Jordansche Normalform/3/Potenz/Formel/Aufgabe/Lösung

Es ist

${}{\begin{aligned}{\begin{pmatrix}{\frac {1}{2}}&1&0\\0&{\frac {1}{2}}&1\\0&0&{\frac {1}{2}}\end{pmatrix}}^{n}&={\left({\begin{pmatrix}{\frac {1}{2}}&0&0\\0&{\frac {1}{2}}&0\\0&0&{\frac {1}{2}}\end{pmatrix}}+{\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}\right)}^{n}\\&={\begin{pmatrix}{\frac {1}{2}}&0&0\\0&{\frac {1}{2}}&0\\0&0&{\frac {1}{2}}\end{pmatrix}}^{n}+n{\begin{pmatrix}{\frac {1}{2}}&0&0\\0&{\frac {1}{2}}&0\\0&0&{\frac {1}{2}}\end{pmatrix}}^{n-1}{\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}+{\binom {n}{2}}{\begin{pmatrix}{\frac {1}{2}}&0&0\\0&{\frac {1}{2}}&0\\0&0&{\frac {1}{2}}\end{pmatrix}}^{n-2}{\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}^{2}\\&={\begin{pmatrix}{\frac {1}{2^{n}}}&0&0\\0&{\frac {1}{2^{n}}}&0\\0&0&{\frac {1}{2^{n}}}\end{pmatrix}}+n{\begin{pmatrix}{\frac {1}{2^{n-1}}}&0&0\\0&{\frac {1}{2^{n-1}}}&0\\0&0&{\frac {1}{2^{n-1}}}\end{pmatrix}}{\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}+{\binom {n}{2}}{\begin{pmatrix}{\frac {1}{2^{n-2}}}&0&0\\0&{\frac {1}{2^{n-2}}}&0\\0&0&{\frac {1}{2^{n-2}}}\end{pmatrix}}{\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{2^{n}}}&0&0\\0&{\frac {1}{2^{n}}}&0\\0&0&{\frac {1}{2^{n}}}\end{pmatrix}}+n{\begin{pmatrix}0&{\frac {1}{2^{n-1}}}&0\\0&0&{\frac {1}{2^{n-1}}}\\0&0&0\end{pmatrix}}+{\binom {n}{2}}{\begin{pmatrix}0&0&{\frac {1}{2^{n-2}}}\\0&0&0\\0&0&0\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{2^{n}}}&{\frac {n}{2^{n-1}}}&{\frac {\binom {n}{2}}{2^{n-2}}}\\0&{\frac {1}{2^{n}}}&{\frac {n}{2^{n-1}}}\\0&0&{\frac {1}{2^{n}}}\end{pmatrix}}\\&={\frac {1}{2^{n}}}{\begin{pmatrix}1&2n&2n(n-1)\\0&1&2n\\0&0&1\end{pmatrix}}.\end{aligned}}$

Zur gelösten Aufgabe