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In fact, for any scalar product, $A$ is never self-adjoint. Let $f$ be a non-degenerate symmetric bilinear form on $\mathbb{R}^n$ and $M=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ be its matrix in the canonical basis. The adjoint $\tilde{A}$ of $A$ is defined by $f(x,\tilde{A}y)=f(Ax,y)$, that is $M\tilde{A}=A^TM$. Thus $A$ is self-adjoint iff $MA=A^TM$. ...

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A closed densely-defined linear operator on a Hilbert space can have empty spectrum. For example, let $H=L^{2}[0,1]$ and let $A=\frac{d}{dx}$ on the domain consisting of absolutely continuous $f \in L^{2}$ for which $f(0)=0$ and $f' \in L^{2}$. To show that $A$ has no spectrum, it is enough to prove that the resolvent $R(\lambda)=(A-\lambda I)^{-1}$ exists ...

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Assuming this is an operator on $\mathbb{R}^2$ with the standard inner product $\left< x, y \right> = x \cdot y$, then yes, this is the correct reasoning. In general, you need to specify an inner product to know if an operator is self-adjoint, because self-adjointness means $\left< x, A y \right> = \left< A x, y \right>$. In this case, ...

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No, the operator is not self adjoint. Your reasoning is correct. Note that the definition of the adjoint of an operator is not simply that the adjoint of $A$ is $A^T$. However, in the context of real matrices taken with respect to the standard dot product or with respect to the standard dual basis, this is what "adjoint" comes to mean. An operator is ...

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$$A(A + I)^{-1} = (A + I - I)(A + I)^{-1}\\ = (A + I)(A + I)^{-1} - I(A + I)^{-1}\\ = (A + I)^{-1}(A+I) - (A + I)^{-1}I\\ = (A + I)^{-1}(A+I-I)\\ = (A + I)^{-1}A$$ since $(A + I)$ commutes with its own inverse, and $I$ commutes with anything.

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This is only a check! The spectrum is neglible: $$0\leq\lambda_\mathbb{C}(\sigma(H))\leq\lambda_\mathbb{C}(\mathbb{R})=0$$ By functional calculus: ...

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Recall that $0 \leq I - S$ means that $(0x, x) \leq ((I - S)x, x)$ for all $x$. Likewise, $I - S \leq cI$ means that $((I - S)x, x) \leq ((cI)x, x)$ for all $x$. As $(0x, x) = 0$, we see that $$0 \leq ((I - S)x, x) \leq ((cI)x, x) = (cx, x) = c(x, x) = c\|x\|^2.$$ Taking absolute values, we see that $|((I - S)x, x)| \leq c\|x\|^2$. Therefore $$\|I - S\| ... 0 The operator K is called the Volterra operator It is a bounded linear operator on L^2 space and is an indefinite integral. The easiest way to prove the Volterra operator has spectrum \{0\} is showing that the spectral radius is zero. To get you started, we must determine that the limit of K^n under the operator norm to the power 1/n is 0, ie, ... 1 If you're looking at a real (as opposed to complex) Hilbert space, then it depends on what you mean by positive. Note, for example, that if we take$$ A = \pmatrix{1&1\\0&1} $$then we have \langle x, Ax \rangle \geq 0 for all x \in \Bbb R^2. 1 This is true for complex Hilbert spaces (added after valuable comments!). If T is a positive operator you have by definition \langle Tx, x \rangle \ge 0 for all vector x in the considered space E. Hence \langle Tx, x \rangle is real for all x \in E which implies that T is self-adjoint in the complex case. 2 Suppose \mathcal{H}=L^{2}[0,\pi]. Suppose \mathcal{M} is a countably dense subset of compactly supported C^{\infty} functions on (0,\pi). Apply Gram-Schmidt in order to obtain an orthonormal basis \{ f_{k} \}_{k=1}^{\infty} of smooth compactly supported functions on (0,\pi). The following operators L_{\alpha,\beta} are selfadjoint$$ ...

