4
$\begingroup$

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 Answer 1

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:

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

$\endgroup$
2
  • $\begingroup$ 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
    Commented Mar 11, 2016 at 16:33
  • 1
    $\begingroup$ I've added to my answer. I hope it's useful. $\endgroup$ Commented Mar 11, 2016 at 16:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .