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So we want to show that if $A-\lambda I$ is not invertible, then $\lambda\geq0$. There are three ways in which $A_\lambda I$ may fail to be invertible: $\ker(A-\lambda I)\ne\{0\}$. In this case $\lambda$ is an eigenvalue. So there exists a unit vetor $v\in H$ with $Av=\lambda v$. Then $$\lambda=\langle\lambda v,v\rangle=\langle Av,v\rangle\geq0.$$ ...

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It's an easy thing to show that $\sigma(T) \subseteq \{ \lambda : |\lambda| \le \|T\| \}$ because the following inverse series converges in operator norm for $\|T\| < |\lambda|$ \begin{align} (T-\lambda I)^{-1} &= \frac{1}{\lambda}(\frac{1}{\lambda}T-I)^{-1} \\ & = -\frac{1}{\lambda}(I-\frac{1}{\lambda}T)^{-1} \\ ...

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Suppose $T$ is a bounded operator on a Banach space $X$. $\lambda\in\rho(T)$ iff $T-\lambda I$ is a linear bijection. In that case, the inverse $(T-\lambda I)^{-1}$ is automatically continuous by the closed graph theorem. There are three basic things that can stand in the way of $T-\lambda I$ being invertible. $T-\lambda I$ is not injective. Equivalently, ...

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For $T$ to have $\lambda$ as an eigenvalue, $T-\lambda I$ must be non-injective. For $\lambda$ to be in the spectrum of $T$, it must only be non-invertible. These are equivalent when $T$ is an operator on a finite-dimensional space, but not in general! For example, let $T$ be the shift operator $(x_0,x_1,\dots) \mapsto (0,x_0,x_1,\dots)$ on your favorite ...

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Some important tricks/theorems: The spectrum is non-empty for Banach algebras (over $\mathbb{C}$) The spectral radius formula $$r(a) = \lim \|a^n\|^{1/n}$$ tells you that if $a$ is nilpotent, then $\sigma(a) = \{0\}$ If $A$ is commutative, then $$\sigma(a) = \{\tau(a) : \tau \in \Omega(A)\}$$ where $\Omega(A)$ denotes the set of non-zero multiplicative ...

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Actually convolution with $k$ defines a bounded operator on $L^2(G)$ if and only if $\hat k$ is bounded. In this case the spectrum is exactly the essential range of $\hat k$. In fact if $m$ is any bounded measurable function on $\hat G$ then $m$ defines a bounded operator on $L^2(G)$ by $$\widehat{Tf}=m\hat f,$$whether or not there exists $k$ with $m=\hat ... 0 Here is an alternative proof of the fact that$r(T)=0$( I add it because it does not seem to appear in the related posts and I find it quite nice). First show by induction that$|A^n x(t)|\leq \frac{t^n}{n!}\|x\|_\infty$for all$t\in[0,1]$: Basis$n=0$:$|x(t)|\leq \|x\|_\infty$for all$t\in[0,1]$by definition of the supremum norm. Inductive step: For ... 2 We will show that$\sigma(A) = \{0\}$, and that$0$belongs to the residual spectrum. As you have shown,$0$is not an eigenvalue, and as$\operatorname{im} A \subseteq \{x \in C[0,1]: x(0) = 0\}$,$A$does not have dense image. Hence$0 \in \sigma_r(A)$. To see that$A - \lambda$is invertible for$\lambda \ne 0$, let$y \in C[0,1]$be given. We have to ... 0 The operator$T$satisfies $$\|Tx\| \le \frac{1}{2}\|x\|, \\ \|T^2x\| \le \frac{1}{2\cdot 3}\|x\|, \\ \|T^3x\| \le \frac{1}{2\cdot 3\cdot 4}\|x\|.$$ In general$\|T^{n}\| \le 1/(n+1)!$. If$Tx=\lambda x$and$x \ne 0$, then $$|\lambda|^n\|x\|=\|T^nx\| \le \frac{1}{(n+1)!}\|x\|$$ If ... 1 The operator$L$maps$\{ a_n \}_{n=-\infty}^{\infty}$to$\{ a_{n+1} \}_{n=-\infty}^{\infty}$. If$e_n$is the standard basis element defined as a sequence of all$0$'s except for a$1$in the$n$-th place, then$Le_n = e_{n-1}$, which is why$L$would be considered to be a left shift. And this follows from the definition$L\{ a_n \} = \{ a_{n+1} \}$: the ... 2 More elementary way: a. If$\lvert\,\lambda\rvert>1$, then$\lambda\not\in\sigma(\mathcal L)$. This is clear as the series$(\lambda-\mathcal L)^{-1}=\lambda^{-1}\sum_{n=0}^\infty (\lambda^{-1}\mathcal L)^n$converges. b. If$\lvert\,\lambda\rvert<1$, then$\lambda\not\in\sigma(\mathcal L)$. This is clear as the series$(\lambda-\mathcal ...

