Show that $\psi:\mathbb{R}^3\rightarrow\mathbb{R}$ given by $\psi(x,y,z) = z$ is linear Show that $\psi:\mathbb{R}^3\rightarrow\mathbb{R}$ given by $\psi(x,y,z) = z$ is linear
I know that to be linear it must satisfy:
(i) $\psi(x+y+z)=\psi(x)+\psi(y)+\psi(z)$
(ii) $\psi(ax)=a\psi(x)$

So, this is how far I've gotten:
(i) I let $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3),$ and $z=(z_1,z_2,z_3)$ to satisfy $\mathbb{R}^3$
and so $\psi(x+y+z) = ((x_1,x_2,x_3)+(y_1,y_2,y_3)+(z_1,z_2,z_3))$
which can be written as $\psi((x_1,y_1,z_1)+(x_2,y_2,z_2)+(x_3,y_3,z_3))$
and applying $\psi$ I got $(z_1+z_2+z_3)$
But this is where I got stuck, I'm not sure where to go from here. Did I approach this problem incorrectly?
(ii) I let $t=(x,y,z)$ so
$\psi(at)=\psi(a(x,y,z))=\psi(ax,ay,az)=a\psi(x,y,z)=a\psi(t)$
so $\psi(at)=a\psi(t)$
Is that a sufficient way to prove that?

You are the right way. But note that you can simplify what you have done. To show that it is linear you need to check 
$$\psi((x_1,y_1,z_1)+(x_2,y_2,z_2))=\psi(x_1,y_1,z_1)+\psi(x_2,y_2,z_2)$$ and 
$$\psi(a(x,y,z))=a\psi(x,y,z).$$ Now, it is
$$\psi((x_1,y_1,z_1)+(x_2,y_2,z_2))=\psi(x_1+x_2,y_1+y_2,z_1+z_2)=z_1+z_2=\psi(x_1,y_1,z_1)+\psi(x_2,y_2,z_2),$$ and
$$\psi(a(x,y,z))=\psi(ax,ay,az)=az=a\psi(x,y,z).$$
