New answers tagged

1

Assume for simplicity that $A$ is unital. If you had that $K=H^{(n)}$, then you know that the projections onto each copy of the direct sum would be $E_{jj}$, where $E_{kj}$ denote the canonical matrix units in $M_n(A)$ (this where "unital" is useful, but it can be dispensed with). Also, your last formula suggests what you need to do: you can define ...


2

Identifying $A$ as a $C^*$-subalgebra of $B(\mathcal H)$ for some Hilbert space $\mathcal H$, and hence identifying $M_n(A)$ as a $*$-subalgebra of $M_n(B(\mathcal H))$, observe that $\sum_{i,j}y_i^*x_{ij}y_j$ can be identified as an operator in $B(\mathcal H)$. So the question boils down to prove that if $\sum_{i,j}y_i^*x_{ij}y_j\in B(\mathcal H)$ is ...


3

Yes, this is standard. Let $T\in A$ selfadjoint, i.e. with $T=T^*$. Note first that $\phi(T)$ is real: since $T+\|T\|\,\text{id}$ is positive, we have that $$ \phi(T)+\|T||\in\mathbb R, $$ so $\phi(T)\in \mathbb R$. Now, as $-T+\|T\|\,\text{id}\geq0$, we get $-\phi(T)+\|T\|\geq0$, so $$\phi(T)\leq\|T\|.$$ Since $-T$ is also selfadjoint, we can also get ...


0

Since $V$ is compact, $C^*(V)$ is the set of norm limits of polynomials in $V$ and $V^*$. All these polynomials are of the form $\lambda I+K$, with $K$ compact. Now, if $T\in C^*(V)$, then $T=\lim \lambda_n I+K_n$. If $T$ is compact, we have nothing to prove; so we assume that $T$ is not compact. This implies that the sequence $\lambda_n$ does not ...


0

The lower bound is zero. Consider the matrices $$ A = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix}$$ and $$ B_\lambda = \begin{pmatrix} \lambda & 0 \\ 0 & 1-\lambda \end{pmatrix}.$$ For $\lambda \ne 1/2$ you have $A \ne B_\lambda$ and $\|A - B_\lambda\|_1=2 \, |1/2 - \lambda|$. For $\lambda \to 1/2$, this converges towards $0$.


0

One may check that: No sub-basic open set in the strong operator topology is an open set in the weak operator topology.


0

Yes. Let $1$ be the unit of $A^{**}$. Let us put $$e=p\wedge q~~~,~~~f_1=(1-p\wedge q)\wedge p~~~,~~~f_2=(1-p\wedge q)\wedge q$$ Then $\{e,f_1,f_2\}$ are pairwise orthogonal and $$p=e+f_1~~~,~~~q=e+f_2$$ There are bounded nets $\{a_i\}$, $\{b_i\}$ and $\{c_i\}$ in $A$ with $$e=w^*-\lim a_i~~,~~f_1=w^*-\lim b_i~~,~~f_2=w^*-\lim c_i$$ such that $a_i\leq ...


0

For any $x\in\mathbb H $ we have $Ax=\lim A_nx$, with $A_nx\in\text {ran }\,A_j $. Because a Hilbert space is metric, we can do this with sequences. Thus $$ \text {ran}\,A\subset \text {cl}\left (\bigcup_n\text {ran}A_j\right). $$


0

\begin{split} \langle0^{B}|\rho^{AB}|0^{B}\rangle=\frac{1}{2}\Big(\langle0^{B}|[{|0^{A}\rangle \otimes|0^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|+|1^{A}\rangle \otimes|1^{B}\rangle\langle0^{A}|\otimes\langle0^{B}|+|0^{A}\rangle \otimes|0^{B}\rangle\langle1^{A}|\otimes\langle1^{B}|+|1^{A}\rangle ...


1

For every normed space $X$, the dual $X^*$ is $1$-complemented in $X^{***}$. Indeed, let $i:X\to X^{**}$ be the canonical embedding; then its adjoint $i^*$ is a projection of norm $1$ of $X^{***}$ to $X^*$. Simply put, it takes a functional $\phi:X^{**}\to \mathbb{C}$ and composes it with $i$. In particular, the above applies to $\mathbb{B}(\mathbb{H})$, ...


0

This is like Lemma 5.1.4 in Kadison & Ringrose. The key is to use the weak-operator compactness of the unit ball of $\mathcal{B}(\mathbb{H})$, Theorem 5.1.3 (p. 306). You then use the general topological fact that compactness is equivalent to "every net has a convergent subnet." Alas, I'm not sure I can shed much light on the proof of compactness ... ...


1

As Murphy says, the unit ball in $B(H)$ is separable and metrizable in the strong topology (see Remark 4.4.2 on pages 133-4 for details). Since a subspace of a separable metric space is separable, the unit ball in $A$ is separable in the strong topology. Now just take a countable strongly dense subset of the unit ball of $A$ and let $B$ be the C*-algebra ...


0

It can be easily shown that $B(X)$ has always nontrivial zero divisor. Thus every Banach algebra which does not have zero divisor is another example. For example the disc algebra, the algebra of all holomorphic functions on the unit disk with continuous extension to the boundary.


0

Yours answer is negative. For example consider the C*-algebra $C[0,1]$. Let us consider the (Borel) charactrisitic function $\chi_{\{Q^c\cap[0,1]\}}$ (which is in the second dual $C[0,1]^{**}$). By your formula this projection does not obtain.


