Etwa so
\(mdot\left(\left(\begin{array}{rr}a1&a2\\a2&a3\\\end{array}\right), \left(\begin{array}{rr}b1&b2\\b2&b3\\\end{array}\right) \right)\)
\( \rightarrow 3 \) a1 b1 \( + \) a2 b2 \( + \) a3 b3
o_1:=e1/sqrt(mdot(e1,e1))
\( \rightarrow o_{1}:=\left(\begin{array}{cc}\frac{1}{\sqrt{7}} & 0 \\ 0 & \frac{2}{\sqrt{7}}\end{array}\right) \)
e_2:=e2 - mdot(o_1,e2) o_1
\( \rightarrow \mathrm{e}_{2}:=\left(\begin{array}{rr}0 & -3 \\ -3 & 0\end{array}\right) \)
o_2:=(e_2) / sqrt(mdot(e_2,e_2))
\( \rightarrow \quad \mathrm{e}_{3}:=\left(\begin{array}{cc}\frac{-2}{7} & 0 \\ 0 & \frac{3}{7}\end{array}\right) \)
e_3:e3 - mdot(o_2,e3) o_2 - mdot(o_1,e3) o_1
o_3:=( e_3) / sqrt(mdot(e_3,e_3))