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übungsblatt 2 fortsetzung

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Daniel Bahrdt 2008-07-10 11:48:03 +02:00
parent 74bedfe3e3
commit 87d8942a0a
1 changed files with 4 additions and 4 deletions
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ueb2.tex
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@ -28,9 +28,9 @@
&= a_n^\ast \braket{a_n}{a_m} a_m \\ &= a_n^\ast \braket{a_n}{a_m} a_m \\
&= e^{-\i \alpha_n} \cdot e^{\i \alpha_m} \cdot \braket{a_n}{a_m} \\ &= e^{-\i \alpha_n} \cdot e^{\i \alpha_m} \cdot \braket{a_n}{a_m} \\
&= e^{\i ( \alpha_m - \alpha_n)} \braket{a_n}{a_m} \\ &= e^{\i ( \alpha_m - \alpha_n)} \braket{a_n}{a_m} \\
&\Rightarrow \\
\braket{a_n}{a_m} &= 0
\end{align} \end{align}
Da dies für alle Eigenvektoren gelten muss, also auch für Eigenvektoren, für die $e^{\i ( \alpha_m - \alpha_n)} \neq 1$, folgt:
$\braket{a_n}{a_m} = 0$
\subsection*{b)} \subsection*{b)}
@ -116,9 +116,9 @@ Dann ist $T^{-1}AT$ die Basistransformation von der A-Basis in die T-Basis.
Zu zeigen: $\det(e^A = e^{\tr(A))$ Zu zeigen: $\det(e^A = e^{\tr(A))$
\begin{align} \begin{align}
g(t) &= \det(e^{At}) \\ g(t) &= \det(e^{At}) \\
&\stackrel{tailor}{} \det(1 + At + \bigO(t^2)) \\ &\stackrel{tailor}{=} \det(1 + At + \bigO(t^2)) \\
&= 1 + \tr(A) + \bigO(t^2) \\ &= 1 + \tr(A) + \bigO(t^2) \\
&\stackrel{tailor ``rückwärs''}{} e^{\tr(A)t} &\stackrel{tailor ``rückwärs''}{=} e^{\tr(A)t}
\end{align} \end{align}
\begin{align} \begin{align}