Lemma 41.3.6.label Let $A$ be a unital $C^{*}$-algebra, then for any $x \in A$ and $y \in A_{sa}$, $x^{*}x \le y$ if and only if
\[\begin{bmatrix}1&x \\ x^{*}&y\end{bmatrix} \ge 0\]
in $M_{2}(A)$. In particular, $\norm{x}_{A} \le 1$ if and only if
\[\begin{bmatrix}1&x \\ x^{*}&1\end{bmatrix} \ge 0\]
in $M_{2}(A)$.
Proof, [Lemma 3.1, Po02]. Using the Gelfand-Naimark Theorem, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$,
\[\angles{ \begin{bmatrix}1&x \\ x^{*}&y\end{bmatrix}\begin{bmatrix}\xi \\ \eta\end{bmatrix}, \begin{bmatrix}\xi \\ \eta\end{bmatrix}}_{H^2}= \norm{\xi}_{H}^{2} + \dpn{x\eta, \xi}{H}+ \dpn{\xi, x\eta}{H}+ \dpn{y\eta, \eta}{H}\]
($\Rightarrow$): If $y \ge x^{*}x$, then by the Cauchy-Schwarz inequality,
\begin{align*}\angles{ \begin{bmatrix}1&x \\ x^{*}&y\end{bmatrix}\begin{bmatrix}\xi \\ \eta\end{bmatrix}, \begin{bmatrix}\xi \\ \eta\end{bmatrix}}_{H^2}&\ge \norm{\xi}_{H}^{2} + \dpn{x\eta, \xi}{H}+ \dpn{\xi, x\eta}{H}+ \dpn{x\eta, x\eta}{H}\\&\ge \norm{\xi}_{H}^{2} - 2\norm{\xi}_{H}\norm{x\eta}_{H} + \norm{x\eta}_{H}^{2} \ge 0\end{align*}
($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$,
\begin{align*}0&\le \angles{ \begin{bmatrix}1&x \\ x^{*}&y\end{bmatrix}\begin{bmatrix}-x\eta \\ \eta\end{bmatrix}, \begin{bmatrix}-x\eta \\ \eta\end{bmatrix}}_{H^2}= \norm{x\eta}_{H}^{2} -2\dpn{x\eta, x\eta}{H}+ \dpn{y\eta, \eta}{H}\\&\le \dpn{y\eta, \eta}{H}- \dpn{x\eta, x\eta}{H}= \dpn{(y - x^*x)\eta, \eta}{H}\end{align*}
so $y \ge x^{*}x$.$\square$
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