Proposition 6.1.13 | The Eigenspace

Proposition 6.1.13.label Using the same notation as in Lemma 6.1.7, if a $C^{*}$-algebra $A$ is not unital, then there exists a norm in $A_{I}$ which makes $A_{I}$ a $C^{*}$-algebra.

Proof. Since $A$ is an ideal of $A_{I}$, we put $L_{x}y=x$ for $x\in A_{I}$ and $y\in A$. By Proposition 6.1.4, we have $\norm{x}{}=\norm{L_x}{}$ for each $x\in A$ (by taking an equivalent norm). Suppose $\norm{L_x}{}=0$ for some $x=x'+\lam 1$, $x'\in A,\lam \in\com,\lam\neq 0$. For every $y\in A$, we have

\begin{align*}0=\frac{1}{\lam}xy=\frac{1}{\lam}x'y+y\end{align*}

so $-\frac{1}{\lam}x'$ is the left identity of $A$; hence $\parens{-\frac{1}{\lam}x'}^{*}$ is a right identity of $A$. Therefore

\begin{align*}-\frac{1}{\lam}x'=\parens{-\frac{1}{\lam}x'}\parens{-\frac{1}{\lam}x'}^{*}=\parens{-\frac{1}{\lam}x'}^{*}\end{align*}

which means that $-\frac{1}{\lam}x'$ must be the identity of $A$, contradicing that $A$ is non-unital. Thus the norm from Proposition 6.1.4 is also a norm for $A_{I}$. Since $A$ is complete and of codimension $1$ in $A_{I}$, by Lemma 6.1.12 $A_{I}$ is complete. For any $x\in A_{I}$ and $\e>0$, there exists $y\in A$ with

\begin{align*}\norm{xy}{}\geq (1-\e)\norm{x}{}&&\norm{y}{}\leq 1\end{align*}

But $xy\in A$ thus

\begin{align*}\norm{x^*x}{}\geq\norm{y^*(x^*x)y}{}=\norm{(xy)^*(xy)}{}=\norm{xy}{}^{2}\geq (1-\e)^{2}\norm{x}{}^{2}\end{align*}

Therefore

\begin{align*}\norm{x}{}^{2}\leq\norm{x^*x}{}\leq\norm{x^*}{}\norm{x}{}=\norm{x}{}^{2}\end{align*}

so $\norm{x^*x}{}=\norm{x^2}{}$$\square$

The Eigenspace

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The Eigenspace

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