Answer
Proof given below.
Work Step by Step
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.
