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Understanding the Spectral Theorem for (Bounded) Operators

Your $A$ can be written as $\sum \lambda_i P_i,$ where $P_i$ is the projection onto the eigenspace of $\lambda_i.$ A sum is a finite version of an integral.
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Is a holomorphic Function of a matrix invertible?

There is this theorem in spectral theory that says (if $g$ is continuous and defined on the spectrum of linear bounded $A$) $$ \sigma(g(A)) = g(\sigma(A)). $$ $\sigma$ of course denotes the spectrum. ...
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Relaxation time and poincare inequality

We assume the chain in question is reversible, so $\lambda_1=1-\gamma$ is real. Consider $f$ which is a nonzero eigenfunction for $\lambda_1$. Note that $E_\pi(f)=0$. Then $$\mathcal E(f,f)=\langle (I-...
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Non-commutative spectrum in Banach Algebras

You have that $T^*T=I$, the identity operator, and it is trivial to check that $\sigma(I)=\{1\}$. Meanwhile, $TT^*=I-P$, where $P$ is the projection onto the first coordinate. Then $\sigma(TT^*)=\{0,1\...
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Why does from $\|Tf_n\| \rightarrow 0$ follow that $\lambda \in [0,1]$ are approximate eigenvalues?

Here is another way to show that $\sigma (T)= [0,1]$. If $\lambda \notin [0,1]$ then for every $f \in L^2(0,1)$, the function $g=\tfrac{1}{\lambda-x} f \in L^2$ and $$ (λ-T)g=f. $$ This shows that $\...
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Why does from $\|Tf_n\| \rightarrow 0$ follow that $\lambda \in [0,1]$ are approximate eigenvalues?

There is a typo in the first part. Actually, $\|(T-\lambda)f_n\|^{2}=2n\int_{\lambda-\frac 1 n}^{\lambda+\frac 1 n} (x-\lambda)^{2}dx =\frac 4 {3n^{2}} \to 0$.
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Spectrum of direct sum and direct integral included in the union of the spectra of its constituents

In the case where the operators are normal everything becomes easier, so let us only think about the non-normal case (and the normal case will follow from it). In the non-normal case one good tool is ...
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Show there exists $\lambda \in \sigma(A)$ such that $|\lambda - \lambda_0|<\epsilon$

Just to add two comments: The result holds for any normal operator $A$. If $A$ is normal, so is $(A-\lambda_0)^{-1}$ (if $\sigma_0\not\in \sigma(A)$), and $\|(A-\lambda_0)^{-1}\|$ is equal to its ...
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Show there exists $\lambda \in \sigma(A)$ such that $|\lambda - \lambda_0|<\epsilon$

It suffices to use the fact that $$\|(\lambda_0-A)^{-1}\|=\frac1{d(\lambda_0,\sigma(A))}$$ for $\lambda_0\in\rho(A),$ where $d(\lambda_0,\sigma(A))=\inf_{\lambda\in\sigma(A)}|\lambda-\lambda_0|$. This ...
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What is "spectral development" of a linear operator?

This is a compact operator with no eigenvalues: the spectrum consists of $0$ alone. So the decomposition is trivial: the generalized eigenspace for $0$ is the whole space.
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Prove that $L_0 = (-i)^m \frac{d^m u}{dx^m}$ is a positive operator

When $m$ is even, let $m=2n$ and it follows $$\langle L_0u,u\rangle=(-1)^n\langle u,u^{(2n)}\rangle=(-1)^{2n}\langle u^{(n)},u^{(n)}\rangle=\|u^{(n)}\|^2\geq0.$$ This equals zero iff $u^{(n)}\equiv0$, ...
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A question on two self-adjoint operators with same spectrum

A self-adjoint operator $T$ admitting a cyclic vector is, as you said, necessarily the multiplication operator on $L^2(\sigma(T),\mu)$ given by $$ T(f)x=xf(x). $$ So, once $\sigma(T)$ is fixed, ...
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Singularity of solutions of a differential equation as functions of its eigenvalue

Let $u(x,\lambda)$ be the unique solution of $Lu=\lambda u$ such that $$ u(0,\lambda)=0,\;\; u_{x}(0,\lambda)=1. $$ For a fixed $x\in [0,1]$, the function $\lambda\rightarrow u(x,\lambda)...
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What does it mean that the spectrum is discrete?

"Discrete" here means "not continuous". If the spectrum of an operator is "all integers" then its spectrum is discrete. If the spectrum of an operator is "all real ...
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Finding $T^n$, where $T: l^{1} \rightarrow l^{1}$

It is useful to embed $\ell^1$ into the ring of formal power series $K[[t]]$ via the map $(x_0, x_1, \dots) \mapsto \sum_m x_mt^m$ where $K$ is the coefficient field (e.g. $K = \Bbb R$). The map $T$ ...
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Finding $T^n$, where $T: l^{1} \rightarrow l^{1}$

I prefer the following explanation. Assume $Tx=\lambda x$ for $x\neq 0.$ Then $S^2x=(\lambda -1)x.$ As $S^2$ is an isometry, then $\lambda\neq 1.$ Denote $\mu=\lambda-1\neq 0.$ Then $\mu^{-1}S^2x=x$ ...
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