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Answer. $S=\{z\in\mathbb C: \lvert\,z\rvert=1\}$. Explanation. The space $\ell^2(\mathbb Z)$ is isometric to $L^2(\mathbb T)$, where $\mathbb T$ is the unit circle (equivalently, the domain of $2\pi-$periodic functions), and this is done by the transfomation $$\varPhi\big((a_k)_{k\in\mathbb Z}\big)=\sum_{k\in\mathbb Z}a_k\mathrm{e}^{ikx},$$ as ...

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HINT.If $a_{n+1}-z a_n=b_n$ then $a_2=z a_1+b_1$, $a_3=z a_2+b_2=z^2 a_1+z b_1+b_2$ ,$a_4=z a_3+b_3=z^3a_1+z^2b_1+z b_2+b_3$, etc. If $|z|=1$ and $b_n= z^{n}/(1+|n|)$, what happens to $|a_n-z^{n-1}a_1|$ for large positive $n$?

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If the characteristic polynomial of $f$ has no multiple roots, then there exist distinct complex numbers $\lambda_1, \dots, \lambda_n$ and a basis $v_1, \dots, v_n$ of $\mathbb{C}^n$ such that $$fv_k = \lambda_k v_k$$ for $k = 1, \dots, n$. Now \begin{align*} f(gv_k) &= g(fv_k) \\ &= \lambda_k (gv_k), \end{align*} which means that $gv_k$ is in the ...

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I presume this is on a Hilbert space, so that your "scalar function" $\chi$ is a self-adjoint projection (which by the spectral theorem is unitarily equivalent to multiplication by a scalar function with values $\{0,1\}$ on $L^2$ of some measure). No, we can't say much about $\rho(\chi A)$ compared to $\rho(A)$. Consider the $2 \times 2$ case $$A = ... 1 First, all functions in the span of u and v are critical points. Second, consider u_0 = \sin(\pi x)\sin(\pi y)\ge 0, u=t\cdot u_0, i.e., eigenfunctions to the smallest eigenvalue of the Laplacian. Then$$ I[u] = \int_\Omega \frac12|Du|^2 - \frac{5\pi^2}2|u|^2 = \frac12\int_\Omega u(-\Delta u - 5\pi^2u) = \frac12t^2 \int_\Omega u_0(2\pi^2 - ...

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Suppose $\lambda$ is an isolated point of the spectrum of $T$. Let $F$ be a continuous function that is identically $1$ near $\lambda$, and is $0$ on the remaining part of the spectrum $\sigma(T)\setminus\{\lambda\}$. Then $F$ is either $1$ or $0$ on the spectrum, which makes $P=F(T)$ an orthogonal projection. Clearly $(T-\lambda I)P=0$ because ...

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You are looking at the following eigenvalue problem on $[0,\infty)$: $$Lf= - \frac{d}{dx}\frac{1}{1+x}\frac{df}{dx}+(x+1)f=\lambda\frac{1}{x+1}f$$ If you can explicitly solve for any particular $\lambda$, then you can determine if the equation is in the limit point or limit circle case at $\infty$. If you are in the limit circle case, then all ...

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If you add the condition that the eigenspaces of $A_1$ and $A_2$ are orthogonal, you will have better luck proving these sorts of bounds for the Schatten norms. If you just want to prove it for the spectral norm, you should only need the eigenspaces of the extreme eigenvalues to be orthogonal.

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Theorem: Suppose $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is a surjective linear operator on a complex Hilbert space $X$ for which $(Ax,x) \ge \|x\|^{2}$ holds for all $x\in\mathcal{D}(A)$. Then $A$ is densely-defined and selfadjoint. Proof: To show that $A$ is densely-defined suppose otherwise. Then there exists $y \perp \mathcal{D}(A)$. However, ...

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