Let $P_z : \mathbb R^3 \to \mathbb R^3$ be the function which orthogonally projects onto the $x$-$y$ linearity plane (that is, the set of vectors $(x, y, z) \in \mathbb R^3$ with $z = 0$). Is $P_z$ linear? If so, prove it; if not, explain why not.
I understand I have to show that the linear map has the 3 properties:
- $P_z(0)=0$
- $P_z(u+w)=P_z(u) + P_z(w)$
- $P_z(cu)=cP_z(u)$
But, how to do it in this problem? Thanks!!!
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$\begingroup$The projection takes any vector $(x, y, z)$ and gives back the vector $(x, y, 0)$. Now you just check. Take vectors, do the vector operation then apply the transformation. Then try again, byt apply the transformation first, then do the vector operations. See that you get the same answer.
For instance, $P_z(cu) = cP_z(u)$: Take a vector $u = (x, y, z)$ and a real number $c$. Then $$ P_z(cu) = P_z(c(x, y, z)) = P_z((cx, cy, cz)) = (cx, cy, 0) $$ while $$ cP_z(u) = cP_z((x, y, z)) = c(x, y, 0) = (cx, cy, 0) $$ and we see that they are the same. Now do a similar thing for sum of two vectors (zero vector comes for free from this one by picking $c = 0$), and you're done.
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