I am reading this book and in the 2nd chapter (II.1.6.4 Corrollary) the author proved the following:
But in the book the author didn't defined explicitly the terms "semisimple algebra"/"radical". The definitions in this context what I know is the one given in the book Simmons, Introduction to Topology and Modern Analysis, (in Chapter Twelve; page-314). So I assumed the definition as follows:
Definition (radical). Let $A$ is a Banach algebra then the radical of $A$ is defined to be the intersection of all "left maximal ideals".
Semisimple. Radical of $A$ is $(0)$.
But in the Corollary II.1.6.4 above, why do $x^*x\in R$ imply $x^*x$ is nilpotent?
I cannot prove it assuming the definition of radical given in Simmons.
Any help is appreciated. Thanks
Well, the result is true - everything in the Jacobson radical of a unital C*-algebra is nilpotent because it's just zero - but I'm not following Blackadar's reasoning.
One characterization of the radical of a unital C*-algebra (or general Banach algebra) $A$ is as the set of those elements $a\in A$ for which $1 + xay$ is invertible for all $x, y\in A$. In particular, for $\lambda\in\mathbb{C}\backslash\{0\}$, $1 - a/\lambda$ is invertible, so that $\lambda\not\in\sigma(a)$, whence $\sigma(a) = \{0\}$. Since $x^*x\in R$, $\sigma(x^*x) = 0$, and $\|x\|= \|x^*x\|^{1/2} = \text{spr}(x^*x)^{1/2} = 0$.