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Yes, this can be done without making the $C^*$-algebra concrete in $B(H)$. All we need is that $z=0$ iff $z^*z=0$, which follows from the $C^*$-identity $\|z^*z\|=\|z\|^2$. Lemma Let $a\in A$. The following assertions are equivalent. 1 - $p=a^*a$ is idempotent, i.e. $p^2=p$. 2 - $aa^*a=a$ 3 - $a^*aa^*=a^*$ 4 - $q=aa^*$ is idempotent. Proof ...

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Note that $M_n(\mathcal{A})$ is a $C^*$-algebra, so we have $$A\leq B\implies C^*AC\leq C^*BC$$ for all $A,B,C\in M_n(\mathcal{A})$. Since $A\leq \Vert A\Vert 1_{M_n(\mathcal{A})}$ for all $A\in M_n(\mathcal{A})_+$ we get $$C^*AC\leq \Vert A\Vert C^*C$$ Now consider $$A= \begin{pmatrix} a^*a & 0 & \ldots & 0\\ 0 & a^*a & \ldots & ... 4 That has to be some of the most awkward notation for the Spectral Theorem that I've seen. The author could state that E is the spectral measure for a, thereby eliminating a from the decorations added to E. Who decorates the eigenfunctions or eigenvalues for an operator with the name of that operator, unless there is some special reason to do so? The ... 3 Is there a result in C^*-algebras that if b satisfies b = b^* and \sigma(b) \subset [\alpha,\beta] where \alpha,\beta > 0, then \alpha I \le b \le \beta I? Then you could prove it as follows. Consider b = a^* a. Then a short argument shows that \|b\| = \|b^{-1}\| = 1. But b is positive, and \inf\sigma(b) = \sup\sigma(b) = 1. Hence ... 3 Define a map$$ \varphi : A\times M_n(\mathbb{C}) \to M_n(A) $$given by$$ \varphi(a,(\lambda_{i,j})) := (\lambda_{i,j}a) $$This is bilinear, and so induces a linear map from the algebraic tensor product$$ \varphi : A\otimes M_n(\mathbb{C}) \to M_n(A) $$It is easy to see that this map is bijective, and a *-homomorphism. 2 Consider the left shift L on \ell^2(\mathbb N),$$ L((a_1,a_2,a_3,\dots)) = (a_2,a_3,\dots) .$$Suppose for contradiction that L = U|A| where U is unitary. Then L^* L = |A|^2 = \text{diag}(0,1,1,\dots) which implies \sigma(|A|^2) = \{0,1\}. But LL^* = U|A|^2 U^* = I which implies \sigma(|A|^2) = \{1\}. Hence contradiction. 2 Hint: the following statement solves your problem: Statement: If P_1,\dots,P_n\in\mathcal H are orthogonal projections such that \sum_{i=1}^n P_i is again an orthogonal projection, then P_i are pairwise orthogonal, i.e. P_iP_j=0,\ i\neq j. Edit. Proof. If Q,R are projections such that Q\geq R then Q\supseteq R. Indeed, let x\in Ran R. ... 2 With the adjoint in the place you put it, the assertion is false: consider M=B(H), a\in M the unilateral shift. Then the range projection of a is I-E_{11}, while the range projection of (a^*a)^{1/2}=I is I. But it is true that a^* and (a^*a)^{1/2} have the same range projection. The fastest way to see it is probably to use the polar ... 2 Use$$ \sigma(ab) \cup \{0\} = \sigma(ba) \cup \{0\} .$$See Do X(X'X)^{-1}(X'X)^{-1}X' and (X'X)^{-1} have the same non-zero eigenvalues? for the proof of this (although if a or b is invertible, as it would be if a and b are positive, then \sigma(ab) = \sigma(ba) can be proved much more easily.) Then \sigma(ab) = ... 1 Note that the matrices$$A = \begin{pmatrix}x + iy & 0 \\ 0 & x-iy\end{pmatrix}\ \ \ \ \text{and} \ \ \ \ \ B = \begin{pmatrix}x & -y \\ y & x\end{pmatrix}$$are similar through the following$$\begin{pmatrix}i & -i \\ 1 & 1\end{pmatrix}\begin{pmatrix}x + iy & 0 \\ 0 & x-iy\end{pmatrix}\begin{pmatrix}-\frac{i}{2} & ...

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It is not hard to see that it is as you say when $\mathcal A$ is a von Neumann algebra. But there are C$^*$-algebras with few to no projections, and so none of those masas can live there. Consider for instance $C^*_r(\mathbb F_2)$, the reduced C$^*$-algebra of the free group on two generators $\mathbb F=\langle a,b\rangle$. This algebra is known to be ...

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Spectral theorem for a self-adjoint operator is often formulated as $a=\int\lambda dE_\lambda,$ where $E_\lambda:\mathbb R\to B(H)$ is the spectral resolution of $a=a^*\in B(H).$ In your book $E(\cdot\ ;a)$ is the spectral measure (function of Borel subsets $\mathbb R$) associated with $a.$ The mapping $\lambda\mapsto E_\lambda=E((-\infty,\lambda];a)$ is now ...

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I suppose $P=E$ in your notation. Only need to check $TE=E$ where $T$ is the central carrier of $G+(I-C_G)E$. Toward this, first use definition of $T$' we have $T(G+(I-C_G)E)=G+(I-C_G)E$, multiply both side by $G$, we get $TG=G$, so $C_G\leq T$, then use this relation and $TG=G$ to simplify $T(G+(I-C_G)E)=G+(I-C_G)E$, you get what we want, i.e., $TE=E$

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It is more or less similar to how we show division Banach algebra over complex number is $\mathbb{C}$. Suppose we can find a nonscalar element $a$ in the center of the prime C star algebra, say $A$, consider the ring homomorphism from $A$ to $A(a-\lambda)$ by sending $x$ to $x(a-\lambda)$, where $\lambda$ is one spectrum point of $a$, then if $ker$ is ...

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Let $C(D)$ denote the cone on a C* algebra $D$. Recall that $C(D)$ is contractible. Consider the natural quotient map $\pi : T\to C(M/B)$, then $$\ker(\pi) = \{f \in T : f(t) \in B \quad\forall t\} = \{f \in C([0,1],B) : f(0) \in A\cap B\}$$ Then one has $$0 \to C([0,1],A\cap B) \to \ker(\pi) \to C(B/A\cap B)\to 0$$ And inclusion map \$C([0,1],A\cap B) ...

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