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It makes a little more sense if I think about it as: Our goal is to create an operator that shifts the entire wave function over by an amount (translation). We have an operator that is essentially the derivative (momentum operator). Let's say we want to shift it over by an amount $\Delta x$. A first approximation would be to create a function that ...

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This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Skoro below. Since I cant comment, I will leave this thought here. Since $||\cdot||_1$ and $||\cdot||$ (as you defined them) are dual norms, it must be that tr$((A+B)^TX)\leq||A+B||_1$ for any $X$ such that $||X||\leq 1$. Therefore, tr$(A^TB)\leq ||A+B||_1 ... 0 Adjoint Formula It holds the relations: $$\mathrm{ad}_\varepsilon(A)=\delta\tau_{+\varepsilon}[A]e^{i\varepsilon H}=e^{i\varepsilon H}\delta\tau_{-\varepsilon}[A]$$ They are derivations: $$\mathrm{ad}_\varepsilon(AB)=\mathrm{ad}_\varepsilon(A)B+A\mathrm{ad}_\varepsilon(B)$$ And they vanish on: $$\mathrm{ad}_\varepsilon(e^{itH})=i[\delta ... 1 Suppose T is selfadjoint with \|T\| =1. Then |(Tx,x)| \le \|T\|\|x\|=1 for all unit vectors x. If |(Tx,x)|=1 for some unit vector x, then the above gives$$ 1=|(Tx,x)|\le \|Tx\|\|x\|\le \|T\|\|x\|^{2}\le 1 $$Therefore, |(Tx,x)|=\|Tx\|\|x\| with x\ne 0 which, by Cauchy-Schwarz, forces Tx=\alpha x for some scalar \alpha and unit vector ... 5 I don't know of any such operator, however, you could multiply your vector by the matrix \begin{bmatrix} 0 & 0 &1 \\ 0& 1 &0 \\ 1&0 &0 \end{bmatrix} and then take the transpose. 1 A linear operator A in a Hilbert space H is of finite rank if and only if A^* is too. This is easy: As vector spaces we have H/\ker(A)=R(A), and so \dim(\ker(A)^\perp) \leq \dim(R(A)) <\infty, but \ker(A)^\perp = \overline{R(A^*)} so that A^* has finite rank whenever A does. The converse is obvious. Now if \ker(A)^\perp is finite ... 1 The inverse operator does not need to be continuous, so we cannot conclude that the range is closed. In fact, if the range is closed, this immediately implies that the inverse operator is continuous by the open mapping theorem. 1 I wouldn't expect a straightforward answer to your question. Suppose that A is finitely generated, i.e. A=F_0'' for some finite F_0. Then, for any sequence \{A_n\}, for any B\subseteq A, for any n, you can take F=F_0 and then B\subseteq A=F_0''=(A_n\cup F_0)''. The problem is that it is not known if every von Neumann algebra is finitely ... 0 Let A and B be ant two sets. Then f:A\rightarrow B is said to be a \textbf{function}, if every element of A is mapped to a unique element of B. Here requirement for A and B is only arbitrary sets. Let V_1,V_2 be any two vector spaces. A map or a function T:V_1 \rightarrow V_2 is an \textbf{operator}. Here minimum requirement of ... 