Show that every matrix $R \in SO(n)$ is connected to a block matrix the form

\begin{bmatrix}

1 & 0 \\

0 & R_1

\end{bmatrix}

where $R_1 \in SO(n-1)$.

This is a part of problem #13 from chapter one of Brian Halls book in Lie Theory. I started off by showing that for any unit vector $v \in \mathbb{R}^2$, there is a path in $R(t) \in SO(2)$ such that $R(0)=I$ and $R(v)=e_1$. This was easy to do because I knew the explicit form of matrices in $SO(2)$.

Then I did the same thing in $\mathbb{R}^3$. That is, for a unit vector $v \in \mathbb{R}^3$ there is a path $R(t) \in SO(3)$ such that $R(0)=I$ and $R(1)v = e_1$. For this, I let $\theta$ be the angle between $v$ and $e_1$ and then I argued that by rotating $v$ by an angle of $\theta$ about $\frac{v \times e_1}{|v \times e_1|}$ then you would land on $e_1$. Using this idea, it is easy to make the required path in $SO(3)$.

I want to do the same thing for a unit vector in $SO(n)$ and $e_1 \in \mathbb{R}^n$. I could wave my hands a bit and argue that the group of rotations of $S^{n-1}$ is transitive. I wouldn't know a more explicit way to do it since In the general case I do not know the general form a matrix in $SO(n)$ and also do not have the cross product on $\mathbb{R}^n$ (like I did for $n=2$ and $n=3$).

Anyway, assuming that for any $v \in \mathbb{R}^n$ with $|v|=1$ there exists a path $R(t) \in SO(n)$ with $R(0)=I$ and $R(1)v = e_1$, I'm trying to figure out what to do next.

## Best Answer

A result from linear algebra is that if $A \in O(n)$, then there is an orthonormal basis of $\mathbb{R}^n$ in which the matrix representation of $A$ contains blocks of the form $$ \begin{pmatrix} c & -s \\ s & c \\ \end{pmatrix}, \hspace{20pt} c^2 + s^2 = 1 $$ and perhaps a $p \times p$ identity matrix block $I_p$ and a $q \times q$ negative identity matrix block $-I_q$. See page 113-114 of https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/linalg.pdf for a proof.

With this characterization of $O(n)$, it is simple to show that the matrix exponential $\text{Exp} : \text{Skew}(n) \to SO(n)$ is surjective (do each block separately; use Euler's formula). This then gives you an obvious smooth path $\gamma : [0, 1] \to SO(n)$ between $I = \text{Exp}(0)$ and $R = \text{Exp}(B) \in SO(n)$. Details are below: