,z)=\sum_{k_x+k_y+k_z=k} f_{k_x,k_y,k_z}x^{k_x}y^{k_y}z^{k_z}$$where the sum runs over all (k_x,k_y,k_z)\in\mathbb{N}^3 whose total is k. Introducing i=k-k_z, we can rewrite this in order to pick out the z-dependence:$$f(x,y,z)=\sum_{k_x+k_y-i=0} f_{k_x,k_y,k-i}x^{k_x}y^{k_y}z^{k-i} =\... 0 It seems to me that they are considering the polynomialf$as a polynomial in$z$for a moment (looking at$x,y$and their powers as coefficients) so that this is indeed the expansion of a polynomial of degree$k$in$z$. Clearly the degree cannot exceed$k$in$z$for that would result in$f \notin P^k$, the lesser terms are always accompanied by powers$x^...

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$W_C = \{u + iv \in X_C \mid u, v \in W\}$ is invariant under $T_C$, so we may as well forget about the real Banach spaces and consider an operator $T$ on a complex Banach space $X$ with a finite-codimensional invariant subspace $W$, such that $\sigma(T) \subseteq \mathbb R$. The claim is that $\sigma(T|_W) \subseteq \mathbb R$. If $\lambda \notin \sigma(T)... 1 Denote by$N_{D_p}(\lambda)$the number of eigenvalues for the Dirichlet Laplacian on the rectangle$D_p$that are less than$\lambda$. Then, and this is what you seem to be missing, since the sequence$\mu_1\leq\mu_2\leq\cdots$enumerate all eigenvalues on the different rectangles, the quantity$M(\lambda)$that you define in the question is just the sum ... 0 I think you are right. If$Tf=\lambda f$, we have$ \lambda f = f + f(1) - f(0), $or $$\tag{1} (\lambda - 1) \, f= f(1)-f(0).$$ If$\lambda=1$, then any$f$with$f(1)=f(0)$satisfies the equation, so$1\in\sigma_p(T)$. In particular, as you mention, constant functions are eigenfunctions for the eigenvalue$1$. When$\lambda\ne1$, the equation$(1)$has ... 1 You are looking in the wrong volume. You should look in Vol I, Chapter VI, Section 4, Subsection 4 starting on page 436. The title of that section is "Asymptotic Distribution of Eigenvalues for an Arbitrary Domain". Good luck! 0 Let$\{\lambda_k\}$be the eigenvalues of$M_1\cup M_2$; similarly, let$\{\lambda_k^{(j)}\}$be the eigenvalues of$M_j$,$j=1,2$; in both cases arrange them in non-decreasing order. The domain monotonicity of Dirichlet eigenvalues implies that $$\lambda_k\ge \max(\lambda_k^{(1)}, \lambda_k^{(2)})\tag1$$ for every$k$. Equality holds in$(1)$when$M_1=M_2$.... 1 The right inequality implies $$\frac{\lambda_n}{n}\geq \frac{4\pi}{ab}.$$ We use that in the left inequality (for the square root term only), $$n\geq \frac{\lambda_n ab}{4\pi}-C\sqrt{\lambda_n} \geq \frac{\lambda_n ab}{4\pi}-C\sqrt{\frac{4\pi n}{ab}}.$$ Rearranging, $$\frac{\lambda_n}{n}\leq\frac{4\pi}{ab}+C\Bigl(\frac{4\pi}{ab}\Bigr)^{3/2}\frac{1}{\... 1 I think it's a bit confusing what they wrote, bordering on incorrect. I'm sure you can find a better reference for these things. The idea is to use the theorem that \lambda_1(M) \leq E(u) / \langle u, u \rangle = \| \nabla u \|^2 / \| u \|^2 for any sufficiently nice u (it does not have to be C^\infty ). Let \varphi_1 \in C^\infty(M_1) be ... 2 The operator consists of two unrelated parts: reflection in the plane (x_1,x_2), which contributes norm 1 and eigenvalues \pm 1, and scaled backward shift, which contributes the eigenvalue 0. There are no other eigenvalues because if \lambda \notin \{0,\pm 1\} then x_3,x_4,\dots must be nonzero and satisfy x_{n+1} = n\lambda x_n for all n. ... 1 The eigenvalues$$ \lambda_{ab} = \pi^2\bigg(\frac{1}{a^2} + \frac{1}{b^2}\bigg) $$for a,b \in \Bbb{N} are squared lengths of vectors in the lattice (\pi/a)\mathbb{N}\times(\pi/b)\mathbb{N}, where I let \Bbb{N} = \{1,2,3,\ldots\} be the positive integers. So the counting function N(\lambda) counts lattice points (in the upper quadrant) whose length ... 1 I cannot contribute references. But, regarding question 1, note that for \lambda\in\mathbb R, you have that T-\lambda I is invertible if and only if T_{\mathbb C}-\lambda I is invertible. Then$$ \sigma_\mathbb{R}(T)=\mathbb{R}\cap\sigma(T). $$To show the above invertibility, let us write T and T_{\mathbb C}. If ST=TS=I, then$$ S_{\mathbb C}... 1 If$T$is invertible, then so is$T'$: indeed, if$ST=TS=I$, then $$S'T'(x+iy)=STx+iSTy=x+iy.$$ Conversely, if$KT'=T'K=I$, then define an operator$S$on$X$by$Sx=K(x+i0)$. Then $$STx=K(Tx+iT0)=KT'(x+i0)=x+i0.