0 Let D \subset H be dense subspace and assume T: D \to H is bounded on D by say M \in \mathbb R^+. This means$$ \sup_{ x \in D \setminus \{0\}} \frac{\Vert Tx \Vert} {\Vert x \Vert } = M. $$We claim that there exists a linear map \tilde T: H \to H which extends T and has norm M. Now, let y \in H be arbitrary. Approximate y by elements ... 0 Construction Given the Hilbert space \ell^2(\Lambda). Construct the spectral measure:$$E(A)\varphi:=\sum_{\lambda\in\Lambda}\delta_{\lambda\in A}\langle\sigma_\lambda,\varphi\rangle\sigma_\lambda\quad(\sigma_\lambda:=\chi_{\{\lambda\}})$$Then one obtains:$$\sigma_0(N)=\Lambda\quad\sigma(N)=\overline{\Lambda}$$Concluding the construction. Reference ... 0 Domain For the domain:$$P_\lambda N\subseteq NP_\lambda\implies P_\lambda\mathcal{D}\subseteq\mathcal{D}$$So one obtains$$\varphi\in\mathcal{D}:\quad\varphi=\sum_\lambda P_\lambda\varphi\in\sum_\lambda\mathcal{S}_\lambda\cap\mathcal{D}$$(That allows to continue!) Operator Regard index sets:$$\#\Lambda<\infty:\quad\lambda_0\in\Lambda$$Then one ... 1 Embedding Consider the embeddings:$$J_\alpha:\mathcal{S}_\alpha\to\mathcal{H}:\quad J_\beta^*J_\alpha=\delta_{\beta\alpha}1_\alpha\quad J_\alpha J_\alpha^*=P_\alpha$$Construct the unitary map:$$U\varphi:=(J_\alpha^*\varphi)_\alpha\quad V(\varphi_\alpha)_\alpha:=\sum_\alpha J_\alpha\varphi_\alpha$$Indeed they are inverses:*$$VU\varphi=\sum_\alpha ... 1 Unitary Map Construct the unitary map: $$V:\mathcal{L}^2(\nu_0)\to\mathcal{H}:\quad Vh:=h(N)\varphi_0$$ By measurable calculus: $$h\in\mathcal{L}^2(\nu_0):\quad\|h(N)\varphi_0\|^2=\int|h|^2\mathrm{d}\nu_0$$ Especially one has: $$V\chi_A=E(A)\varphi_0:\quad\mathcal{H}=\overline{\{V\chi_A:A\in\mathcal{B}(\mathbb{C})\}}$$ Concluding unitarity. ... 1 Consider the POSET: $$\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}=\mathcal{S}_\varphi$$ $$\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}\perp\mathcal{S}'$$ It is nonempty: $$\mathcal{S}_0=(0):\quad\{\mathcal{S}_0\}\in\Lambda$$ And admits upper bounds: ... 2 Yes, you are correct. For reference: The Spectral Theorem For a Pair of Commuting Operators http://www.mi.ras.ru/~snovikov/78.pdf 1 For every positive element$a$one has$a\leq\Vert a\Vert I$-2 It is the adjoint of right shift operator which is a isometry, so it's norm is 1. 1 No, Consider$L: \ell_2 \to \ell_2$. Since$\vert \vert L \vert \vert_{op} = \sup_{\vert \vert x \vert \vert_2 =1} \vert \vert L(x) \vert \vert_2 $Since$\vert \vert L(x) \vert \vert_2^2 = \sum_{i \geq 2} x_i^2$and$\vert\vert x \vert\vert_2 = \sum_{i \geq 1} x_i^2$it is clear that$ \vert \vert L(x) \vert \vert_2 \leq \vert \vert x \vert \vert_2^2 $... 0 If$\xi\in l^p$with$\|\xi\|_p = 1$, then $$\|T\xi\|_p^p = \sum_{i=0}^\infty |\xi_{2i+1}|^p\leqslant\sum_{i=0}^\infty |\xi_i|^p=1,$$ so$T$is bounded and$\|T\|\leqslant 1$. Let$\xi_0=(1,0,0,\ldots)$, then$\|\xi_0\|_p =1$and$T\xi_0=\xi_0$, so$\|T\xi_0\|_p=1$and$\|T\|\geqslant 1$. 