Cayley-Hamilton/3x3-Matrix/3/Überprüfe/Aufgabe/Lösung

a) Das charakteristische Polynom ist

${}{\begin{aligned}&\,\,\,\,\,\,\,\det \left({\begin{pmatrix}X&0&0\\0&X&0\\0&0&X\end{pmatrix}}-{\begin{pmatrix}2&0&-2\\5&3-{\mathrm {i} }&0\\{\mathrm {i} }&3&4\end{pmatrix}}\right)\\&=\det {\begin{pmatrix}X-2&0&2\\-5&X-3+{\mathrm {i} }&0\\-{\mathrm {i} }&-3&X-4\end{pmatrix}}\\&={\left(X-2\right)}{\left(X-3+{\mathrm {i} }\right)}{\left(X-4\right)}+2{\left(15+{\mathrm {i} }{\left(X-3+{\mathrm {i} }\right)}\right)}\\&=X^{3}+{\left(-2-3+{\mathrm {i} }-4\right)}X^{2}+{\left(6-2{\mathrm {i} }+8+12-4{\mathrm {i} }+2{\mathrm {i} }\right)}X+8{\left(-3+{\mathrm {i} }\right)}+30+2{\mathrm {i} }{\left(-3+{\mathrm {i} }\right)}\\&=X^{3}+{\left(-9+{\mathrm {i} }\right)}X^{2}+{\left(26-4{\mathrm {i} }\right)}X+4+2{\mathrm {i} }.\,\end{aligned}}$

b) Es ist

${}{\begin{aligned}{\begin{pmatrix}2&0&-2\\5&3-{\mathrm {i} }&0\\{\mathrm {i} }&3&4\end{pmatrix}}^{2}&={\begin{pmatrix}2&0&-2\\5&3-{\mathrm {i} }&0\\{\mathrm {i} }&3&4\end{pmatrix}}{\begin{pmatrix}2&0&-2\\5&3-{\mathrm {i} }&0\\{\mathrm {i} }&3&4\end{pmatrix}}\\&={\begin{pmatrix}4-2{\mathrm {i} }&-6&-12\\25-5{\mathrm {i} }&8-6{\mathrm {i} }&-10\\15+6{\mathrm {i} }&21-3{\mathrm {i} }&16-2{\mathrm {i} }\end{pmatrix}}.\end{aligned}}$

Es ist

${}{\begin{aligned}{\begin{pmatrix}2&0&-2\\5&3-{\mathrm {i} }&0\\{\mathrm {i} }&3&4\end{pmatrix}}^{3}&={\begin{pmatrix}2&0&-2\\5&3-{\mathrm {i} }&0\\{\mathrm {i} }&3&4\end{pmatrix}}{\begin{pmatrix}4-2{\mathrm {i} }&-6&-12\\25-5{\mathrm {i} }&8-6{\mathrm {i} }&-10\\15+6{\mathrm {i} }&21-3{\mathrm {i} }&16-2{\mathrm {i} }\end{pmatrix}}\\&={\begin{pmatrix}-22-16{\mathrm {i} }&-54+6{\mathrm {i} }&-56+4{\mathrm {i} }\\90-50{\mathrm {i} }&-12-26{\mathrm {i} }&-90+10{\mathrm {i} }\\137+13{\mathrm {i} }&108-36{\mathrm {i} }&34-20{\mathrm {i} }\end{pmatrix}}.\end{aligned}}$

c) Es ist

${}{\begin{aligned}&\,\,\,\,\,\,\,{\left(X^{3}+{\left(-9+{\mathrm {i} }\right)}X^{2}+{\left(26-4{\mathrm {i} }\right)}X+4+2{\mathrm {i} }\right)}(M)\\&=M^{3}+{\left(-9+{\mathrm {i} }\right)}M^{2}+{\left(26-4{\mathrm {i} }\right)}M+{\left(4+2{\mathrm {i} }\right)}E_{3}\\&={\begin{pmatrix}-22-16{\mathrm {i} }&-54+6{\mathrm {i} }&-56+4{\mathrm {i} }\\90-50{\mathrm {i} }&-12-26{\mathrm {i} }&-90+10{\mathrm {i} }\\137+13{\mathrm {i} }&108-36{\mathrm {i} }&34-20{\mathrm {i} }\end{pmatrix}}+{\left(-9+{\mathrm {i} }\right)}{\begin{pmatrix}4-2{\mathrm {i} }&-6&-12\\25-5{\mathrm {i} }&8-6{\mathrm {i} }&-10\\15+6{\mathrm {i} }&21-3{\mathrm {i} }&16-2{\mathrm {i} }\end{pmatrix}}+{\left(26-4{\mathrm {i} }\right)}{\begin{pmatrix}2&0&-2\\5&3-{\mathrm {i} }&0\\{\mathrm {i} }&3&4\end{pmatrix}}+{\left(4+2{\mathrm {i} }\right)}{\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}}\\&={\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}}.\,\end{aligned}}$

Zur gelösten Aufgabe