Bei ${\displaystyle {}a=0}$ wird das Gleichungssystem zu

${\displaystyle {}5x+z=0\,,}$

${\displaystyle {}3z=0\,.}$

Also ist

${\displaystyle {}x=z=0\,}$

und ${\displaystyle {}y}$ beliebig, somit ist

${\displaystyle {}L_{0}={\left\{y{\begin{pmatrix}0\\1\\0\end{pmatrix}}\mid y\in K\right\}}\,.}$

Sei also ${\displaystyle {}a\neq 0}$. Wir rechnen ${\displaystyle {}II-aI}$ und erhalten

${\displaystyle {}-3ax+(3-a(1-a))z=0\,}$

bzw.

${\displaystyle {}x={\frac {3-a+a^{2}}{3a}}z\,.}$

Die erste Gleichung liefert

${\displaystyle {}{\begin{aligned}y&={\frac {1}{a}}{\left(-5x+(a-1)z\right)}\\&={\frac {1}{a}}{\left(-5{\frac {3-a+a^{2}}{3a}}z+(a-1)z\right)}\\&={\frac {1}{3a^{2}}}{\left(-5(3-a+a^{2})+3a(a-1)\right)}z\\&={\frac {1}{3a^{2}}}{\left(-2a^{2}+2a-15\right)}z\\&={\frac {-2a^{2}+2a-15}{3a^{2}}}z.\end{aligned}}}$

Somit ist

${\displaystyle {}L_{0}={\left\{z{\begin{pmatrix}{\frac {3-a+a^{2}}{3a}}\\{\frac {-2a^{2}+2a-15}{3a^{2}}}\\1\end{pmatrix}}\mid z\in K\right\}}\,.}$