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3

False. The closed span of $e_k$ for $k \ge n$ is invariant.


2

Since $T - \lambda I$ is also normal, we have $$ \| T - \lambda I \| = \text{spr} (T - \lambda I) = 0, $$ showing that $T = \lambda I$. (I recently asked basically the same question (Self-adjoint operator with single point spectrum), but your formulation is more general so I thought it might be worth sharing the answer here.)


1

To your first question, let $$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}. $$ Then $$ p\vee q=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ p+q=\begin{bmatrix}3/2&1/2\\1/2&1/2\end{bmatrix}, $$ so $$ p+q-p\vee q=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix}, $$ which is ...


1

Note that the two equalities $A=AA^*A$ and $A^*=A^*AA^*$ are the same, since you can obtain one from the other by taking adjoints. Assume first that $A=AA^*A$. By multiplying by $A^*$ on the left, we get $$ A^*A=(A^*A)^2.$$ It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible. Conversely,...


1

(You don't say how you got the second equality; since it is not trivial, I'm not sure how you did it and so it is done below) Since $A^*A$ is positive and compact, it is orthogonally diagonalizable (spectral theorem): $A^*A=U^*D^2U$ for some unitary $U$ and $D$ diagonal with diagonal $s_1(A),s_2(A),\ldots$ Assume $s_1(A)\geq s_2(A)\geq \cdots$ Since $U$ ...


1

Consider $u\in B(H_1\oplus H_2, H_1\oplus H_2)$ and since $u^*u$ is projection, then $u$ is partial isometry, which means that $uu^*u=u.$


1

using spectral theorem since $T$ is normal it exist a spectral measure $E$ such that $$ Tx=\int_{\sigma(T)}tdE(x)=\int_{\{\lambda\}}tdE(x)=\lambda E(\{\lambda\})(x)=\lambda E(\sigma(T))(x)=\lambda I (x)=\lambda x $$


1

If $\psi$ is a state vector, meaning $\|\psi\|=1$, then the expected value of an observable $A$ for the system in the state $\psi$ is $$ (A\psi,\psi). $$ That's what you're trying to show. One must assume that $\psi$ is in the domain of the observable $A$. A selfadjoint operator $A : \mathcal{D}(A)\subseteq H\rightarrow H$ on a Hilbert space $H$ ...


1

Using Binet's formula: $$F_n=\frac{1}{\sqrt{5}}\left(\varphi^n-\psi^n\right)$$ where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi = \frac{1-\sqrt{5}}{2}$. Then $$(Df)(x) = \frac{1}{\sqrt{5}}\left(f(\varphi x)-f(\psi x)\right)$$ If you write the scaling operators $(S_\alpha f)(x)=f(\alpha x)$, then you can write the above as: $$D=\frac{1}{\sqrt{5}}\left(S_\...


1

$1\iff4$ : Assume $1$. Given $x\in N_+$, there exists nonzero $y\in N_\tau^+$ with $y\leq x$. Now use Zorn to find a maximal ordered family $\{y_j\}\subset N_\tau^+$ with $y_j\leq x$ for all $j$. As the net is bounded, it has a sup, say $y=\lim_{sot}y_j$. Then $y=x$, because otherwise a nonzero element of $N+\tau^+$ below $y-x$ contradicts the maximality. ...



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