Answer
a.
$cu \in W$
b.
Sample: $u=\left[\begin{array}{l}
2\\
3
\end{array}\right]$ and $v=\left[\begin{array}{l}
-3\\
-2
\end{array}\right]$
Work Step by Step
a.
If $\mathrm{u}=\left[\begin{array}{l}
u_{1}\\
u_{2}
\end{array}\right]\in W$, then $u_{1}u_{2} \geq 0$,
For a scalar c, $cu=\mathrm{u}=\left[\begin{array}{l}
cu_{1}\\
cu_{2}
\end{array}\right]$, and
$(c u_{1})(cu_{2})=c^{2}(u_{1}u_{2}) \geq 0$,
(the product of nonnegative numbers is nonnegative,
and $c^{2}$ and $u_{1}u_{2}$ are both nonnegative),
so $cu \in W.$
b.
Select $u=\left[\begin{array}{l}
u_{1}\\
u_{2}
\end{array}\right]$ and $\mathrm{v}=\left[\begin{array}{l}
v_{1}\\
v_{2}
\end{array}\right]$ such that one the sums
$u_{1}+v_{1}, \quad u_{2}+v_{2} $
is negative, while the other is positive.
Sample: $u=\left[\begin{array}{l}
2\\
3
\end{array}\right]\in W$ and $v=\left[\begin{array}{l}
-3\\
-2
\end{array}\right]\in W,$
$u+v=\left[\begin{array}{l}
-1\\
1
\end{array}\right]\not\in W$, because $(-1)(1) < 0 $
