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.
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.
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.
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:
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).