# 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?

• Ok, I was just thinking that since it was in $\mathbb{R}^3$ that I would have to add $z$ to the equation. – Lindsey G Jan 27 '16 at 18:18
• Stop tagging these with "covering-spaces". – Alex Provost Jan 27 '16 at 22:33

$$\psi((x_1,y_1,z_1)+(x_2,y_2,z_2))=\psi(x_1,y_1,z_1)+\psi(x_2,y_2,z_2)$$ 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
$$\psi(a(x,y,z))=\psi(ax,ay,az)=az=a\psi(x,y,z).$$
A more expedient way to show linearity is do it in 1 shot: i.e., show $$\psi(\alpha u+\beta v)=\alpha\psi(u)+\beta\psi(v), \forall u,v\in\mathbb{R}^3, \forall \alpha,\beta\in\mathbb{R}.$$ Let $u=(x_1,y_1,z_1)'$ and $v=(x_2,y_2,z_2)'$, then $$\alpha u+\beta v=(\alpha x_1+\beta x_2,\alpha y_1+\beta y_2,\alpha z_1+\beta z_2)'$$ and so $$\psi(\alpha u+\beta v)=\alpha x_1+\beta x_2=\alpha\psi(u)+\beta\psi(v)$$ where we have used $x_1=\psi(u)$ and $x_2=\psi(v)$.