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Let $f_1,f_2$ be two orthogonal non zero vectors that span a plane $\pi$. The projection of a vector $w$ onto the plane $\pi$ is given by $Proj_{\pi}(w)= $$\frac{w\cdot n}{f_1\cdot f_1}\cdot f_1 + \frac{w\cdot n}{f_2\cdot f_2}\cdot f_2$. Prove that the function $T : R^3 \to R^3$ defined by $T(w) = Proj_{\pi}(w) $ is a linear transformation.

Things that I know:
A transformation $T : R^m →R^m$ is called a linear transformation if and only if
1)$T(u + v) = T(u) + T(v)$ for all $u$ and $v$ in $R^n$.
2) $T(cv) = cT(v)$ for all $ v$ in $R^n$ and all scalars $c$

So I think I need to prove these two things, but I am not sure how to proceed with that.

Edit for the second part of the proof

For Scaler Multiplication. Proof: for another vector $w ∈ R^n$ and a scalar c, it $proj_v(u+w) = proj_v(u)+proj_v(w) and proj_v(cu) = c(proj_v(u)$

twest
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    Start with $Proj_\pi (u+v) = \frac{(u+v)\cdot f_1}{f_1\cdot f_1} f_1 + \frac{(u+v)\cdot f_2}{f_2\cdot f_2} f_2$ and see if you can manipulate it until you get $Proj_\pi(u) + Proj_\pi (v)$. Then do the analogous proof with $Proj_\pi(cv)$. –  Oct 30 '16 at 16:18
  • Has this question not already been asked some hours ago ? – Astyx Oct 30 '16 at 16:20
  • @ Bye_World what would be the first step in manipulating this formula? Just so I can get started – twest Oct 30 '16 at 16:21
  • @twest Use the bilinearity of the dot product to separate each term on the right into two terms. –  Oct 30 '16 at 16:21
  • @ Bye_World so I got this much so far $Proj_{\pi}(u+v)= $$\frac{(u+v)\cdot f_1}{f_1\cdot f_1}\cdot f_1 + \frac{(u+v)\cdot f_2}{f_2\cdot f_2}\cdot f_2 = $ $\frac{u\cdot f_1+v\cdot f_1}{f_1\cdot f_1}\cdot f_1 + \frac{(u\cdot f_2+v\cdot f_2)\cdot }{f_2\cdot f_2} \cdot f_2 $ but where do I go from here? – twest Oct 30 '16 at 16:37
  • @ Bye_World for the second part of the proof, is this logic correct For Scaler Multiplication. Proof: for another vector $w ∈ R^n$ and a scalar c, it $proj_v(u+w) = proj_v(u)+proj_v(w) and proj_v(cu) = c(proj_v(u)$ – twest Oct 30 '16 at 17:12
  • @ Astyx I have made a edit to the post, does this logic look correct? – twest Oct 30 '16 at 17:36
  • @twest I suspect that your messages aren't pinging users because you're putting a space between @ and the username. See my answer below for how to precede with this proof. –  Oct 30 '16 at 18:47

1 Answers1

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Part the First (Additivity):

Let $u,v \in \Bbb R^3$. Then

$$\begin{align}\operatorname{Proj}_\pi (u+v) &= \frac{(u+v)\cdot f_1}{f_1\cdot f_1} f_1 + \frac{(u+v)\cdot f_2}{f_2\cdot f_2} f_2 \\ &= \frac{u\cdot f_1+v\cdot f_1}{f_1\cdot f_1} f_1 + \frac{u\cdot f_2+v\cdot f_2}{f_2\cdot f_2} f_2 \\ &= \left[\frac{u\cdot f_1}{f_1\cdot f_1} + \frac{v\cdot f_1}{f_1\cdot f_1}\right] f_1 + \left[\frac{u\cdot f_2}{f_2\cdot f_2} + \frac{v\cdot f_2}{f_2\cdot f_2}\right] f_2 \\ &= \frac{u\cdot f_1}{f_1\cdot f_1} f_1 + \frac{v\cdot f_1}{f_1\cdot f_1} f_1 + \frac{u\cdot f_2}{f_2\cdot f_2} f_2 + \frac{v\cdot f_2}{f_2\cdot f_2} f_2 \\ &= \left[\frac{u\cdot f_1}{f_1\cdot f_1} f_1 + \frac{u\cdot f_2}{f_2\cdot f_2} f_2\right] + \left[\frac{v\cdot f_1}{f_1\cdot f_1} f_1 + \frac{v\cdot f_2}{f_2\cdot f_2} f_2\right] \\ &= \operatorname{Proj}_\pi (u) + \operatorname{Proj}_\pi (v)\end{align}$$

Thus $\operatorname{Proj}_\pi$ is additive.

Part the Second (Homogeneity):

Try this on your own using the above as a guide.