Orthogonal matrix of determinant $-1$

I read somewhere that for an orthogonal matrix in $\mathbb{R}^n$, that if $P(u) = -u$ for any unit vector $u$ and $P(v) = v$ with $\langle u, v \rangle = 0$, then this matrix has determinant $-1$. I'm having trouble making sense of this and piece everything together.

Solution 1:

If $P(u)=-u$ for some unit vector $u\in \mathbb{R}^n$, and for any $v\in \mathbb{R}^n$ with $\langle u,v\rangle=0$, we have $P(v)=v$ then the determinant of $P$ is indeed $-1$.

Pick a basis for $\langle u\rangle^\perp=\{v\vert\,\, \langle u,v\rangle=0\}$ and add $u$ to it. The result is a basis for $\mathbb{R}^n$, as $$\langle u\rangle^\perp\cap\langle u\rangle=0, \qquad\langle u\rangle^\perp + $\langle u\rangle=\mathbb{R}^n.$$

With respect to this basis $P$ is represented by a matrix which differs from the identity in only the diagonal element on the row and column corresponding to $u$, which is $-1$. The determinant of $P$ is therefore $-1$.