1

Let $V$ be an euclidean vector space with $a \in V$ of length 1. I showed that $$f: V \rightarrow V, x \rightarrow 2\langle a,x\rangle a -x $$ is an orthogonal transformation and hence either a rotation, reflection or a combination of both.

How excatly am I supposed to figure out, what kind of transformation is taking place?

Hekri
  • 53
  • 1
    I suppose that $x$ is a vector and $\langle a,x \rangle$ is the scalar product, so it is a scalar. How you define the sum of a scalar and a vector? – Emilio Novati Jun 09 '17 at 14:13
  • corrected it. Thanks – Hekri Jun 09 '17 at 14:45
  • 1
    $x\mapsto x-2\langle x,a\rangle a$ is the orthogonal reflection w.r.t. the hyperplane $H$ that has $a$ as its normal. So your $f$ is a reflection followed be negation. The reflection has determinant $-1$, but the determinant of negation depends on the parity of the dimension ov $V$. – Jyrki Lahtonen Jun 09 '17 at 15:20
  • Thanks, thats the answer, but I can't upvote you. How exactly did you figure that out? – Hekri Jun 09 '17 at 15:53
  • The formula for the reflection is kinda well-known. And I have studied groups generated by reflections in my time. – Jyrki Lahtonen Jun 09 '17 at 16:23
  • @Hekri You're getting the story a bit backwards, I suspect.. The formula didn't come from divine revelation and then us mortals struggled to interpret it; rather, someone at some point wanted to figure out the formula for reflecting one vector $x$ across another $a$, and then successfully figured out the formula. – anon Jun 10 '17 at 06:15
  • Here's how: every vector may be decomposed into parallel and perpendicular components (with respect to the unit vector $a$) as $x=\lambda a+b$, where $\lambda$ is some scalar component and $b$ is orthogonal to $a$. Taking the dot product yields $\langle a,x\rangle=\lambda$. The reflection of $x$ across $a$ must preserve its parallel component and negate its perpendicular component (visualize this in 3D to see this makes sense), and thus is given by $$\lambda a-b=2\lambda a-(\lambda a+b)=2\langle a,x\rangle a-x.$$ – anon Jun 10 '17 at 06:15

3 Answers3

1

Hint:

By the definition of orthogonal transformation you have to prove that: $\langle f(x),f(y)\rangle=\langle x,y\rangle \quad \forall x,y \in V$.

Calculate: $$ \langle (2\langle a,x\rangle a -x),(2\langle a,y\rangle a -y) \rangle $$ using the properties of the inner product, and remember that $\langle a,a\rangle=1$

This transformation can be interpreted as a reflection in the plane ( $2-$dimensional subspace) spanned by $a$ and $x$.

A sketch of this interpretation:

without loss of generality, we can find a basis in $span\{a,x\}$ such that $a=(1,0)$ and $x=(x_1,x_2)$

So we have $f(x)=(2x_1,0)-(x_1,x_2)=(x_1,-x_2)$

That is a reflection.

Emilio Novati
  • 62,675
0

If $f$ is a linear transformation, write $f$ in matricial form. This can be helpful.

Bruno
  • 503
0

$\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$A natural approach (whether you know the answer or not) is to try to compute $$ f(x) = 2\Brak{a, x} a - x $$ for as many vectors as you can. Here, the primary impediment is our (lack of) ability to evaluate $\Brak{a, x}$.

  • Since $\|a\| = 1$, we have $\Brak{a, a} = \|a\|^{2} = 1$. Consequently, $$ f(a) = 2\Brak{a, a} a - a = a. $$ Success!

  • If $\Brak{a, x} = 0$, i.e., $x$ is orthogonal to $a$, then $$ f(x) = -x. $$ Again, success!

But this answers the question: $a$ and the set of vectors orthogonal to $a$ span $V$, so we've found the eigenspace decomposition of $f$.

Andrew D. Hwang
  • 78,195