# Tag Info

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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