add stupid calculations in la4 and update theo5
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@@ -3,6 +3,8 @@
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\title{Lineare Algebra 2: Übungsblatt 4}
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\title{Lineare Algebra 2: Übungsblatt 4}
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\author{Dominik Daniel, Christian Merten}
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\author{Dominik Daniel, Christian Merten}
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\usepackage[]{gauss}
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\begin{document}
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\begin{document}
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\punkte[16]
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\punkte[16]
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@@ -57,4 +59,155 @@
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\end{enumerate}
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\end{enumerate}
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\end{aufgabe}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}[(a)]
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\item Kurze Rechung ergibt
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\begin{align*}
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P_A &= \begin{gmatrix}[p] t-10 & 11 & 11 & 32 \\
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1 & t & 2 & - 4 \\
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-1 & 1 & t-1 & 4 \\
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-2 & 2 & 2 & t+6
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\rowops
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\swap{0}{1}
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\end{gmatrix}
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\sim
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\begin{gmatrix}[p]
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1 & t & 2 & - 4 \\
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t-10 & 11 & 11 & 32 \\
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-1 & 1 & t-1 & 4 \\
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-2 & 2 & 2 & t+6
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\rowops
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\add[-(t-10)]{0}{1}
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\add{0}{2}
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\add[2]{0}{3}
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\colops
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\add[-t]{0}{1}
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\add[-2]{0}{2}
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\add[4]{0}{3}
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\end{gmatrix} \\
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&\sim
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\begin{gmatrix}[p]
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1 & 0 & 0 & 0 \\
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0 & 11-t^2 + 10t & -8 + 4t \\
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0 & t+1 & t+1 & 0 \\
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0 & 2+2t & 6 & t-2
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\rowops
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\swap{1}{3}
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\colops
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\swap{1}{2}
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\end{gmatrix}
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\sim
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\begin{gmatrix}[p]
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1 & 0 & 0 & 0 \\
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0 & 6 & 2+2t & t-2 \\
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0 & t+1 & t+1 & 0 \\
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0 & 31-2t & 11-t^2 + 10t & -8+4t
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\rowops
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\mult{1}{\cdot \frac{1}{6}}
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\end{gmatrix} \\
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&\sim
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\begin{gmatrix}[p]
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1 & 0 & 0 & 0 \\
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0 & 1 & \frac{1}{3} + \frac{1}{3}t & \frac{1}{6}t - \frac{1}{3} \\
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0 & t+1 & t+1 & 0 \\
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0 & 31-2t & 11-t^2+10t & -8+4t
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\rowops
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\add[-(t+1)]{1}{2}
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\add[-(31-2t)]{1}{3}
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\colops
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\add[-\left(\frac{1}{3} + \frac{1}{3}t \right)]{1}{2}
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\add[-\left(\frac{1}{6}t - \frac{1}{3} \right)]{1}{3}
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\end{gmatrix}
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\sim
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\begin{gmatrix}[p]
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E_2 & 0 & 0 \\
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0 & -\frac{1}{3} t^2+\frac{2}{3} + \frac{1}{3}t & -\frac{1}{6}t^2 + \frac{1}{6}t + \frac{1}{3} \\
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0 & \frac{2}{3} + \frac{1}{3}t - \frac{1}{3}t^2 & \frac{1}{3}t^2 - \frac{11}{6}t + \frac{7}{3}
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\rowops
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\add[-1]{1}{2}
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\colops
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\add[-\frac{1}{2}]{1}{2}
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\end{gmatrix} \\
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&\sim
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\begin{gmatrix}[p]
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E_2 & 0 & 0 \\
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0 & -\frac{1}{3}t^2 + \frac{1}{3}t + \frac{2}{3} & 0 \\
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0 & 0 & \frac{1}{2}t^2 - 2t + 2
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\rowops
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\mult{1}{\cdot (-3)}
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\mult{2}{\cdot 2}
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\end{gmatrix}
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\sim
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\begin{gmatrix}[p]
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E_2 & 0 & 0 \\
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0 & (t+1)(t-2) & 0 \\
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0 & 0 & (t-2)^2
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\rowops
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\add{1}{2}
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\colops
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\add[-1]{1}{2}
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\end{gmatrix} \\
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&\sim
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\begin{gmatrix}[p]
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E_2 & 0 & 0 \\
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0 & (t+1)(t-2) & -(t+1)(t-2) \\
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0 & (t+1)(t-2) & -3t + 6
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\rowops
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\swap{1}{2}
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\colops
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\swap{1}{2}
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\end{gmatrix}
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\sim
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\begin{gmatrix}[p]
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E_2 & 0 & 0 \\
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0 & -3t + 6 & (t-2)(t+1) \\
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0 & -(t-2)(t+1) & (t-2)(t+1)
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\rowops
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\mult{1}{\cdot \left( -\frac{1}{3} \right)}
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\colops
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\mult{2}{\cdot 3}
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\end{gmatrix} \\
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&\sim
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\begin{gmatrix}[p]
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E_2 & 0 & 0 \\
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0 & t - 2 & -(t-2)(t+1) \\
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0 & -(t-2)(t+1) & 3(t-2)(t+1)
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\rowops
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\add[t+1]{1}{2}
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\colops
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\add[t+1]{1}{2}
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\end{gmatrix}
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\sim
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\begin{gmatrix}[p]
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E_2 & 0 & 0 \\
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0 & t-2 & 0 \\
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0 & 0 & (t-2)^2(t+1)
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\end{gmatrix}
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.\end{align*}
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Damit folgen als Invariantenteiler: $c_1 = c_2 = 1$, $c_3 = t- 2$ und $c_4 = (t-2)^2(t+1)$.
