# 4-Zahlen-Matrix/Nullstelle/Eigenvektor/Aufgabe/Lösung

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Es ist

{\displaystyle {}{\begin{aligned}{\begin{pmatrix}-1&1&0&0\\0&-1&1&0\\0&0&-1&1\\-1&0&0&1\end{pmatrix}}{\begin{pmatrix}{\frac {1}{(1+\lambda )^{3}}}\\{\frac {1}{(1+\lambda )^{2}}}\\{\frac {1}{(1+\lambda )}}\\1\end{pmatrix}}&={\begin{pmatrix}-{\frac {1}{(1+\lambda )^{3}}}+{\frac {1}{(1+\lambda )^{2}}}\\-{\frac {1}{(1+\lambda )^{2}}}+{\frac {1}{(1+\lambda )}}\\-{\frac {1}{(1+\lambda )}}+1\\-{\frac {1}{(1+\lambda )^{3}}}+1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {-1+1+\lambda }{(1+\lambda )^{3}}}\\{\frac {-1+1+\lambda }{(1+\lambda )^{2}}}\\{\frac {-1+1+\lambda }{1+\lambda }}\\{\frac {-1+(1+\lambda )^{3}}{(1+\lambda )^{3}}}\end{pmatrix}}\\&={\begin{pmatrix}{\frac {\lambda }{(1+\lambda )^{3}}}\\{\frac {\lambda }{(1+\lambda )^{2}}}\\{\frac {\lambda }{1+\lambda }}\\{\frac {\lambda ^{3}+3\lambda ^{2}+3\lambda }{(1+\lambda )^{3}}}\end{pmatrix}}\\&=\lambda {\begin{pmatrix}{\frac {1}{(1+\lambda )^{3}}}\\{\frac {1}{(1+\lambda )^{2}}}\\{\frac {1}{1+\lambda }}\\{\frac {\lambda ^{2}+3\lambda +3}{(1+\lambda )^{3}}}\end{pmatrix}}\\&=\lambda {\begin{pmatrix}{\frac {1}{(1+\lambda )^{3}}}\\{\frac {1}{(1+\lambda )^{2}}}\\{\frac {1}{1+\lambda }}\\{\frac {\lambda ^{2}+3\lambda +3}{\lambda ^{3}+3\lambda ^{2}+3\lambda +1}}\end{pmatrix}}\\&=\lambda {\begin{pmatrix}{\frac {1}{(1+\lambda )^{3}}}\\{\frac {1}{(1+\lambda )^{2}}}\\{\frac {1}{1+\lambda }}\\{\frac {\lambda ^{2}+3\lambda +3}{\lambda ^{2}+3\lambda +3}}\end{pmatrix}}\\&=\lambda {\begin{pmatrix}{\frac {1}{(1+\lambda )^{3}}}\\{\frac {1}{(1+\lambda )^{2}}}\\{\frac {1}{1+\lambda }}\\1\end{pmatrix}}.\end{aligned}}}