1 Because$V = TNT^{-1}$is normal and$N$is normal, then $$V = \int \lambda dE_{V}(\lambda),\;\;\; N=\int\lambda dE_{N}(\lambda),$$ and$E_{V}(S) = TE_{N}(S)T^{-1}$for all Borel subsets$S$. Hence, $$TE_{N}(S)T^{-1} = E_{V}(S)=E_{V}(S)^{\star}=(T^{-1})^{\star}E_{N}(S)T^{\star}.$$ Therefore, $$... 0 Regard the sets:$$\Omega_n:=\{|f|\leq n\}\in\mathcal{B}(\mathbb{C})$$Denote for shorthand:$$1_n:=\chi_{\Omega_n}\quad f_n:=f1_n$$By dominated convergence:$$\langle f(E)\varphi,\psi\rangle=\lim_n\langle f_n(E)\varphi,\psi\rangle=\lim_n\langle\varphi,\overline{f_n}(E)\psi\rangle=\langle\varphi,\overline{f}(E)\psi\rangle$$They extend by: ... 0 Remind the domain:$$\mathcal{D}f(E)g(E)=\mathcal{D}(fg)(E)\cap\mathcal{D}g(E)$$One has the bound:$$\int|f|^2\mathrm{d}\nu_\varphi\leq\int(1+|f|^4)\mathrm{d}\nu_\varphi=\int|f|^4\mathrm{d}\nu_\varphi+\|\varphi\|^2$$So for the domain:$$\mathcal{D}|f|^2\subseteq\mathcal{D}f(E)=\mathcal{D}\overline{f}(E)$$Concluding the assertion. -1 Here is an interesting observation: if v \in \ker(N) then T^*T v \in \ker(N) Proof: keep in mind that \ker(N)=\ker(N^*) and that this holds for all normal operators. Let v \in \ker(N) be any vector. Since T is invertible there is \tilde{v} such that v = T^{-1}\tilde{v}. Then T N T^{-1} \tilde{v} = 0. But being T N T^{-1} normal we get ... 0 We have N^*=N^{-1} and {T^*}^{-1}N^*T^*=TN^*T^{-1}. Multiply by T^{-1} to the left and {T^*}^{-1} to the right. One gets$$T^{-1}{T^*}^{-1}N^*=N^*T^{-1}{T^*}^{-1}$$Bearing in mind that T^{-1}{T^*}^{-1}=\left(T^*T\right)^{-1}, one has$$\left(T^*T\right)^{-1}N^{-1}=N^{-1}\left(T^*T\right)^{-1}$$And we're done because ... 0 There is a direct analog of closed Graph theorem of non-archimedean fields. see page 61, theorem 3.5 in Non-archimedean functional analysis. A. C. M. van Rooij. 1 An alternative is to use this identity for quadratic forms:$$ Q(\lambda x+(1-\lambda)y) = \lambda Q(x) + (1-\lambda) Q(y) - \lambda (1-\lambda) Q(x-y). $$It's then pretty obvious that \lambda (1-\lambda) Q(x-y)>0 if 0<\lambda<1 and Q>0, so the convexity is obvious. Now, you may be concerned about this identity applying here. ... 2 To complete the proof you can observe that since T is self-adjoint, positive, and bounded it has a square-root. Then \begin{eqnarray*} \Phi(tx+(1-t)y) &=& t^2\Phi(x)+t(1-t)((Tx,y)+(Ty,x))+(1-t)^2\Phi(y) \\ &=& t^2\Phi(x)+t(1-t)((T^{1/2}x,T^{1/2}y)+(T^{1/2}y,T^{1/2}x))+(1-t)^2\Phi(y) \\ &\leq& ... 1 Suppose one has:$$\vartheta:=\|f(E)\|<\infty$$Regard the sets:$$\Omega_n:=\{\vartheta+\frac{1}{n}\leq|f|\leq\vartheta+n\}\in\mathcal{B}(\mathbb{C})$$And their union:$$\Omega:=\bigcup_{n=1}^\infty\Omega_n=\{|f|>\vartheta\}\in\mathcal{B}(\mathbb{C})$$Regard the functions:$$1_n:=\chi_{\Omega_n}:\quad 1_n(E)=E(\Omega_n)$$Denote for readability: ... 1 Meanwhile I got it... Support For the spectral radius:$$\|Z\|\leq1\implies r(Z)\leq1$$So the support lies in:$$\mathrm{supp}E=\sigma(Z)\subseteq\overline{\mathbb{D}}$$By the previous threads:$$g(E_Z)=\sqrt{1-Z^*Z}=\sqrt{Q}$$So the boundary is trivial:$$\mathcal{N}g(E_Z)=(0)\implies E_Z\{g=0\}=(0)$$(That allows to continue!) Preoperator By ... 