Answer
Proof given below.
Work Step by Step
Proof of the first statement:
By Property 4 of the Dot Product (boxed on p. 613)
${\bf u}\cdot{\bf u}=|{\bf u}|^{2}\geq 0$,
which proves the first statement.
Proof of the second statement:
($\Rightarrow$)
If ${\bf u}={\bf 0}$, then $|{\bf u}|=0$, and ${\bf u}\cdot{\bf u}=0.$
Thus, ${\bf u}={\bf 0}\Rightarrow{\bf u}\cdot{\bf u}=0$
($\Leftarrow$)
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, ${\bf u}\cdot{\bf u}=0\Rightarrow{\bf u}={\bf 0}$.
Both implications are true, so
${\bf u}\cdot{\bf u}=0\quad\Leftrightarrow\quad {\bf u}={\bf 0}$
This proves that dot multiplication is positive definite.
