# Every matrix $R \in SO(n)$ is connected to a matrix the form $\begin{bmatrix} 1 & 0 \\ 0 & R_1 \end{bmatrix}$ Where $R_1 \in SO(n-1)$

linear algebramatrices

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.

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:
Let $$\text{Exp}$$ denote the matrix exponential. Note that $$\text{Exp} \colon \text{Skew}(n) \to SO(n)$$. I claim that $$\text{Exp} : \text{Skew}(n) \to SO(n)$$ is surjective. Let $$A \in SO(n)$$. It suffices to show that each block of $$A$$, as described above, is the exponential of some skew symmetric matrix. Euler's formula shows you how to get the two by two blocks as exponentials of $$i\theta$$, which is skew symmetric. The fact that $$\det(A) > 0$$ implies that $$q$$ is even. Since $$q$$ is even, the $$-I_q$$ block can be split into two by two blocks with $$c = -1, s = 0$$, so Euler's formula works here too. The $$I_p$$ block is just $$\text{Exp}(0)$$. Putting these blocks together gives us $$B \in \text{Skew}(n)$$ with $$\text{Exp}(B) = A$$.
Now set $$\gamma(t) = \text{Exp}(tB)$$ for $$t \in [0, 1]$$. Since $$tB \in \text{Skew}(n)$$, it follows that $$\gamma(t) \in SO(n)$$, so $$\gamma : [0, 1] \to SO(n)$$. Since $$\gamma(0) = I$$, $$\gamma(1) = A$$, this gives the desired path.