0 By the calculus:$$f(E)E\{f=0\}=(f1_{\{f=0\}})(E)=0$$Thus one has:$$\varphi\in\mathcal{R}E\{f=0\}\implies f(E)\varphi=0$$Conversely regard:$$f^{-1}(\lambda):=\frac{1}{f(\lambda)}\quad(E(\lambda)\neq0)$$By the calculus:$$f^{-1}(E)f(E)\subseteq(f^{-1}f)(E)=1(E)=1$$But the domains agree: ... 0 Regard dense elements:$$\overline{\mathcal{D}N^*}=\mathcal{H}:\quad\varphi\in\mathcal{D}N^*$$A calculation gives:$$ZZ^*\varphi=NQN^*\varphi=NN^*Q\varphi$$The operator is closed:$$Q\in\mathcal{B}(\mathcal{H}):\quad NN^*=\overline{NN^*}\implies NN^*Q=\overline{NN^*Q}$$By closed graph theorem:$$\mathcal{D}(NN^*Q)=\mathcal{H}\implies ... 2 Denote$[T]=(a_{ij})$and$[T^*]=(b_{ij})$. Denote further$X=(x_{ij})$with$x_{ij}=\langle e_i,e_j\rangle$and$Y=(y_{ij})$with$y_{ij}=\langle b_i,b_j\rangle$. Then$ T(e_j)= \sum_k a_{kj} b_k $and $$\langle Te_j,b_i\rangle = \sum_k\langle a_{kj} b_k,b_i\rangle = \sum_k a_{kj} y_{ki} .$$ Analogously$T^*(b_i) = \sum_k b_{ki} e_k\$, and $$\langle ... 2 That follows is a particular constructible solution in T, an upper triangular matrix with non-negative diagonal. There is a unique hermitian \geq 0 H s.t. A=H^2; let rank(H)=rank(A)=r. We consider the standard (Maple) QR decomposition of H: H=QR where Q\in M_{n,r},R\in M_{r,n}, the diagonal of R being positive; moreover Q^*Q=I_r. The ... 0 You are right, my previous attempt was wrong: here are examples that answer your question: take A=C(X) to be the Banach algebra of all continuous function on some compact space (say X=[0,1], or take X to be a point so that A=\mathbb{C}) and define a new norm on A by \|f\|_r:= r \sup_{x\in X} |f(x)|, where r>1 is some (fixed) number. Then ... 2 No. Let$$ S=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ T=\begin{bmatrix}0&0\\1&0\end{bmatrix}. $$Then$$ \text{Tr}(TS)=1,\ \ \text{ and } \|T\|\,|\text{Tr}(S)|=0. $$1 One approach is to consider \dot{A} : X/\mathcal{N}(A)\rightarrow Y, which is continuous and a linear bijection between X/\mathcal{N}(A) and Y. So the inverse of \dot{A} is continuous, and the statement of that continuity gives you what you want. 1 If A and B are positive definite densely-defined selfadjoint linear operators on a complex Hilbert space H, then AB^{-1} and BA^{-1} are bounded if \mathcal{D}(A)=\mathcal{D}(B). Then A^{s}B^{-s} and B^{s}A^{-s} are bounded for 0 < s \le 1 with$$ \|A^{s}B^{-s}\| \le \|AB^{-1}\|^{s} \\ \|B^{s}A^{-s}\| \le ...

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Evolution Regard dense elements: $$\overline{\mathcal{D}(H)}=\mathcal{H}:\quad\varphi\in\mathcal{D}(H)$$ By the previous thread: $$\mathrm{ad}(A)\in\mathcal{B}(\mathcal{H})\implies\frac{\mathrm{d}}{\mathrm{d}t}\tau^t[A]\varphi=\tau^t[\mathrm{ad}(A)]\varphi$$ That gives the identity: ...

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