Reflection relating two subspaces

Question

Let $S_1, S_2 \subseteq \mathbb{R}^n$ be two linear $k$-dimensional subspaces. Does there always exist a hyperplane $H$ such that $S_1 = R_H S_2$, where $R_H$ denotes the orthogonal reflection across $H$?

Geometric intuition leads me to suspect that this is obviously true, but a proof is eluding me.

Edit: As @zvbxrpl pointed out below, my intuition was wrong and the claim is false. However, the claim may still be true with "hyperplane" replaced by some linear subspace -- @zvbxrpl attempted to show this, but unfortunately there was an error in the proof.

Is the claim true if "hyperplane" is replaced by some linear subspace?

Edit 2: Studiosus has given an affirmative answer to the above revised question using Riemannian geometry. This is nice, but I am interested in a constructive, elementary answer using only linear algebra.

Answer 1

Let $E$ be an Euclidian vector space and $S_1,S_2$ two subspaces of same dimension $k$. Assume that $E = S_1 \oplus S_2$. Then there exists $f$ an orthogonal symetry such that $f(S_1)=S_2$ and $f(S_2)=S_1$.

Proof: Let $P$ be the matrix of the scalar product on $E$, and choose a basis of $S_1$ and a basis of $S_2$ such that $P$ has the form : $$ P = \begin{bmatrix} I_k & C \\ {}^tC & I_k \end{bmatrix}.$$ Let $A$ be a $k \times k$ invertible matrix and $f$ be an endomorphism of $E$ such that its matrix is : $$ f = \begin{bmatrix} 0 & A^{-1} \\ A & 0 \end{bmatrix}.$$ Then $f$ is a symetry such that $f(S_1)=S_2$ and $f(S_2)=S_1$. But $f$ is orthogonal iff $${}^t f.P.f = P,$$ that is, $$\begin{bmatrix} {}^tA.A & {}^tA.^tC.A^{-1} \\ {}^tA^{-1}.C.A & {}^tA^{-1}.A^{-1} \end{bmatrix} = \begin{bmatrix} I_k & C \\ {}^tC & I_k \end{bmatrix}.$$ Which means that $A$ is orthogonal and $CA$ is symetric. So if $A$ is the inverse of the orthognal part of $C$ in the polar decomposition, then $f$ is an orthogonal symetry.

In the general case, if $S_1$ and $S_2$ have an intersection. Denote $S = S_1 \cap S_2$, and write $S_1 = S \oplus^{\perp} T_1$ and $S_2 = S \oplus^{\perp} T_2$. Then $E = S \oplus^{\perp} (T_1 \oplus T_2)$. Now choose $f : E \rightarrow E$ such that $f$ switchs $T_1$ and $T_2$ on $T_1 \oplus T_2$ as above and such that $f$ is the identity on $S$.

Answer 2

What you seek to prove is false if you require the subspace $H$ to be a hyperplane.

To see this, let $S_1,S_2$ two planes in $\mathbb{R}^4$ that only meet in the origin (e.g. $S_1=span\{e_1,e_2\},$ $S_2=span\{e_3,e_4\}$), then for every hyperplane $H\subseteq\mathbb{R}^4$ we have $L\subseteq H\cap S_2$ for some line $L$ in $\mathbb{R}^4$, which means that $L\subseteq R_HS_2$, so $R_HS_2\neq S_1$.

What is true is that for two equidimensional subspaces $S_1,S_2\subseteq\mathbb{R}^n$ there is a subspace $T$ such that $R_TS_2=S_1$.

Let $\{u_1,\dots,u_h\}$ an orthonormal basis of $S_1\cap S_2$, $\{u_1,\dots,u_h,v_1,\dots,v_k\}$ an orthonormal basis of $S_1$, and $\{u_1,\dots,u_h,w_1,\dots,w_k\}$ an orthonormal basis of $S_2$. We put $T=span\{v_1+w_1,\dots,v_k+w_k\}$.

It is easy to see that $R_TS_2=S_1$. In fact we have $R_T(u_i)=-u_i$ since $T\perp(S_1\cap S_2)$ and $$R_T(w_i)=\Pi_T(w_i)-\Pi_{T^\perp}(w_i)=\frac{w_i+v_i}{2}-\frac{w_i-v_i}{2}=v_i$$ where $\Pi_T$ and $ \Pi_{T^\perp}$ are the orthogonal projections on $T$ and $T^\perp$, respectively. This concludes the proof. EDIT: This paragraph is wrong, see comments.

It's clear that the subspace $T$ is very much not unique. The presented choice for $T$ gives a subspace of minimal dimension with the desired property. If you are interested in a subspace of maximal dimension, a possible choice is $T'=T\oplus(S_1\cap S_2)\oplus(S_1+S_2)^\perp$

Answer 3

Here is a proof of the claim made by zvbxrpl using Riemannian geometry; it is a bit over the top as there should be just a linear algebra argument, I just do not have to to work out an elementary proof:

Theorem. For every pair of $k$-dimensional linear subspaces $P, Q$ in $E^n$ there exists an isometric involution $s=s_R$ whose fixed point set is a $k$-dimensional subspace $R$ in $E^n$ such that $s(P)=Q, s(Q)=P$.

Proof. The space of $k$-dimensional linear subspaces in $E^n$ can be equipped with the structure of a smooth manifold, called the Grassmannian $Gr_k(E^n)$. The latter admits a Riemannian metric making it into a symmetric space: For every $P\in Gr_k(E^n)$ there exists an isometric involution $s_P$ fixing $P$ pointwise such that the derivative of $s_P$ at $P$ is $-1$. More specifically, $s_P$ is the unique isometric (with respect to the euclidean metric on $E^n$) linear involution of $E^n$ fixing $P$ pointwise and having $-1$-eigenspace equal to the orthogonal complement to $P$.

See for instance pages 1-5 of these notes by Eschenburg where he works out this in detail.

Now, given two elements $P, Q\in Gr_k(E^n)$, there is a (possibly nonunique) minimizing geodesic $c$ connecting $P$ to $Q$. Let $R$ be the midpoint of this geodesic. Then the involution $s_R$ will send $c$ to itself and reverse its orientation, thus sending $P$ to $Q$ and $Q$ to $P$. qed