Tag Info

This is indeed only true in infinite dimensions. Suppose your compact operator $A$ has finite spectrum. Then there are only finitely many eigenvectors with nonzero eigenvalue. Let $F$ be the subspace spanned by those eigenvectors and let $E$ be its orthogonal complement. Since $F$ is finite dimensional, $E$ is infinite dimensional, and in particular $E ... 4 More is true, in fact. If$H$is a Hilbert space, and$T\colon H \to H$a compact normal operator, then$H$is the closure of the direct sum of the eigenspaces of$T$. For$\lambda \in \sigma_P(T)$, let$E_\lambda$be the eigenspace of$T$corresponding to the eigenvalue$\lambda$, and let$S = \bigoplus\limits_{\lambda\in\sigma_P(T)} E_\lambda$. Then$S$... 3 Consider$H=L^2(\Omega,\mu)$where$\Omega\subseteq\mathbb C$is compact, and multiplication operator$T:H\to H,\ (Tf)(z)=zf(z).$Then$T$is bounded and normal. By the spectral theorem$T=\int_{\mathbb C}\lambda dE(\lambda).$In this case you can give an explicit formula for the spectral measure$E:$$$(E(X)f)(z)=\chi_X(z)f(z),$$ where$\chi_X$is the ... 2 Note that $$\langle DUv,DUv\rangle = (DUv)^*DUv = (Uv)^*D^*D(Uv)^* = (Uv)^*|D|^2(Uv)^*$$ Setting$w = Uv$, this is just$w^*|D|^2w$. So, noting that$\langle w,w \rangle = \langle v,v \rangle$the question can be rephrased as "when does$w^*|D|^2w = \langle w,w \rangle$"? Now, suppose$w$is given by$w = \pmatrix{w_1 & \cdots & w_n}^T$and$D$... 2 Suppose that$\lambda\not\in\{0,1\}$. Then $$a(a-\lambda)=a^2-\lambda a=a-\lambda a=(1-\lambda)a.$$ Similarly, $$(1-a)(a-\lambda)=(1-a)a-(1-a)\lambda=-(1-a)\lambda.$$ Let $$b=\frac1{1-\lambda}\,a-\frac1\lambda\,(1-a).$$ Then $$b(a-\lambda)=ba-\lambda b=\frac1{1-\lambda}\,a-\frac\lambda{1-\lambda}\,a+1-a=1,$$ and one can also check that ... 1 As proposed in (http://math.stackexchange.com/questions/685680/how-to-use-parseval-s-plancherel-s-identity), You can use that $$\int_{-n}^ne^{itx} f(t+α)\,dt=\int_{-n+α}^{n+α}e^{i(t-α)x}f(t)\,dt =e^{-iαx}\int_{-n+α}^{n+α}e^{itx}f(t)\,dt$$ etc. to reduce the expression for$H_n$to integrals over the segments$[\pm n-α,\pm n+α]$. Then apply the asymptotic ... 1$\mathbb{E}x[n+\tau]y[n] := R_{xy}[\tau]$is the expected value of the cross-correlation at lag$\tau$. This is the average over the signal of the degree to which the signal$y$"now" can be used to estimate the signal$x$at "now +$\tau$". Since these signals are assumed stochastic (but stationary enough for these expectations to be stationary) we cannot ... 1 The eigenfunctions$u_j$of the Laplacian in$\Omega$, corresponding to the eigenvalues$\lambda_j$, constitute an orthogonal (and WLOG orthonormal) basis of$L^2(\Omega)$. Thus if$v_0(x)=v(x,0)$, where$u$is a solution of the Heat Equation (with homogeneous boundary conditions) then $$v_0=\sum_j \langle v_0,u_j\rangle u_j,$$ and as ... 1 Suppose$X$is a Banach space and suppose that$T : X\rightarrow X$is linear. Then the following are true: 1. If$X$is finite-dimensional, then$T$is compact, regardless of whether or not$0 \in \sigma(T)$. 2. If$X$is infinite-dimensional and$T$is compact, then$0 \in \sigma(T)$. 1 There is a more general result that can be found in Rudin's Functional Analysis in the chapter on bounded operators on a Hilbert space. Suppose$T\in\mathcal{B}(H)$is normal and$E$is its spectral decomposition. If$f\in C(\sigma(T))$and if$\omega_{0}=f^{-1}(0)$, then$\mathcal{N}(f(T))=\mathcal{R}(E(\omega_{0}))$where$\mathcal{N}(f(T))$denotes the ... 1 This is a typo. The second$x$should be a different letter. Since$x\in M=\mathrm{im}\,E(\omega)$, there is a$y\in H$such that$E(\omega)y=x$. Therefore $$Tx=TE(\omega)y=E(\omega)Ty$$ so$Tx\in M$. It is possible that$E(\omega)=0$, but the point is that there exists$\omega$such that$E(\omega)\not=0$, because otherwise$E(\omega)=0\$ for all Borel sets ...