Existence of orthogonal transform that switches a linear subspace and a complement of it

Question

Suppose $V$ is an $m$ dimensional linear subspace of $\mathbb{R}^{2m}$, $W$ is a complement of $V$, ($V$ and $W$ might not be orthogonal). Does there exist a orthogonal transform $Q \in \mathbb{R}^{2m \times 2m}$ s.t. $Q(V)=W, Q(W)=V$?

Intuitively I believe $Q$ does exist. For $m=1$, $V$ and $W$ are just lines through the origin, so we can take the angle bisector of $V,W$ and define $Q$ to be the mirror reflection across the angle bisector. But I don't know how to argue for $m>1$. It's difficult for me to define a high dimensional analogy of angle bisector of two linear subspaces with equal dimension.

Answer 1

For $m=2$ the idea of @Qiaochu Yuan works, but the isometric isomorphism $T$ has to be chosen appropriately. The transformation $Q(v+w)=Tv+T^{-1}w$ is an isometry iff $$\langle Tv,T^{-1}w\rangle =\langle v,w\rangle\quad (*)$$ Let $v_1,v_2$ and $u_1,u_2$ be orthonormal bases of $V$ and $W$ respectively. Define $T$ by $$Tv_1=(\cos\alpha) u_1+(\sin\alpha) u_2=:w_1,\quad Tv_2=-(\sin\alpha )u_1+(\cos\alpha) u_2=:w_2$$ Then $w_1,w_2$ form an orthonormal basis of $W.$ Therefore the transformation $T$ extends to an isometric isomorphism from $V$ onto $W.$ The condition $(*)$ will be satisfied if $$\langle w_i,v_j\rangle=\langle Tv_i,T^{-1}w_j\rangle =\langle v_i,w_j\rangle\quad (**)$$ Clearly the latter holds for $i=j.$ Therefore it suffices to consider $i\neq j.$ WLOG we may assume that $i=1$ and $j=2.$ Then $$\langle w_1,v_2\rangle=(\cos\alpha) \langle u_1,v_2\rangle+ (\sin\alpha) \langle u_2,v_2\rangle\\ \langle v_1,w_2\rangle=-(\sin\alpha) \langle v_1,u_1\rangle+ (\cos\alpha) \langle v_1,u_2\rangle$$ The condition $(**)$ is satisfied if $$\sin\alpha\,[\langle v_1,u_1\rangle+\langle u_2,v_2\rangle]= \cos\alpha\,[\langle v_1,u_2\rangle-\langle u_1,v_2\rangle]$$ The latter is satisfied for an angle $\alpha\in (-\pi/2,\pi/2].$

Answer 2

Let $E$ be a $2m$-dimensional real inner product space such that $E = V\oplus W$ where $\dim(V)=\dim(W)=m$ but $V$ and $W$ are not assumed to be orthogonal.

Claim: There is a self-adjoint $Q\in \mathrm{O}(E)$ with $Q(V)=W$ and $Q(W)=V$. $Q$ is not unique unless $V\cap W^{\perp}=\{0\}$.

Proof: If $Q\in \mathrm{O}(E)$ is an isometry such that $Q(V)=W$, then its adjoint $Q^{\intercal}$ is its inverse, and $Q(V^{\perp})=W^{\perp}$. Moreover, the inner product restricts to give a pairing $b_V$ on $V$ and $b_W$ on $W$, but also a pairing $$ b_{V,W}=\langle -,- \rangle \colon V\times W \to \mathbb R $$ and if $v \in V, w\in W$ we must have $b_{V,W}(v,w) = \langle v, w\rangle = \langle Q(w), Q(v) \rangle = b_{V,W}(Q(w),Q(v))$. In particular since the left-radical of $b_{V,W}$ is $V\cap W^{\perp}$ while the right-radical is $V^{\perp}\cap W$ the since $Q(V)=W$ and $Q$ is an isometry, $Q(V\cap W^{\perp}) = W\cap V^{\perp}$. Since $b_V$ is an inner product on $V$, it follows that $V = (V\cap W^{\perp}) \oplus (V\cap W^{\perp})^{\perp_V}$, where $(-)^{\perp_V}$ denotes the subspace of vectors in $V$ orthogonal to $(-)$, that is $(-)^{\perp_V} = (-)^{\perp} \cap V$. Thus we obtain $$ V = (V\cap W^{\perp}) \oplus (V\cap(W+V^{\perp})), $$ an orthogonal direct sum. Similarly, we find that $W$ decomposes as the orthogonal direct sum $(W\cap V^{\perp})\oplus (W\cap(V+W^{\perp}))$. The pairing $b_{V,W}$ restricts to a nondegenerate pairing between $V\cap (W+V^{\perp})$ and $W\cap (V+W^{\perp})$, so that we have $$ E = (V\cap W^{\perp})\oplus \left(V\cap(W+V^{\perp}) \oplus (W\cap (V+W^{\perp})\right) \oplus (V^{\perp}\cap W) $$ an orthogonal direct sum decomposition of $E$ into $3$ subspaces, the second of which has two summands which are nondegenerately paired by the inner product.

Recall that a nondegenerate bilinear pairing $b\colon U_1\times U_2\to \mathbb R$ on two finite-dimensional vector spaces $U_1$ and $U_2$ induces isomorphisms $\theta_1\colon U_1\to U_2^*$ and $\theta_2\colon U_2 \to U_1^*$ (where $U_1$ and $U_2$ may or may not be distinct) defined by $\theta_1(u_1)(u_2) = b(u_1,u_2)$.

To construct $Q$, one may choose any isometry $q\colon (V\cap W^{\perp}) \to (W\cap V^{\perp})$. On the other hand, the bilinear pairings $b_V, b_W,b_{V,W}$ give maps $\theta_V\colon V\to V^*, \theta_W \colon W\to W^*$ and $\theta_{V,W}\colon V \to W^*$, which restrict to give the following diagram:

$$ \require{AMScd} \begin{CD} V\cap (W+V^{\perp}) @>{\theta_{V,W}}>> (W\cap (V+W^{\perp}))^* \\ @V{\theta_{V}}VV @A{\theta_{W}}AA \\ (V\cap (W+V^{\perp})^* @<{\theta_{W,V}}<< (W\cap(V+W^{\perp})) \end{CD} $$ Thus we obtain a canonical isometry $c:=\theta_W^{-1}\circ \theta_{V,W}\colon V\cap (V^{\perp}+W)\to W\cap (W^{\perp}+V)$. on $V\cap (W+V^{\perp})$, and its adjoint $c^{\intercal}\colon \theta_{W,V}\circ\theta_V^{-1}\colon W\cap (W^{\perp}+V) \to V\cap(V^{\perp}+W)$ is its inverse. Cashing out what this means from the definitions is that the condition that $$ \langle v,w\rangle = \langle c^\intercal(w),c(v)\rangle \quad \forall v\in V, w\in W $$ in fact uniquely specifies what $c^{\intercal}(w)$ and $c(v)$ have to be!

Combining with $q$ we obtain the map $Q$ given by $(q\oplus c)\oplus (q^{\intercal}\oplus c^{\intercal})$, an isometry interchanging $V$ and $W$. The uniqueness assertion is just the statement that $c$ is canonical.