Typo fixes.
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Bokuan Li
2026-10-20 10:10:31 -04:00
parent f41c62ed3a
commit 8f4bc16df1
3 changed files with 5 additions and 5 deletions
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@@ -94,7 +94,7 @@ The following is a slight generalisation of one form of the above example, but s
\norm{\phi}_{\text{cb}} &\le \norm{\phi}_{S^*}
\end{align*}
(2): If $\xi = \eta$, then
(2): If $\xi = \eta$, then for any $a \in A$ with $a \ge 0$,
\[
\sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0
\]
@@ -221,7 +221,7 @@ The following is a slight generalisation of one form of the above example, but s
\begin{example}[Reduction Map]
\label{example:matric-reduction-map}
Let $\phi: M_n(\complex) \to M_n(\complex)$ be defined by
Let $n \ge 2$ and $\phi: M_n(\complex) \to M_n(\complex)$ be defined by
\[
\phi(x) = \text{Tr}(x)1_{M_n(\complex)} - x
\]
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@@ -308,7 +308,7 @@
so $(K, U)$ is a unitary power dilation of $T$.
(U): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $(\pi, V, K)$ is minimal if and only if $\bigcup_{n \in \integer}U^n H$ is dense in $K$, if and only if $U$ admits no non-trivial reducing subspace.
(U): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $(\pi, V, K)$ is minimal if and only if the span of $\bigcup_{n \in \integer}U^n H$ is dense in $K$, if and only if $U$ admits no non-trivial reducing subspace.
By \hyperref[uniqueness of minimal Stinespring representations]{proposition:stinespring-minimal-uniqueness}, such a dilation is unique.
\end{proof}
@@ -444,7 +444,7 @@
\normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n} = f_{\xi, \eta}(z_1, \cdots, z_n)
\]
By maximality, $f_{\xi, \eta}$ achieves its maximum modulus at $(z_1, \cdots, z_n)$, so (1) implies that $(z_1, \cdots, z_n) \in \partial B$. At which point, $|z_j| = 1$ for all $1 \le j \le n$, and $\text{diag}(z_1, \cdots, z_n)$ is unitary. Therefore $T(z_1, \cdots, z_n) = U\text{diag}(z_1, \cdots, z_n)V$ is also unitary.
By maximality, $f_{\xi, \eta}$ achieves its maximum modulus at $(z_1, \cdots, z_n)$, so (1) implies that $|z_j| = 1$ for all $1 \le j \le n$ and $\text{diag}(z_1, \cdots, z_n)$ is unitary. Therefore $T(z_1, \cdots, z_n) = U\text{diag}(z_1, \cdots, z_n)V$ is also unitary.
(3): Since $(s_1, \cdots, s_n) \in B$, (2) implies that
\begin{align*}
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@@ -116,7 +116,7 @@
\begin{align*}
A &= (I - R)^{-1} + (I - R^*)^{-1} - I \\
&= (I - R)^{-1}(I - R^* + I - R - (I - R)(I - R)^*)(I - R^*)^{-1} \\
&= (I - R)^{-1}(I - RR^*)(1 - R^*)^{-1} \ge 0
&= (I - R)^{-1}(I - RR^*)(I - R^*)^{-1} \ge 0
\end{align*}
\end{proof}