Es ist

${\displaystyle {}{\begin{pmatrix}1&1\\1&0\end{pmatrix}}{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}1\\1\end{pmatrix}}\,,}$

${\displaystyle {}{\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{2}{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{\begin{pmatrix}1\\1\end{pmatrix}}={\begin{pmatrix}2\\1\end{pmatrix}}\,,}$

${\displaystyle {}{\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{3}{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{\begin{pmatrix}2\\1\end{pmatrix}}={\begin{pmatrix}3\\2\end{pmatrix}}\,,}$

${\displaystyle {}{\begin{pmatrix}1&1\\1&0\end{pmatrix}}^{4}{\begin{pmatrix}1\\0\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{\begin{pmatrix}3\\2\end{pmatrix}}={\begin{pmatrix}5\\3\end{pmatrix}}\,.}$

b) Wir behaupten

${\displaystyle {}y_{n+1}=x_{n}\,}$

und die rekursive Beziehung

${\displaystyle {}x_{n+1}=x_{n}+x_{n-1}\,}$

mit den Anfangsbedingungen ${\displaystyle {}x_{0}=0}$ und ${\displaystyle {}x_{1}=1}$. Beides beweisen wir durch Induktion über ${\displaystyle {}n}$, wobei der Induktionsanfang klar ist. Wegen

${\displaystyle {}{\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}}=M{\begin{pmatrix}x_{n}\\y_{n}\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\end{pmatrix}}{\begin{pmatrix}x_{n}\\y_{n}\end{pmatrix}}\,}$

ist

${\displaystyle {}x_{n+1}=x_{n}+y_{n}=x_{n}+x_{n-1}\,}$

und

${\displaystyle {}y_{n+1}=x_{n}\,.}$

c) Das charakteristische Polynom zu ${\displaystyle {}M}$ ist

${\displaystyle {}\det {\begin{pmatrix}x-1&-1\\-1&x\end{pmatrix}}=(x-1)x-1=x^{2}-x-1={\left(x-{\frac {1}{2}}\right)}^{2}-{\frac {1}{4}}-1={\left(x-{\frac {1}{2}}\right)}^{2}-{\frac {5}{4}}\,.}$

Somit sind

${\displaystyle {}x_{1},x_{2}={\frac {\pm {\sqrt {5}}}{2}}+{\frac {1}{2}}\,}$

die Nullstellen und nach Fakt die Eigenwerte von ${\displaystyle {}M}$. Der Kern zu

${\displaystyle {}{\begin{pmatrix}{\frac {{\sqrt {5}}+1}{2}}-1&-1\\-1&{\frac {{\sqrt {5}}+1}{2}}\end{pmatrix}}={\begin{pmatrix}{\frac {{\sqrt {5}}-1}{2}}&-1\\-1&{\frac {{\sqrt {5}}+1}{2}}\end{pmatrix}}\,}$

wird von

${\displaystyle {\begin{pmatrix}1\\{\frac {{\sqrt {5}}-1}{2}}\end{pmatrix}}}$

und der Kern zu

${\displaystyle {}{\begin{pmatrix}{\frac {-{\sqrt {5}}+1}{2}}-1&-1\\-1&{\frac {-{\sqrt {5}}+1}{2}}\end{pmatrix}}={\begin{pmatrix}{\frac {-{\sqrt {5}}-1}{2}}&-1\\-1&{\frac {-{\sqrt {5}}+1}{2}}\end{pmatrix}}\,}$

wird von

${\displaystyle {\begin{pmatrix}-1\\{\frac {{\sqrt {5}}+1}{2}}\end{pmatrix}}}$

erzeugt. Die Eigenvektoren sind also

${\displaystyle {\begin{pmatrix}1\\{\frac {{\sqrt {5}}-1}{2}}\end{pmatrix}}}$

und

${\displaystyle {\begin{pmatrix}-1\\{\frac {{\sqrt {5}}+1}{2}}\end{pmatrix}}.}$