0

Let $\{e_n\}$ be an orthonormal sequence in $H$ and denote $p_n$ by the rank one projection onto $\mathbb{C}e_n$. Then the positive element $x=\sum \frac{1}{n}p_n$ works. To see it, assume $x$ is a convex combination of projections $q_1,\cdots,q_k$, namely $x=t_1q_1+\cdots t_kq_k$. Then there is $j$ such that $q_jH$ contains an infinite subset $E$ of the ...


0

The Kadison Transivity theorem is just your answer. See for example (Murphy page 150).


0

Your problem is solved by the following important fact which has been proved by Arens & Kadison 1n 1968 (see their paper: Pure state and approximate identities) Theorem (Arense & Kadison). Let $a$ be an strictly positive in a C*-algebra $A$. The $\{a^{\frac{1}{n}}\}$ forms an approximate identity for $A$. In your problem, you assumed $\sigma(a)$ ...


2

First, we establish a simple result. Suppose we have two operators $A,B$ with $[A,B]=1$. Then one can easily prove via induction that more generally $[A,B^k]=k B^{k-1}$ for any natural $k$. Therefore the map $d_B$ defined as $d_B(\cdot)=[A,\cdot]$ acts as a derivative on all such terms $B^k$, and since this map is linear we furthermore have that this extends ...


1

An irreducible representation $\pi:M\to B(K)$ has range dense in $B(K)$; but, if $\pi$ is normal, the range is all of $B(K)$. As $\ker\pi$ is a weakly closed ideal of $M$, it is of the form $pM$ for some central projection $p\in M$, and we get $(1-p)M\simeq B(K)$. In other words, $M$ has sufficiently many normal irreducible representations if and only if ...


0

First of all, let us note that the minimal projections in $zA^{**}$ are just minimal projections in $A^{**}$, because as you said $z$ is the supremum of minimal projections. Let $L$ be a maximal closed left ideal in $A$. As you mentioned, the weak star closure of $L$ denoted by $L^{\circ\circ}$ is in the form of $A^{**}e$ for some projection in $A^{**}$ ...


0

As for Q1, the uniform bounded principle should be applied. Let $\{u_n\}$ be a net of operators in $B(H)$ which is also weakly convergent. Any vector $y\in H$, induces the sequence of bounded linear functionals $\phi_n$ on $H$ given by $x\to \langle u_nx,y\rangle$. Since $u_n$ is weakly convergent then $\phi_n$ is point-wise convergent and so it is also ...


3

This is far from being true, indeed, every symmetric operator has only real eigenvalues: If $\psi\in\ker(T-\lambda),\,\psi\neq 0$, then $$ \lambda\|\psi\|^2=\langle T\psi,\psi\rangle=\langle \psi,T\psi\rangle=\bar\lambda\|\psi\|^2, $$ hence $\lambda=\bar\lambda$. Now every symmetric operator that is not self-adjoint yields a counterexample to your ...


0

You will find complete descriptions of all closed ideals in the papers M. Daws, Closed ideals in the Banach algebra of operators on classical non-separable spaces, Math. Proc. Cambridge Philos. Soc. 140 (2006), no. 2, 317–332. W.B. Johnson, T. Kania, G. Schechtman, Closed ideals of operators on and complemented subspaces of Banach spaces of ...


1

Yes, it is. Condition $u^*u\ge uu^*$ is called hyponormality. Proposition: Let $M$ a subspace of $H$ and $T\in B(H)$. If $M$ is $T$-invariant, then $(T|_M)^*=PT^*|_M$, where $P$ is the orthogonal projection onto $M$. Now, if $S$ is subnormal in $H$, there is a superspace $K\ge H$ and a normal operator $N\in B(K)$ such that $N|_H=S$ and $H$ is ...


1

I detail the way WishBeLeibniz proposes. Since $U$ is normal, there is a unitary $R$ s.t. $R^*UR=diag(\lambda_j)$. Since $U$ is unitary, $\lambda_j=e^{i\theta_j}$ where $\theta_j\in\mathbb{R}$; then $R^*UR=\exp(idiag(\theta_j))$ and $U=\exp(iRdiag(\theta_j)R^*)$; finally, $H=Rdiag(\theta_j)R^*$ is hermitian and satisfies $U=e^{iH}$. Of course, $H$ is not ...


0

You can see the ideal of compact operators as the ideal generated by the finite-rank operators. You can play the same game and consider the ideal generated by those operators which have separable range.


0

Take any $0\neq x\in \mathbb{H} $ and cosider ideal $\mathcal{M} =\{ T\in \mathcal{B} (\mathbb{H} ) : T(x) =0\}.$


1

What you do is to define $\phi $ by linearity on $$\mathcal M_\phi=\text {span}\,\{a\in A^+:\ \phi (a)<\infty\}. $$ Then the inequality from your comment shows that $b^*a\in \mathcal M_\phi $ whenever $a,b\in\mathcal N_\phi $.


2

In my experience, the topic of subalgebras of $M_n(\mathbb C)$ is not part of the usual linear algebra curriculum. What you need to understand first is the form that finite-dimensional C$^*$-algebras have. A finite-dimensional C$^*$-algebra $A$ is always a finite direct sum $$\bigoplus_{k=1}^m M_{n(k)}(\mathbb C).$$ The "blocks" can be identified via the ...


2

Unless I'm misreading your definition, this is an antichain of projections in $M_2(\mathbb C)$: $$ P_t=\begin{bmatrix}t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix},\ \ t\in[0,1]. $$ This idea can be made to work easily in any type I von Neumann algebra. For types II and III, a deeper idea is needed, namely halving: given any projection ...


1

Without being too familiar with the material, I am not 100% certain, but I think that Goodearl's von Neumann regular rings discusses what you want. It definitely covers the types of VNR rings and their idempotents.



Top 50 recent answers are included