$$ In both cases the induced inverse is bounded by the Open Mapping Theorem. It follows that if$\lambda\in\mathbb R$, then$T-\lambda I$is ... 1 The notation can also be interpreted as follows: order the eigenvalues$\lambda_1\le\lambda_2\le \ldots$. Consider$\lambda_N$for a large$N$, and now follow Willie's outline to approximately determine$N$: The number of smaller eigenvalues is one fourth the number of lattice points in the ellipse $$x^2/a^2 + y^2/b^2 = \lambda_N/\pi^2 ,$$ and this is ... 1 Your proof is fine. What you are missing to work at points other than zero is the following lemma: Lemma. Let$f:X\to\mathbb C$with$X$a compact subset of$\mathbb R$,$\varepsilon>0$and$R>0$. Then there exists$\delta=\delta(\varepsilon,R)>0$such that if$a,b\in A^+$, with$\|a\|+\|b\|<R$, with$\sigma(a)\cup\sigma(b)\subset X$, and such ... 0 Is the operator defined by (Tf)(x) = x f(x) one linear ? Taking (Tf)(x+y) = (x+y)f(x+y) = xf(x+y) + yf(x+y), that is different from (Tf)(x) + (Tf)(y). If I am correct, then is the example still valid ? 0 If$A$is symmetric, then$A$has an orthonormal basis of eigenvectors. The eigenvectors associated with different eigenvalues are automatically orthogonal. But you have to perform Gram-Schmidt on the eigenvectors with the same eigenvalue in order to get an orthonormal basis of the eigenspace. Once you have the orthonormal basis of eigenvectors, you put them ... 1 I suggest the book by Pazy on Semigroups of Linear Operators. It's one of the most elegant and readable books in Functional Analysis I've encountered. On page 14, there is a theorem: Theorem: A Linear operator is dissipative if and only if $$\|(\lambda I-A)x\| \ge \lambda \|x\|,\;\; x\in\mathcal{D}(A),\; \lambda >0.$$ His setting for ... 4 Consider the Laplacian matrix of the cycle on$n$vertices, given by \begin{equation*} L = \begin{bmatrix} 2 & - 1 & 0 & \cdots & 0 & -1\\ -1 & 2 & -1 & \cdots & 0 & 0\\ 0 & -1 & 2 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \... 1 The Laplacian is given by: $$L = \begin{pmatrix} 2 & -1 & 0 & \cdots & -1 \\ -1 & 2 & -1 & \cdots & 0 \\ 0 & -1 & 2 & \cdots & 0 \\ \vdots & & & \vdots & \\ -1 & 0 & 0 & \cdots & 2 \end{... 1 Yes. In fact, if A is m \times n and B is n \times m, m \ge n, the characteristic polynomial of AB is \lambda^{m-n} times the characteristic polynomial of BA. 0 That statement isn't quite true; what they mean (as reflected in the consistent use of weak inequalities in the preceding derivations) is the weak version of the statement: An additional constraint cannot decrease the value of the maximin. This is true because eliminating some of the options in the minimisation cannot lead to a lower minimum. For example, ... 2 Let's assume we're on a bounded domain \Omega\subset\mathbb{R}^N and we've fixed boundary conditions. Recall that the Laplacian \Delta induces an orthogonal decomposition of L^2(M) into eigenspaces E_k, where E_k is associated to the k^{th} eigenvalue \lambda_k, and \Delta acts by scaling on E_k. Write 0\leq \lambda_1\leq\lambda_2\leq\... 0 Hint: D ia the diagonal matrix that has as diagonal elements the eigenvalues of A and M is a matrix that has as columns the corresponding eigenvectors. Do you know how to find these? ( see here) 1 Any one-to-one quasinilpotent operator will do (quasinilpotent are the operators T with \sigma(T)=\{0\}); the Volterra operator from user3808066's answer is one example. Any finite set K\subset\mathbb C can be realized this way: if K=\{k_1,\ldots,k_n\} and T is the Volterra operator, then$$ \bigoplus_{j=1}^nT+k_jI $$has spectrum k_1,\ldots,k_n... 0 Since A\subseteq A^* (because A is symmetric), it's enough to prove that D(A^*)\subseteq D(A). Take y\in D(A^*)=D(A^*\pm\mathrm i). As (A^*\pm\mathrm i)y\in\mathcal R(A^*\pm\mathrm i)\subseteq \mathcal{H}=\mathcal R(A\pm \mathrm i), there exists x\in D(A\pm\mathrm i) such that$$(A^*\pm\mathrm i)y=(A\pm\mathrm i)x=(A^*\pm\mathrm i)x$$and thus ... 1 A counter example is the Volterra operator, it is a bounded linear operator between Hilbert spaces, with no eigenvalues while the spectrum is \{0\}. It is defined as$$ V:L^2(0,1)\to L^2(0,1), f\mapsto \left(t\mapsto \int_0^t f(x)\mathrm{d}x\right).$$1 Normally the singular values of a matrix$A$are defined as the (positive) square roots of the eigenvalues of$A^*A\$.

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