gramsche matrix

Meiner Meinung nach müßte die Gram-Matrix so aussehen:

${\displaystyle {\mathrm{G<mspace\; width="0.12em"></mspace>}}_{\mathrm{B<mspace\; width="0.12em"></mspace>}}\left(\mathrm{\beta <mspace\; width="0.12em"></mspace>}\right)=\left(\begin{array}{ccc}\begin{array}{c}\mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_1,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_1)\\ \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_2,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_1)\end{array}& \begin{array}{c}\mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_1,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_2)\\ \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_2,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_2)\end{array}& \begin{array}{c}\begin{array}{cc}\mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_1,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_3)& \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_1,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_4)\end{array}\\ \begin{array}{cc}\mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_2,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_3)& \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_2,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_4)\end{array}\end{array}\\ \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_3,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_1)& \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_3,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_2)& \begin{array}{cc}\mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_3,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_3)& \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_3,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_4)\end{array}\\ \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_4,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_1)& \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_4,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_2)& \begin{array}{cc}\mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_4,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_3)& \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_4,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_4)\end{array}\end{array}\right)}$

Um die einzelnen Elemente zu berechnen setzt man ${\displaystyle \mathrm{v<mspace\; width="0.12em"></mspace>}={\mathrm{v<mspace\; width="0.12em"></mspace>}}_1}$ und ${\displaystyle \mathrm{w<mspace\; width="0.12em"></mspace>}={\mathrm{v<mspace\; width="0.12em"></mspace>}}_1}$.

Wegen ${\displaystyle \mathrm{v<mspace\; width="0.12em"></mspace>}={\mathrm{a<mspace\; width="0.12em"></mspace>}}_1\cdot {\mathrm{v<mspace\; width="0.12em"></mspace>}}_1+{\mathrm{a<mspace\; width="0.12em"></mspace>}}_2\cdot {\mathrm{v<mspace\; width="0.12em"></mspace>}}_2+{\mathrm{a<mspace\; width="0.12em"></mspace>}}_3\cdot {\mathrm{v<mspace\; width="0.12em"></mspace>}}_3+{\mathrm{a<mspace\; width="0.12em"></mspace>}}_4\cdot {\mathrm{v<mspace\; width="0.12em"></mspace>}}_4}$ ist dann ${\displaystyle {\mathrm{a<mspace\; width="0.12em"></mspace>}}_1=1,{\mathrm{a<mspace\; width="0.12em"></mspace>}}_2={\mathrm{a<mspace\; width="0.12em"></mspace>}}_3={\mathrm{a<mspace\; width="0.12em"></mspace>}}_4=0}$.

Analog erhält man für ${\displaystyle \mathrm{w<mspace\; width="0.12em"></mspace>}}$: ${\displaystyle {\mathrm{b<mspace\; width="0.12em"></mspace>}}_1=1,{\mathrm{b<mspace\; width="0.12em"></mspace>}}_2={\mathrm{b<mspace\; width="0.12em"></mspace>}}_3={\mathrm{b<mspace\; width="0.12em"></mspace>}}_4=0}$. Diese Werte setzt man in die Formel für ${\displaystyle \mathrm{\beta <mspace\; width="0.12em"></mspace>}}$ ein (die meisten Terme sind Null), und man bekommt heraus ${\displaystyle \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_1,{\mathrm{v<mspace\; width="0.12em"></mspace>}}_2)=3}$.

Analog verfährt man für die anderen ${\displaystyle \mathrm{\beta <mspace\; width="0.12em"></mspace>}({\mathrm{v<mspace\; width="0.12em"></mspace>}}_{\mathrm{i<mspace\; width="0.12em"></mspace>}},{\mathrm{v<mspace\; width="0.12em"></mspace>}}_{\mathrm{k<mspace\; width="0.12em"></mspace>}})}$, wobei immer ${\displaystyle {\mathrm{a<mspace\; width="0.12em"></mspace>}}_{\mathrm{i<mspace\; width="0.12em"></mspace>}}={\mathrm{b<mspace\; width="0.12em"></mspace>}}_{\mathrm{k<mspace\; width="0.12em"></mspace>}}=1}$ ist und alle anderen

Koeffizienten Null.