A vollen Rang => A'A spd

Hallo,

symmetrisch: ${\displaystyle {\left({\mathrm{A<mspace\; width="0.12em"></mspace>}}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}\mathrm{A<mspace\; width="0.12em"></mspace>}\right)}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}={\mathrm{A<mspace\; width="0.12em"></mspace>}}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}{\mathrm{A<mspace\; width="0.12em"></mspace>}}^{{\mathrm{T<mspace\; width="0.12em"></mspace>}}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}}={\mathrm{A<mspace\; width="0.12em"></mspace>}}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}\mathrm{A<mspace\; width="0.12em"></mspace>}}$

pos. def.: ${\displaystyle {\mathrm{x<mspace\; width="0.12em"></mspace>}}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}{\mathrm{A<mspace\; width="0.12em"></mspace>}}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}\mathrm{A<mspace\; width="0.12em"></mspace>}\mathrm{x<mspace\; width="0.12em"></mspace>}={\left(\mathrm{A<mspace\; width="0.12em"></mspace>}\mathrm{x<mspace\; width="0.12em"></mspace>}\right)}^{\mathrm{T<mspace\; width="0.12em"></mspace>}}\mathrm{A<mspace\; width="0.12em"></mspace>}\mathrm{x<mspace\; width="0.12em"></mspace>}=\left|\right|\mathrm{A<mspace\; width="0.12em"></mspace>}\mathrm{x<mspace\; width="0.12em"></mspace>}|{|}^2\ge 0}$. Wegen vollem Rang von A ist ker A={0} und somit ${\displaystyle \mathrm{A<mspace\; width="0.12em"></mspace>}\mathrm{x<mspace\; width="0.12em"></mspace>}=0\iff \mathrm{x<mspace\; width="0.12em"></mspace>}=0}$. Somit gilt ${\displaystyle \left|\right|\mathrm{A<mspace\; width="0.12em"></mspace>}\mathrm{x<mspace\; width="0.12em"></mspace>}|{|}^2=0\iff \mathrm{x<mspace\; width="0.12em"></mspace>}=0}$.

Gruß

Stephan