Small fixes.
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Bokuan Li
2026-10-18 22:37:26 -04:00
parent 96fa212256
commit 8543c25b4e
2 changed files with 2 additions and 2 deletions
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@@ -373,7 +373,7 @@ A significant property of Hilbert spaces is that every closed subspace is comple
\label{definition:hilbert-quadratic-function} \label{definition:hilbert-quadratic-function}
Let $H$ be a complex Hilbert space and $q: H \to \complex$, then $q$ is \textbf{quadratic} if: Let $H$ be a complex Hilbert space and $q: H \to \complex$, then $q$ is \textbf{quadratic} if:
\begin{enumerate} \begin{enumerate}
\item For each $\lambda \in \complex$ and $\xi \in H$, $q(\lambda x) = |\lambda|^2 q(\xi)$. \item For each $\lambda \in \complex$ and $\xi \in H$, $q(\lambda \xi) = |\lambda|^2 q(\xi)$.
\item (\textbf{Parallelogram Law}) For each $\xi, \eta \in H$, \item (\textbf{Parallelogram Law}) For each $\xi, \eta \in H$,
\[ \[
q(\xi + \eta) + q(\xi - \eta) = 2q(\xi) + 2q(\eta) q(\xi + \eta) + q(\xi - \eta) = 2q(\xi) + 2q(\eta)
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@@ -70,7 +70,7 @@
(T2) $\Rightarrow$ (T1): For each $n \in \natp$, (T2) $\Rightarrow$ (T1): For each $n \in \natp$,
\begin{align*} \begin{align*}
0 &\le \braks{\sum_{j = 1}^n E_{1j} \otimes z^j}^*\braks{\sum_{j = 1}^n E_{1j} \otimes z^j} = \sum_{j, k = 1}^n E_{jk} \otimes z^{j-k} \\ 0 &\le \braks{\sum_{j = 1}^n E_{j1} \otimes z^j}\braks{\sum_{j = 1}^n E_{j1} \otimes z^j}^* = \sum_{j, k = 1}^n E_{jk} \otimes z^{j-k} \\
0 &\le \phi\braks{\sum_{j, k = 1}^n E_{jk} \otimes z^{j-k}} = \sum_{j, k = 1}^n E_{jk} \otimes A_{j - k} 0 &\le \phi\braks{\sum_{j, k = 1}^n E_{jk} \otimes z^{j-k}} = \sum_{j, k = 1}^n E_{jk} \otimes A_{j - k}
\end{align*} \end{align*}
\end{proof} \end{proof}