TU Wien Nav:Mathematik für Informatik (Buch)/3.25

3.25 Man bestimme die Eigenwerte der Matrix A:

$(a)\quad A={\begin{pmatrix}3&-1\\-1&3\end{pmatrix}}\quad (b)\quad A={\begin{pmatrix}1&-1\\-1&1\end{pmatrix}}$
$(c)\quad A={\begin{pmatrix}0&{\frac {1}{2}}&{\frac {1}{2}}\\{\frac {1}{2}}&0&{\frac {1}{2}}\\{\frac {1}{2}}&{\frac {1}{2}}&0\end{pmatrix}}\quad (d)\quad A={\begin{pmatrix}5&-8&10\\-8&11&2\\10&2&2\\\end{pmatrix}}$

Lösungen

(a)

$\det(\lambda \cdot I_{2}-A)=\det({\begin{pmatrix}\lambda &0\\0&\lambda \\\end{pmatrix}}-{\begin{pmatrix}3&-1\\-1&3\end{pmatrix}})=\det({\begin{pmatrix}\lambda -3&1\\1&\lambda -3\end{pmatrix}})=$
$=\lambda ^{2}-6\lambda +8$

Lösen der quadratischen Gleichung:

${\frac {6\pm {\sqrt {4}}}{2}}=3\pm 1=\{2,4\}$

(b)

$\det({\begin{pmatrix}\lambda &0\\0&\lambda \end{pmatrix}}-{\begin{pmatrix}1&-1\\-1&1\end{pmatrix}})=\det({\begin{pmatrix}\lambda -1&1\\1&\lambda -1\end{pmatrix}})=\lambda ^{2}-2\lambda$

${\frac {2\pm {\sqrt {4}}}{2}}=1\pm 1=\{0,2\}$