Fix layout
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@ -1530,7 +1530,7 @@ Angenommen \(\alpha - \lambda \id: V \to V\) nilpotent. Dann besitzt \(\alpha\)
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$\implies (\alpha - \lambda \id)(v) = (\alpha - \lambda \id)^k (\alpha - \lambda \id)(w) \in \im(\alpha - \lambda \id)^k \checkmark$
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\end{itemize}
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\item Es gilt $\dim(V) = \dim(V_1) + \dim(V_2)$ nach der Dimensionsformel. Es genügt also zu zeigen, dass
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$V_1 \cap V_2 = \{0\}$. Sei $v\in V_1 \cap V_2$
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$V_1 \cap V_2 = \{0\}$. \\ Sei $v\in V_1 \cap V_2$
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\begin{align*}
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& \underbrace{\implies}_{v\in V_2} \exists w\in V: v = (\alpha - \lambda \id)^k(w) \\
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& \underbrace{\implies}_{v\in V_1} (\alpha - \lambda \id)^{2k}(w) = 0 \\
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