Invariant subspace under rotation Let $U$ be a $n-1$-dimensional subspace of $\mathbb{R}^n$. Consider 2 vectors $x,y\in U$ with $||x||=||y||$ and assume there is a rotation $\phi$ such that $\phi(x)=y.$
How can we show that $\phi(U)=U$ (assuming this is correct)?
Thank you.
 A: It is false if $n>3$. For example, consider
$$U=\operatorname*{span}\{(1,0,0,0),(0,1,0,0), (0,0, 1,0)\}=\operatorname*{span}\{e_1, e_2, e_3\},$$
which is a subspace of $\mathbb R^4$. The rotation matrix
$$
\phi=\begin{bmatrix} 0& 1 & 0 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 &0 \end{bmatrix}$$
is such that
$$
\phi e_1=e_2,$$
so your condition is satisfied with $x=e_1, y=e_2$. However, $\phi e_3=e_4$, so $\phi(U)\ne U$.
Sometimes our three-dimensional intuition fails in higher dimension.
A: Nevertheless on can find rotations which satisfy the property.

*

*Any reflexion preserves distances, so it is a rotation (=orthogonal matrix), and in your case, you can for example take the reflexion w.r.t the hyperplane $H:=(\mathbf{x}-\mathbf{y})^{\perp}$.

Forgetting the subspace $U$ we have the decomposition $\mathbb{R}^n=(\mathbf{x}-\mathbf{y}) \oplus H$. We can adapt it to $U\subset \mathbb{R}$ by choosing $\mathbf{z}\in U^{\perp}\subset (\mathbf{x}-\mathbf{y})^{\perp} =H $. One has
$$ \mathbb{R}^n=\mathbb{R}(\mathbf{x}-\mathbf{y}) \oplus (H\cap U) \oplus \mathbf{z}\quad \text{and}\quad U= \mathbb{R}(\mathbf{x}-\mathbf{y}) \oplus (H\cap U) $$
In restriction to the line $\mathbb{R}(\mathbf{x}-\mathbf{y})$ the reflexion is $-1$ and so preserves this space and it is identity on $H$ and so also preserves $(H\cap U)$. Hence it preserves $\mathbb{R}(\mathbf{x}-\mathbf{y}) \oplus (H\cap U)=U$.


*Another more intuitive rotation (and one that preserves orientation, i.e. of determinant 1) would be that around the axis $\mathbf{z}$ of angle precisely the angle between $\mathbf{x}$ and $\mathbf{y}$. So in fact use $\mathbf{z}$ to define a rotation on the subspace $U$ and extend the map to the whole $\mathbb{R}^n$ by identity on the last direction $\mathbf{z}$.

