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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 ...

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 ...

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 $D$ ...

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Since $U$ is unitary, $UU^t=Id$, hence your question is equivalent to ask for vectors s.t. $\langle Dv,Dv\rangle=\langle v,v\rangle$. Since $D$ is diagonal, this is the case for vectors that are eigen-vectors of $D$ with eigenvalue $\pm1$.

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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 ...

0

The spectral theorem is a general argument. For a specific normal operators, there may be a concrete "diagonalizing" unitary. Take the shift $S$ on $l^2(\mathbb{Z})$. In this concrete example, the diagonalizing unitary is the Fourier transform $$\mathcal{F} : l^2(\mathbb{Z}) \rightarrow L^2(\mathbb{T}).$$ The operator $$\mathcal{F} S\mathcal{F}^{-1} ... 3 Consider H=L^2(\Omega,\mu) where \Omega\subseteq\mathbb C is compact, and multiplication operator T:H\to H,\ (Tf)(z)=zf(z). Then T is bounded and normal. By the spectral theorem T=\int_{\mathbb C}\lambda dE(\lambda). In this case you can give an explicit formula for the spectral measure E:$$(E(X)f)(z)=\chi_X(z)f(z),$$where \chi_X is the ... 0 In what follows we prove that: If \{\lambda_n\} are eigenvalues (corresponding to linearly independent eigenvectors) of the self-adjoint compact operator K, then \lambda_n\to 0. Suppose not, and in particular, that there exists an \varepsilon>0, such that \lvert\lambda_{j_n}\rvert\ge\varepsilon, for \{\lambda_{j_n}\} a subsequence of ... 1 As proposed in (http://math.stackexchange.com/questions/685680/how-to-use-parseval-s-plancherel-s-identity), You can use that$$\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$$etc. to reduce the expression for H_n to integrals over the segments [\pm n-α,\pm n+α]. Then apply the asymptotic ... 1 There is a more general result that can be found in Rudin's Functional Analysis in the chapter on bounded operators on a Hilbert space. Suppose T\in\mathcal{B}(H) is normal and E is its spectral decomposition. If f\in C(\sigma(T)) and if \omega_{0}=f^{-1}(0), then \mathcal{N}(f(T))=\mathcal{R}(E(\omega_{0})) where \mathcal{N}(f(T)) denotes the ... 0 Hint. There is an orthogonal base in H such that G,P and G+P are diagonal operators. 1 Hint. Using the appropriate Spectral Theorem for unbounded normal operators on Hilbert spaces (do your homework to find the most appropriate version) you should see that G and P are spectrally analyzed simultaneously, and hence 1 is added to the spectrum of G where it was 0. 1 The eigenfunctions u_j of the Laplacian in \Omega, corresponding to the eigenvalues \lambda_j, constitute an orthogonal (and WLOG orthonormal) basis of L^2(\Omega). Thus if v_0(x)=v(x,0), where u is a solution of the Heat Equation (with homogeneous boundary conditions) then$$ v_0=\sum_j \langle v_0,u_j\rangle u_j, $$and as ... 4 More is true, in fact. If H is a Hilbert space, and T\colon H \to H a compact normal operator, then H is the closure of the direct sum of the eigenspaces of T. For \lambda \in \sigma_P(T), let E_\lambda be the eigenspace of T corresponding to the eigenvalue \lambda, and let S = \bigoplus\limits_{\lambda\in\sigma_P(T)} E_\lambda. Then S ... 1 Suppose X is a Banach space and suppose that T : X\rightarrow X is linear. Then the following are true: 1. If X is finite-dimensional, then T is compact, regardless of whether or not 0 \in \sigma(T). 2. If X is infinite-dimensional and T is compact, then 0 \in \sigma(T). 3 If 0 is not in the spectrum of A, then A is invertible with inverse B. Since the product of compact operators is compact, the identity I = AB is compact. But this is impossible if A acts on an infinite-dimensional Banach/Hilbert space, because the closed unit ball is not compact. 5 This is indeed only true in infinite dimensions. Suppose your compact operator A has finite spectrum. Then there are only finitely many eigenvectors with nonzero eigenvalue. Let F be the subspace spanned by those eigenvectors and let E be its orthogonal complement. Since F is finite dimensional, E is infinite dimensional, and in particular E ... 0 Let K be the operator in question. Its spectral decomposition is$$K=∑_{n}λ_{n}P_{n},$$where P_{n} is the eigenprojector associated with the eigenvalue λ_{n}. It is finite dimensional and we can write$$P_{n}=∑_{j_{n}}P_{j_{n}}$$with the P_{j_{n}} a finite set of one-dimensional orthogonal projectors. Hence$$K=∑_{n}∑_{j_{n}}λ_{n}P_{j_{n}}

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