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Die Determinantenteiler sind damit $d_1 = 1$, $d_2 = 1$, $d_3 = t-2$ und $d_4 = (t-2)^{3}(t+1)$.
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\item Hier ist sofort ersichtlich:
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\begin{align*}
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\text{det}(P_B)
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= \begin{gmatrix}[v]
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t+5 & 3 & -5 \\
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0 & t-1 & 1 \\
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8 & 4 & t-7
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\end{gmatrix}
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= (t-1)^{3}
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\neq
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(t-2)(t-1)(t+1)
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=
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\begin{gmatrix}[v]
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t+3 & -8 & -12 \\
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-1 & t+2 & 3 \\
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2 & -4 & t-7
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\end{gmatrix}
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= \text{det}(P_C)
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.\end{align*}
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Damit ist $d_3_{B} \neq d_3_{C}$, also sind nach Invariantenteilersatz
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$B$ und $C$ nicht ähnlich.
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\end{enumerate}
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\end{aufgabe}
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\end{document}
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\end{document}
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Binary file not shown.
@@ -36,7 +36,7 @@
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H &= \frac{p_{\vartheta}^2}{mR^2} - L \\
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H &= \frac{p_{\vartheta}^2}{mR^2} - L \\
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&= \frac{p_{\vartheta}^2}{mR^2}
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&= \frac{p_{\vartheta}^2}{mR^2}
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- \frac{1}{2} mR^2 (\dot{\vartheta}^2 + \omega^2\sin^2\vartheta) + mg R \cos\vartheta \\
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- \frac{1}{2} mR^2 (\dot{\vartheta}^2 + \omega^2\sin^2\vartheta) + mg R \cos\vartheta \\
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&= \frac{p_{\vartheta}^2}{2mR^2} - \frac{1}{2} \omega^2 \sin^2\vartheta + mgR\cos\vartheta
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&= \frac{p_{\vartheta}^2}{2mR^2} - \frac{1}{2} R^2 \omega^2 \sin^2\vartheta + mgR\cos\vartheta
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.\end{align*}
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.\end{align*}
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\item Es liegt hier zwar ein Potential vor, allerdings ist
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\item Es liegt hier zwar ein Potential vor, allerdings ist
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die kinetische Energie des Systems gegeben als
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die kinetische Energie des Systems gegeben als
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@@ -47,8 +47,9 @@
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an der zeitabhängigen Zwangsbedingung.
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an der zeitabhängigen Zwangsbedingung.
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Wegen $\frac{\partial H}{\partial t} = 0$ ist die Hamilton Funktion zeitlich
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Wegen $\frac{\partial H}{\partial t} = 0$ ist die Hamilton Funktion zeitlich
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erhalten. Außerdem ist $\dot{w} = 0$, also ist das System
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erhalten. Da die Zwangskräfte von der Zeit abhängen, ist das
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invariant gegenüber Zeittranslation. Damit folgt Energieerhaltung.
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System nicht invariant gegenüber Zeittranslation, also ist die Energie nicht
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erhalten.
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\item Für die kanonischen Gleichungen folgt
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\item Für die kanonischen Gleichungen folgt
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\begin{align*}
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\begin{align*}
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\frac{\partial H}{\partial p_{\vartheta}} &= \frac{2 p_{\vartheta}}{mR^2} - \frac{p_{\vartheta}}{mR^2}
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\frac{\partial H}{\partial p_{\vartheta}} &= \frac{2 p_{\vartheta}}{mR^2} - \frac{p_{\vartheta}}{mR^2}
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