TITLE: Symmetry of Grassmanians
QUESTION [11 upvotes]: I thought this might be simple (now I'm not sure) but can't solve it: why is it true that for $X,Y$ two linear $n$-subspaces of $\mathbb{R}^{n+k}$ there exists an orthogonal transformation of $\mathbb{R}^{n+k}$ that takes $X$ to $Y$ and $Y$ to $X$?
(This is excercise 5-D in Milnor's Characteristic classes, so the statement should hold.)

REPLY [2 votes]: What follows is inspired by the comments following Greg Kuperberg's answer to a MO question. Let $V$ be a $2n$ dimensional euclidean vector space, and $X,Y$ two subspaces of equal dimension. Without loss of generality, and as explained by Raul Gomez, we may assume $X\oplus Y= V$.
We are going to find a family $P_1,\dots,P_n$ of mutually orthogonal planes such that

$P_1\stackrel{\:\perp}{\oplus}\cdots\stackrel{\:\perp}{\oplus} P_n=V$,
for every $i$, $P_i\cap X$ and $P_i\cap Y$ are both lines.

(the first point simply reiterates the requirement that the $P_i$ be orthogonal). Once this is achieved, we may consider, for every $i$, one of the two reflections $r_i$ of $P_i$ that exchange $X_i$ and $Y_i$. The linear transform $r_1\oplus\cdots\oplus r_n$ is an orthonormal transformation of $V$ that exchanges $X$ and $Y$.

Consider the linear map $\pi:X^{\perp}\to X$ defined as the composite $X^{\perp}\xrightarrow{q} Y \xrightarrow{p} X$ where the first map projects onto $Y$ parallel to $X$ (this is always an isomorphism because both $Y$ and $X^{\perp}$ are complements to $X$), and the second map projects onto $X$ parallel to $X^{\perp}$.

It follows from the singular value decomposition theorem that there exists orthonormal bases $(\chi_1,\dots,\chi_n)$ and $(x_1,\dots,x_n)$ of $X^{\perp}$ and $X$ respectively, aswell as some nonnegative real numbers $0\leq\sigma_n\leq\cdots\leq\sigma_1$, such that for all $i$,
$$\pi(\chi_i)=\sigma_i x_i\,.$$
Now define, for all $i$, $P_i=\mathrm{Vect}(\chi_i,x_i)$. These are, by construction, pairwise orthogonal planes. Furhtermore, $P_i\cap X=\mathbb{R}x_i$ is a line, and so is $P_i\cap Y$. Indeed, $\pi(\chi_i)=\sigma_i x_i$ and $\pi(\chi_i)=p\circ q\,(\chi_i)=p(\chi_i+x)=\chi_i+x+\chi$ for some $x\in X$ and $\chi\in X^{\perp}$ such that $\chi_i+x=y\in Y$. Then
$$\sigma_ix_i-x=\chi_i+\chi\in X\cap X^{\perp}=0$$
implies $\chi=-\chi_i$ and thus $y=\sigma_ix_i-\chi=\sigma_ix_i+\chi_i$ is a nonzero element of $Y\cap P_i$, thus $Y\cap P_i$ is a line (it can't be the plane $P_i$).