## Thomas' Calculus 13th Edition

By Property 4 of the Dot Product (boxed on p. 709) ${\bf u}\cdot{\bf u}=|{\bf u}|^{2}\geq 0$, which proves the first condition. Proof of the second condition: If ${\bf u}={\bf 0}$, then $|{\bf u}|=0$, and ${\bf u}\cdot{\bf u}=0.$ Thus, ${\bf u}={\bf 0}\Rightarrow{\bf u}\cdot{\bf u}=0$ If ${\bf u}\cdot{\bf u}$=$0$, if we assume that ${\bf u}\neq{\bf 0}$, it would follow that $|{\bf u}|^{2}\neq 0,$ which is a contradiction to ${\bf u}\cdot{\bf u}=|{\bf u}|^{2}$=$0$ Thus, ${\bf u}\cdot{\bf u}=0\Rightarrow{\bf u}={\bf 0}$. which proves that dot multiplication is positive definite.