Matrices in $\operatorname{O}(n) \setminus \operatorname{SO}(n)$

Question

Can any matrix $M \in \operatorname{O}(n) \setminus \operatorname{SO}(n)$ be written as $I_n - uu^T$ where $I_n \in \mathbb{R}^{n \times n}$ is the identity matrix and $u \in \mathbb{R}^n$, $||u||_2 = 1$?

I know that any $M = I_n - uu^T$ is a reflector and that $\operatorname{O}(n) \setminus \operatorname{SO}(n)$ is the set of all reflectors, but I'm unable to see if we can write every element in $\operatorname{O}(n) \setminus \operatorname{SO}(n)$ this way.

Answer 1

The first counterexample comes in three dimensions. $M=-I$ is orthogonal with determinant $-1$, and it can't be written as $I-uu^T$ because $M-I$ has rank $3$ instead of rank $1$.

Answer 2

I think no because $I_n-uu^t$ is a symmetric matrix while not all matrix in $O(n)/SO(n)$ are symmetric because if $A\in O(n)$ then

$A^t=A^{-1}$

so $A\in O(n)$ is symmetric if and only if $A=A^{-1}$

For example

https://wikimedia.org/api/rest_v1/media/math/render/svg/d757d0647c657cc0244c525cd824dd1bb6e6a668

is not symmetric (if that matrix $A$ is in $SO(n)$ you can consider $-A$ that it is in $O(n)/SO(n)$ because $n=3$ is odd)

You can also observe that if $n$ is odd and if by contraddiction all the matrix of $O(n)/SO(n)$ can be written as $I_n-uu^t$ then all the matrix of $SO(n)$ can be written as $uu^t-I_n$ because if $A\in SO(n)$ then $-A\in O(n)$ but $n$ is odd so

$det(-A)=(-1)^ndet(A)=-det(A)=-1$

then $-A\in O(n)/SO(n)$ so $-A=I_n-uu^t$ and you have that

$A=uu^t-I_n$

This is a contradiction because not all special orthogonal matrix can be written as $uu^t-I_n$