3
$\begingroup$

Projection matrix: $$ \begin{pmatrix} \cos^2\theta & \cos\theta \sin\theta\\ \cos\theta \sin\theta & \sin^2\theta\\ \end{pmatrix} $$

Reflection matrix: $$ \begin{pmatrix} 2\cos^2\theta -1 & 2\cos\theta sin\theta\\ 2\cos\theta \sin\theta & 2\sin^2\theta -1\\ \end{pmatrix} $$ which is equivalent to $$ \begin{pmatrix} \cos2\theta & \sin2\theta\\ \sin2\theta & -\cos2\theta\\ \end{pmatrix} $$ I can arrive at the second form geometrically, but the problem asks that I do it with the projection matrix and vector addition. Any hints/tips?

It could also be that the problem is poorly worded, in which case a geometrical analysis is all that is necessary.

    "Find a formula for R based on P. Explain your arguments by drawing a 
     graph, using the rules of sums of vectors."

I drew the graph and was able to successfully explain it in terms of sums of vectors.

$\endgroup$
1
$\begingroup$

Hint:

use this fact: if $X$ is a point, $X'$ its projection on the axis and $X''$ the reflection of $X$, than $X'$ is the middle point of $XX''$.


so, if $P$ is the projection matrix and $R$ the reflection, we have:

$$ X'=PX \qquad X''=RX $$ and, we write the fact that $X'$ is the midpoint as: $$ X+X''=2X' $$

so:

$$ X+RX=2PX \quad \Rightarrow \quad RX=2PX-X=(2P-I)X $$

$\endgroup$
  • $\begingroup$ Yeah, it makes sense that the reflection is a projection onto the $\theta$ line, then another projection onto another $\theta$ line, but there's something missing. My textbook highlights it as $2P-I$, but it has no explanation for why the I pops up. $\endgroup$ – Wayfinder Mar 11 '16 at 16:33
  • 1
    $\begingroup$ I've added to my answer. I hope it's useful. $\endgroup$ – Emilio Novati Mar 11 '16 at 16:41

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.