In a complex vector space V, let u, v, w E V. In a complex vector space V, let $u, v, w\in V$. 
a) Show that 
$$
{\rm{Span}}\{u,v,w\}={\rm{Span}}\{u+\sqrt{-2}v+w,v+\sqrt{-2}w+u,w+\sqrt{-2}u+v\}
$$
 A: The key to attack this problem is the same procedure as in showing the equality between two sets $A$ and $B$. You had to show that $A\subseteq B$ and $B\subseteq A$ to conclude that $A=B$. To show $A\subseteq B$, you first pick and arbitrary element $a$ in $A$ and argue that $a$ is also in $B$. The same with $B\subseteq A$.
Let me help you with the inclusion "$\supseteq$". Let $x$ be an arbitrary vector in
$$\operatorname{span}\big(\{u+w+\sqrt{-2}v,u+v+\sqrt{-2}w,v+w+\sqrt{-2}u\}\big)$$
then, $x$ can be written as a linear combination of $u+w+\sqrt{-2}v$, $u+v+\sqrt{-2}w$ and $v+w+\sqrt{-2}u$ :
$$x=a_1(u+w+\sqrt{-2}v)+a_2(u+v+\sqrt{-2}w)+a_3(v+w+\sqrt{-2}u) \tag{1}$$
where $a_1,a_2$ and $a_3$ are in $\mathbb C$. Rearranging $(1)$ we see that
$$x=(a_1+a_2+a_3\sqrt{-2})u+(a_1\sqrt{-2}+a_2+a_3)v+(a_1+a_2\sqrt{-2}+a_3)w$$
that is, $x$ also can be written as a linear combination of the vectors $u$, $v$ and $w$. So, $x\in\operatorname{span}\big(\{u,v,w\}\big)$ as we want to show.
The inclusion "$\subseteq$" is little harder but is the same idea. Luck!
