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0

Kaplansky's density theorem lets you choose $a $ with $\|a\|\leq\|u\|$.


3

Because of the subspaces being selfadjoint, $B(H)V_1\subset V_2$ implies that $V_1B(H)\subset V_2$. If $V_1\ne0$, then $B(H)V_1B(H)\subset V_3$ contains all finite-rank operators, and thus $V_3$, being closed, contains the compact operators. If $V_3$ contains a non-compact operator, then the ideas in this answer show that $I\in V_7$ (I didn't count ...


2

What you need to show is that $\phi $ is one-to-one, and that it is a $*$-homomorphism. That lets you define the norm on $M_n (A) $ and the two properties you want are inherited automatically from the codomain. Then you need to show that $M_n (A) $ is closed. This is done by showing that norm convergence is equivalent to entrywise norm convergence, and as ...


0

1) Yes. 2) CP on a Hilbert space does not make sense. You need $X\subset B (H) $ a C$^*$-algebra, and $Y\subset B (K) $ on the codomain. There, you have to convince yourself that entry wise sot convergence in $M_n (Y) $ implies sot convergence in $M_n (Y) $. Note that $\langle T_jx,x\rangle \to\langle Tx,x\rangle$ is wot convergence, not sot.


0

Concerning 1, yes. This is an example where abstract nonsense is the most useful: An inverse limit is a categorical limit in the category of $C^*$-algebras, and those are unique up to unique isomorphism. If you check that $A_0$ fulfills the universal property, that means automatically that $A_0 \cong \lim{A_I}$. A proof goes like that: Assume there's a ...


2

Suppose that $A_1=\{a\in A: \|a\|<1\}$ and $B=\{ b\in A: \|be\|<1\}$. Obviously $Be\subset A_1$, and $S_e(Be)=T_e(Be)$. Hence $S_e(Be)\subset T_e(A_1)$, and consequently $\overline{S_e(Be)}^w\subset \overline{T_e(A_1)}^w$. But $\overline{T_e(A_1)}^w$ is weakly compact from assumption, and so $\overline{S_e(Be)}^w$ is weakly compact. But ...


2

First write $$ \lambda I-P = \lambda(I-P+P)-P = \lambda(I-P)+(\lambda-1)P $$ If $P^{2}=P$ then $(I-P)^{2}=(I-P)$ and $P(I-P)=(I-P)P=0$, and the inverse of the above can be spotted for $\lambda\in\{0,1\}$: $$ (\lambda I-P)^{-1} = \frac{1}{\lambda}(I-P)+\frac{1}{\lambda-1}P. $$ Therefore $\sigma(P)\subseteq \{0,1\}$. And $0\notin\sigma(P) \iff ...


3

I'm assuming that by $\sigma(\Sigma P)\leq1$ you mean that $\|\Sigma P\|\leq1$. Suppose that $\|P+Q\|\leq1$. So $0\leq P+Q\leq 1$. Then $(P+Q)^2\leq P+Q$ (just conjugate with $(P+Q)^{1/2}$). That is, $$ P+Q+QP+PQ\leq P+Q, $$ or $QP+PQ\leq0$. If we conjugate this inequality with $Q$, we get $QPQ+QPQ\leq0$. But $QPQ\geq0$, so $QPQ=0$. Then $$ ...


-1

Let's enumerate the conditions: (1) $P\perp P'$; (2) $0=PP'=P'P$; (3) $\Sigma P$ is a projection. We can assume that $A\subset B(H)$ for some Hilbert space $H$ (this basically follows from the GNS construction); also, remember that for positive operators $T\in B(H)$ we have $\langle T\xi,\xi\rangle\geq 0$ for all $\xi\in H$. For (1)$\Rightarrow$(2), assume ...


1

Perhaps an easier approach is as follows: Let $R$ be any unital commutative ring, then there is a one-to-one correspondence between ideals in $R$ and ideals in $M_2(R)$. In fact, if $J \subset M_2(R)$ is an ideal, then $$ I = \{a \in R : \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \in J\} $$ is the corresponding ideal in $R$ such that $J = M_2(I)$. ...


1

The category of $C^*$-algebras is complete like Martin Brandenburg explained. However, given an inverse system of $C^*$-algebras, it is often more convenient to consider its limit in the category of all topological $*$-algebras. The limit of a diagram $(A_i,\lVert - \rVert)_{i \in I}$ of $C^*$-algebras is then the topological $*$-subalgebra of the ...


