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Any $X\in O^+(n)$ (orthogonal matrices with positive determinant) is the product of an even number of reflection?

I am not able to prove this. Please help.

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What do you know about $O(n)$ ? –  Belgi Jul 18 '12 at 16:06
    
$O(n)$ is compact. determinants are $\{+1,-1\}$, collumn vectors are orthonormal. –  Bunuelian Trick Jul 18 '12 at 16:08
    
See here math.stackexchange.com/q/138020/30357 for an answer without the positivity assumption. To finish an answer to your question, you only need to know how reflections influence the determinant. –  Simon Markett Jul 18 '12 at 16:11
    
Hint: what is the determinant of (a matrix representing) a reflection? –  Geoff Robinson Jul 18 '12 at 16:39
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2 Answers

I'm imagining you may be interested in the case $K = \mathbb{R}$, $q = x_1^2 + \ldots + x_n^2$, but the result holds more generally:

Theorem (Cartan-Dieudonne): Let $q = q(x_1,\ldots,x_n)$ be a nondegenerate quadratic form over a field $K$ of characteristic different from $2$. Then every element of the orthogonal group $O(q)$ of $q$ is a product of at most $n$ reflections.

For a proof see e.g. $\S 8.4$ in these notes.

Since the determinant of a reflection is $-1$, an element of $O(q)$ has determinant $+1/-1$ according to whether it can be written as a product of an even/odd number of reflections.

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Since the OP asked this as a follow up to one of my answers, I guess I should answer this one.

You need to know some basic facts about orthogonal matrices:

  • that $X\in O(n)$ if and only if the columns of $X$ form an orthonormal system which is again equivalent to the fact that the rows of $X$ are an orthonormal base.
  • that for any two unit vectors $v, w$ in $\mathbb{R}^n$ there exists a reflection $R$ such that $Rv=w$.
  • the product of orthogonal matrices is orthogonal.
  • the inverse of an orthogonal matrix is it's transpose.

If you know that then you will easily see that, if $X =(r_1, \ldots, r_n)$ with an orthonormal base $\{r_i\}$, there exists a reflection $R_1$ such that $R_1 X=(e_1, r_2^', \cdots , r_n^')$, where $e_1=(1,0,\ldots,0)^T$. $R_1 X$ is again orthogonal (third statement above). So the first row of this matrix is $(1,0, ,\ldots,0)$, that is, $R_1X$ is an orthogonal matrix with a $1$ in the upper left corner and zeroes in the other entries of the first row and column. Now a simple induction shows that there exist (at most) $n$ reflections which transform $X$ into the identity matrix, i.e. $$R_n\cdots R_1 X = Id $$

Since the inverse of an orthogonal matrix is simply the transpose you get $$X = (R_n\cdots R_1)^T $$

The four basic propeties I mentionend should be easy to find in any basic text book.

(The fact that the number of reflections is even follows by taking determinants).

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could you tell me how to prove the basic fact number two? which fact you used to write $R_1X=(e_1,r_2',\dots,r_n')$? –  Bunuelian Trick Jul 18 '12 at 18:15
    
@Patience the idea is to use the reflection through the hyperplane which bisects the angle between $v$ and $w$. That is, the reflection through the plane normal to $N=v-w$, which is given explicitly by $$r\mapsto r-2\langle r, \frac{N}{|N|}\rangle\frac{ N}{|N|}$$ (If f$w=v$ take any hyperplane $E$ containing them and reflect through $E$). –  user20266 Jul 18 '12 at 18:50
    
what is $r$? ?? –  Bunuelian Trick Jul 18 '12 at 19:04
    
and I hope you are going to answer my two doubt :( –  Bunuelian Trick Jul 18 '12 at 19:06
    
@Patience $r$ is the free variable, a vector in Euclidean space. The reflection is given by $$Rr = r-2\langle r, n\rangle n$$ with $n=N/|N|$. –  user20266 Jul 19 '12 at 5:37
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