2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Ok, I was just thinking that since it was in $\mathbb{R}^3$ that I would have to add $z$ to the equation. $\endgroup$ – Lindsey G Jan 27 '16 at 18:18
  • $\begingroup$ Stop tagging these with "covering-spaces". $\endgroup$ – Alex Provost Jan 27 '16 at 22:33
0
$\begingroup$

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

$$\psi(a(x,y,z))=\psi(ax,ay,az)=az=a\psi(x,y,z).$$

$\endgroup$
0
$\begingroup$

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 $x_2=\psi(v)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.