3

The key inequality is $|\tau(x)|\leq\tau(|x|)$. I cannot really follow what Jesse is doing in his first inequality, but all we need to do it take the triangle inequality to get a sum of terms $$ |(\tau\otimes\tau)((p_k\otimes1)(v^*\otimes w^*)U(p_k\otimes 1))|. $$ Then, with $x=(p_k\otimes1)(v^*\otimes w^*)U(p_k\otimes 1)$, \begin{align} ...


3

The result is that $\varphi_\tau(M)$ is a von Neumann algebra if and only if $\tau $ is normal. If $\tau$ is normal, then so is $\varphi_\tau$ and so $\varphi_\tau(M)$ is a von Neumann algebra. Conversely, suppose that $\varphi_\tau(M)$ is a von Neumann algebra. Let $\{x_j\}$ be a monotone net of selfadjoints in $M$ and let $x=\sup x_j\in M$. As ...


2

I will write $f_k=r_1+\cdots+r_k$, because otherwise we get $M_{k+1}(\mathbb C)$. The projections $r_j$ are each equivalent to $1_A$, so there exist partial isometries $e_{1,j},\ldots,e_{1,k}$ with $$ e_{1,j}^*e_{1,j}^\vphantom{*}=r_j,\ \ \ \ e_{1,j}^\vphantom{*}e_{1,j}^*=r_1. $$ Now define $$ e_{k,j}^\vphantom{*}=e_{1,k}^*e_{1,j}^\vphantom{*}. $$ This is ...


1

This doesn't seem to be true. Take e.g. the $C^*$ algebra $\mathcal{A} = \mathbb{C}$, the complex numbers. Then the element $-1$ satisfies $$(-1)(-1)^*(-1) = -1$$ but its spectrum (when viewed as an operator) is $\{-1\}$.


1

You don't need the square, which gives you the additional step of justifying that $(1-A)^2\leq 1-A$. You can simply do, since $0\leq A\leq 1$ (because $P\leq A\leq Q$), $$ 0\leq P(1-A)P=P-PAP=0, $$ so $$ 0=P(1-A)P=[(1-A)^{1/2}P](1-A)^{1/2}P, $$ from where $(1-A)^{1/2}P=0$, and then $(1-A)P=0$.


0

If $Q \ge P$, then \begin{align} \|Px\|^{2} & =\|QPx\|^{2}+\|(I-Q)Px\|^{2} \\ & = (PQPx,x)+\|(I-Q)Px\|^{2} \\ & \ge (PPPx,x)+\|(I-Q)Px\|^{2} \\ & = \|Px\|^{2}+\|(I-Q)Px\|^{2} \\ \implies & (I-Q)Px=0. \end{align}


3

The assertion $P\leq Q$ means $Q-P\geq0$. Then $$ 0\leq P(Q-P)P=PQP-P\leq P^2-P=P-P=0. $$ So $P=PQP$. Now we can write this equality as $$0=P-PQP=P(I-Q)P=[(I-Q)P]^*[(I-Q)P],$$ so $(I-Q)P=0$, i.e. $P=QP$. Taking adjoints, $P=PQ$. The converse also holds: if $P=PQ=QP $, then $$Q-P=Q^2-QPQ=Q (I-P)Q\geq0. $$ Edit: for the equivalence $Q-P\geq0$ iff $Q-P$ is a ...


1

Here is an argument for the last part. Assuming we already know that $I=\overline{\bigcup_{B}B\cap I}$: If $I,J$ are ideals in $A$ and $IJ=0$, then for any $B\in S$ we have $0=B\cap IJ=(B\cap I)(B\cap J)$. As $B$ is prime, $B\cap I=0$ or $B\cap J=0$. Suppose $B_1\cap I\ne0$, and $B_2\cap J\ne0$, with $B_1\subset B_2$; then $B_2\cap I\supseteq B_1\cap ...


1

Our main problem is that, in principle, we cannot just "map the question" to the elements of $S$ by taking intersections. To solve this, we can use an argument similar to the one on the proof of Theorem 6.2.6 of the same book (and I will even use the same notation). Given an ideal $I$ of $A$, let $J=\overline{\left(\bigcup_{B\in S}B\cap I\right)}$. Then $J$ ...



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