# Tag Info

3

A maximal ideal is by definition a proper ideal (which happens to not be strictly contained in any other proper ideal). If $I$ is an ideal and an element $x\in I$ is invertible, then $x^{-1}x=e\in I$ so $a=a\cdot e\in I$ for all $a\in A$, so $I=A$. So no proper ideal (and in particular, no maximal ideal) can contain an invertible element.

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 ...

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 ...

3

It's self-adjoint projections in a $C^*$ algebra that must have norm $1$! Which observation actually leads to the example you want: $$\left[\begin{matrix}1&a\\0&0\end{matrix}\right].$$ So there you are, the algebra of $2\times 2$ matrices. For a commutative Banach algebra with identity and large idempotents: Say $M_n$ is the span of ...

2

As Gro-Tsen commented, the fact that $\Gamma$ is an isometry implies its image is closed because $\mathfrak{U}$ is a complete metric space (since it is closed in $C(X)$). Any subset of a metric space that is complete as a metric space is automatically closed, since a sequence that converged to a point outside the set would be a Cauchy sequence with no limit ...

2

Let $A$ be a finite-dimensional Banach algebra. The group $K_0(A)$ only depends on the underlying ring structure of $A$. In what follows, $A$ can be any finite-dimensional algebra over an arbitrary field $k$, well $A\neq 0$. The map $\{\text{finitely generated projective$A$-modules}\}\mapsto \mathbb Z$ given by $P\mapsto\dim_kP$ is well defined (being $A$ ...

2

That $\Gamma(y)$ should be closed doesn't follow from the fact that $\Gamma$ is a closed subalgebra. Example. Let $\mathbb N^+ = \mathbb N \setminus \{0\}$ be the set of positive integers. Consider $X = \ell^2(\mathbb N^+)$ and let $R : X \to X$ be the following weighted right shift:  (Rx)(n) = \begin{cases} \quad 0, & \quad\text{if $n = 1$}; ...

1

Let $\{\phi\}$ be a net of pure states on $A$ and assume that it is $w^*$-convergent to $\phi$. Clearly $\phi$ is a positive linear functional on $A$. To prove $\phi$ is a pure state, it is enough to show that support of $\phi$ is a minimal projection in $A^{**}$. Assume $q$ is a non-trivial projection majorized by the support of $\phi$. We have that ...

1

The spectrum of $a$ is a compact set that does not contain $0$. So there is a disk $D$ around $0$ with $D\cap\sigma(a)=\emptyset$. Thus, on $\sigma(a)$, $f:t\longmapsto 1/t$ is continuous, so $f\in C(\sigma(a))$. Then $f(a)\in C^*(a)$ via the Gelfand transform. Or, even easier, you could check that $f$ is analytic on $\sigma(a)$, and so $a^{-1}$ belongs ...

1


Only top voted, non community-wiki answers of a minimum length are eligible