Householder's transformations of space Let we have two vectors $u, v \in \mathbb{R}^n$ such that the angle between them is acute ($u^Tv > 0$). I want to prove that there are such unitary transformation, that maps this vector to the first octant. Moreover I want to prove that such map can be represented as a product of two Householder reflections.
My idea is that if we choose a 2-dimentional plane containing vectors $(1, 0,0,...,0)^T$ and $(0, \frac{1}{\sqrt{n-1}}, \frac{1}{\sqrt{n-1}}, ..., \frac{1}{\sqrt{n-1}})^T$ and map vectors $u$ and $v$ such way that they lies in this plane and bisetor between $u$ and $v$ maps to vector $(\frac{1}{2}, \frac{1}{2\sqrt{n-1}}, \frac{1}{2\sqrt{n-1}}, ..., \frac{1}{2\sqrt{n-1}})^T$ then we get necessary transform. So because of length are reserved, this map is unitary. This is proof of existing such map. But I still can not find such two Householder matrices that represent it.
Thanks for the help!

Here is a rough idea. You may assume that $u$ and $v$ are unit vectors. Pick any two unit vectors $u''$ and $v''$ inside the first octant subtending the same angle at the origin as $u$ and $v$ do. Consider the angle bisectors $b=\frac{u+v}2$ and $b''=\frac{u''+v''}2$. Apply a Householder reflection $H$ about the plane orthogonal to $b-b''$, so that $b$ is mapped to $b''$.
Let the images of $u'=Hu$ and $v'=Hv$. So, both the plane $P'$ that passes through the origin and $u',v'$ and the plane $P''$ that passes through the origin and $u'',v''$ have $b''$ as an axis. Let $R$ be a plane that contains this axis and bisects the angle subtended by $P'$ and $P''$. Perform a reflection about $R$. Then the ordered pair $(u',v')$ is either mapped to $(u'',v'')$ or $(v'',u'')$.
