Connectedness of Orthogonal Matrices with positive determinant 
Prove that the set of orthogonal matrices $n\times n$ with positive determinant is arc-connected.

A set $X\subset \mathbb{R}^n$ is arc-connected if for any two points,$x,y\in X$ exist a continous map $f: [0,1]\to X$ such that $f(0)=x$ and $f(1)=y$.
We can see a matrix $n\times n$ like a subset of $\mathbb{R}^{n^2}$.
 A: It is enough to find a path between $A \in \operatorname{SO}_n(\mathbb{R})$ and the identity matrix $I_n$ which stays in $\operatorname{SO}_n(\mathbb{R})$. The spectral theorem for orthogonal matrices says we can find an orthogonal $P$ such that
$$ P^TAP = \begin{cases} \operatorname{diag}(R(\theta_1), \dots, R(\theta_k)) & n = 2k \text{ is even}, \\
\operatorname{diag}(R(\theta_1),\dots,R(\theta_k), 1) & n = 2k + 1 \text{ is odd} \end{cases} $$
where 
$$ R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} $$
is a $2 \times 2$ rotation matrix. If $n$ is even then
$$ t \mapsto P\operatorname{diag}(R((1 - t) \theta_1), \dots, R((1 - t) \theta_{k})P^T $$
is then a path between $A$ and $I_n$ and similarly for the odd dimensional case.
A: This can be found in textbooks. All one needs is to join an orthogonal matrix (with determinant 1) to the identity by means of a path in $SO(n)$.
Such a matrix $A$ is conjugate in $SO(n)$ to a diagonal sum of matrices
$\pmatrix{\cos t&\sin t\\-\sin t&\cos t}$ together with an extra $(1)$ when $n$ is odd. Now one can join these 2-by-2 rotation matrices to the identity in $SO(2)$.
