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Notation: $v$// is $v$ parallel symbol, $v\bot$ is $v$ perpendicular, and both are relative to plane $\sqcap$

Let $\sqcap \subseteq$ $\mathbb{R}^3$ be the plane whose equation is $x + y + z = 0$. Let $F\sqcap$ : $\mathbb{R}^3\to \mathbb{R}^3$ be the reflection through the plane $\sqcap$: that is, for a vector $v \in \mathbb{R}^3$, if we write v = v// + v⊥, where $v$// $\in \sqcap$ and $v\bot$ is orthogonal to the plane $\sqcap$, then $F\sqcap(v)$ = $v$// $− v\bot$.

Determine the representing matrix $([F\sqcap]E \to E)$ of $F\sqcap$, where $E = \{e1, e2, e3\}$ is the standard basis of $\mathbb{R}^3$.

I know that $v\bot\bullet v$//$=0$, I don't think that tells me anything useful for this problem.

I know that $v-v\bot=v$//=$F\sqcap(v)+v\bot$, which yields $v-2v\bot=F\sqcap(v)$

the v and $-2v\bot$ term don't tell me enough for a full transformation because I don't have enough information to form a basis in $\mathbb{R}^3$ and define the transformation with respect to $E$.

Any clue is appreciated... I've been stuck on this since last night.

3SAT
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Eric
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  • I outline related material in this older answer. I hope that will help you, because I describe the general formula for a reflection w.r.t. a plane through the origin there. I'm sure there are many answers here, where that formula is derived. – Jyrki Lahtonen Aug 08 '15 at 20:06
  • @Eric I edited $\mathbb{R}^3$ insted of R3, is it ok? – 3SAT Aug 08 '15 at 20:10
  • thank you for the edit,

    @JyrkiLahtonen, I read your older post, and I think the problem is asking me to derive the 'recipe' you posted. I cannot tell from your answer how you arrived at the transformation s(x⃗ )=s(x⃗ )=x⃗ −2(x⃗ ⋅n⃗ )•n⃗ / ||n⃗||^2.. any further help is appreciated.

     – Eric Aug 08 '15 at 20:16
  • @Eric Use "mathjax" language – 3SAT Aug 08 '15 at 20:16
  • Eric, in this answer I derive that formula for a reflection in the plane. The same idea works: split the vector into two components: one parallel to the normal of the plane/line you reflect w.r.t., the other orthogonal to that. One component does not change in the reflection, the other gets multiplied by $-1$. Sorry about the murky picture. Also the asker of that question used programmer's notation for the coordinates of the points, so this may be unreadable. Looking for something better... – Jyrki Lahtonen Aug 08 '15 at 20:22

1 Answers1

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I'm not sure if this helps, but if you can find a basis $V$ by first finding two linearly independent vectors in your plane (i.e. two linearly independent solutions to your plane equation, e.g. just plug in values for $x$ and $y$ to get $z$), and then one vector orthogonal to your plane (i.e. the vector of coefficients that define the plane), then the transformation with respect to this basis $V$is trivial, it's just the diagonal matrix $D$ with $(1,1,-1)$ on the diagonal. So then if you want the matrix of the same transformation, but with respect to the standard basis, it's just $VDV^{-1}$.

user2566092
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  • I'm not sure where that D matrix came from. Could you elaborate on that a bit?

    Your help is appreciated, thank you.

     – Eric Aug 09 '15 at 02:15
  • @Eric Sure, all vectors on your plane are eigenvectors with eigenvalue 1, because the reflection doesn't affect them. On the other hand, a vector that is orthogonal to your plane is an eigenvector with eigenvalue -1 because it is totally reflected, i.e. sent to its opposite. You can only have 3 linearly independent eigenvectors so that completely characterizes the transformation. With respect to the basis $V$ so obtained, any vector is expressed uniquely as a linear combination of your two eigenvectors in the plane and the orthogonal eigenvector, and then the matrix $D$ applies. – user2566092 Aug 09 '15 at 20:24