Setting
Consider two independent orthogonal matrices, which are decomposed into 4 blocks:
\begin{align}
Q_{1}
=
\left[\begin{array}{cc}
A_{1} & B_{1}\\
C_{1} & D_{1}
\end{array}\right]
,
\,Q_{2}=\left[\begin{array}{cc}
A_{2} & B_{2}\\
C_{2} & D_{2}
\end{array}\right]\in \mathbb{R}^{d\times d}
\end{align}
where $A_i \in \mathbb{R}^{r\times r}$ and $D_i \in \mathbb{C}^{(d-r)\times (d-r)}$ and $d\ge 3r$.
Also notice that
\begin{align}
\forall i=1,2:I_{d}=\left[\begin{array}{cc}
I_{r} & 0\\
0 & I_{d-r}
\end{array}\right]
&=
Q_{i}^{\top}Q_{i}=\left[\begin{array}{cc}
A_{i}^{\top}A_{i}+C_{i}^{\top}C_{i} & A_{i}^{\top}B_{i}+C_{i}^{\top}D_{i}\\
B_{i}^{\top}A_{i}+D_{i}^{\top}C_{i} & B_{i}^{\top}B_{i}+D_{i}^{\top}D_{i}
\end{array}\right]
\\
&=Q_{i}Q_{i}^{\top}=\left[\begin{array}{cc}
A_{i}A_{i}^{\top}+B_{i}B_{i}^{\top} & A_{i}C_{i}^{\top}+B_{i}D_{i}^{\top}\\
C_{i}A_{i}^{\top}+D_{i}B_{i}^{\top} & C_{i}C_{i}^{\top}+D_{i}D_{i}^{\top}
\end{array}\right]
\end{align}
Goal
I am trying for a few days to prove (or refute) the following lemma.
In my numerical simulations, it always holds for random $Q$s.
Prove (or refute) that if $v\in\mathbb{C}^{d-r}$ is an eigenvector
of $D_1^\top M$
(with $\lambda=1$), i.e.,
\begin{align}
D_{1}^{\top}
\underbrace{D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)}_{\triangleq M}v&=v
\end{align}
then it is in the nullspace of $C_2^\top M$, i.e.,
\begin{align}
C_{2}^{\top}
\underbrace{D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)}_{=M}v&=0_{r}
\end{align}
Update: Alternative goal
If one could show that any eigenvector such that
$D_{1}^{\top}
D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v=v$
is also in the null space of $B_1$, i.e., $B_1 v = 0_r$, then I know how to prove the rest of the lemma above using it. And again, empirically, this also seems to always hold.
I feel like I may be missing some simple trick here.
Any help would be greatly appreciated.
Best Answer
Using the direction Federico Poloni gave, I was finally able to prove it.
Notice that the matrices are real, so I use $A^\top$ and $A^H$ interchangeably.
Let $v\in\mathbb{\mathbb{C}}^{d-r}$ be a normalized eigenvector as required.
Lemma: The norm of $v$ is preserved under the following transformations $\left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}=\left\Vert D_{2}^{\top}D_{1}v\right\Vert _{2}=\left\Vert D_{1}v\right\Vert _{2}$.
Proof. We use a simple trick using the Euclidean (operator) norms, \begin{align} v&=D_{1}^{\top}D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v\\v^{H}v&=v^{H}D_{1}^{H}D_{2}\left(B_{2}^{H}B_{1}+D_{2}^{H}D_{1}\right)v\\1&=\left\Vert v^{H}D_{1}^{H}D_{2}\left(B_{2}^{H}B_{1}+D_{2}^{H}D_{1}\right)v\right\Vert _{2}\\&\le\underbrace{\left\Vert v^{H}D_{1}^{H}D_{2}\right\Vert _{2}}_{\le1}\underbrace{\left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}}_{\le1}\\&\le\underbrace{\left\Vert v^{H}D_{1}^{H}\right\Vert _{2}}_{\le1}\underbrace{\left\Vert D_{2}\right\Vert _{2}}_{\le1}\underbrace{\left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}}_{\le1}\le1, \end{align} where we used the fact that $\left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)$ is the bottom-right block of the orthogonal matrix $Q_{2}^{\top}Q_{1}$. As a conclusion, we get the following equalities: \begin{align} \left\Vert \left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v\right\Vert _{2}&=1\\\left\Vert v^{H}D_{1}^{H}D_{2}\right\Vert _{2}=\left\Vert D_{2}^{H}D_{1}v\right\Vert _{2}=\left\Vert D_{2}^{\top}D_{1}v\right\Vert _{2}&=1\\\left\Vert v^{H}D_{1}^{H}\right\Vert _{2}=\left\Vert D_{1}v\right\Vert _{2}&=1. \end{align}
Lemma: The vector $v$ is in the null space of $B_{2}^{\top}B_{1}$, i.e., $B_{2}^{\top}B_{1}v=0_{r}$.
Proof. We get back to the equality above and show \begin{align}1&=v^{H}D_{1}^{H}D_{2}\left(B_{2}^{H}B_{1}+D_{2}^{H}D_{1}\right)v\\&=v^{H}D_{1}^{H}D_{2}B_{2}^{H}B_{1}v+v^{H}D_{1}^{H}D_{2}D_{2}^{H}D_{1}v\\&=v^{H}D_{1}^{H}D_{2}B_{2}^{H}B_{1}v+\underbrace{\left\Vert D_{2}^{H}D_{1}v\right\Vert _{2}}_{=1}\\0&=v^{H}D_{1}^{H}D_{2}B_{2}^{H}B_{1}v. \end{align} We use the fact that $\left(B_{2}^{T}B_{1}+D_{2}^{T}D_{1}\right)v$ is a normalized vector to obtain \begin{align} 1&=\left\Vert \left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v\right\Vert _{2}^{2}=\left\Vert B_{2}^{\top}B_{1}v\right\Vert _{2}^{2}+\underbrace{2v^{\top}D_{1}^{\top}D_{2}B_{2}^{\top}B_{1}v}_{=0}+\underbrace{\left\Vert D_{2}^{\top}D_{1}v\right\Vert _{2}^{2}}_{=1}\\0&=\left\Vert B_{2}^{\top}B_{1}v\right\Vert _{2}^{2}\\0_{r}&=B_{2}^{\top}B_{1}v. \end{align}
Lemma: The vector $v$ is in the null space of $C_{2}^{\top}D_{1}$.
Proof. We use the above results, \begin{align} 1&=v^{H}D_{1}^{H}D_{2}D_{2}^{H}D_{1}v=v^{H}D_{1}^{H}\left(I_{d-r}-C_{2}C_{2}^{H}\right)D_{1}v \\&=\underbrace{v^{H}D_{1}^{H}D_{1}v}_{=1}-v^{\top}D_{1}^{H}C_{2}C_{2}^{H}D_{1}v\\0&=\left\Vert C_{2}^{\top}D_{1}v\right\Vert _{2}^{2}\\0_{r}&=C_{2}^{\top}D_{1}v. \end{align}
Finally, we are ready to conclude the theorem we wanted to prove.
\begin{align} C_{2}^{\top}D_{2}\left(B_{2}^{\top}B_{1}+D_{2}^{\top}D_{1}\right)v &=C_{2}^{\top}D_{2}D_{2}^{\top}D_{1}v \\&=C_{2}^{\top}\left(I_{d-r}-C_{2}C_{2}^{T}\right)D_{1}v \\ &=\underbrace{C_{2}^{\top}D_{1}v}_{=0_{r}}-C_{2}^{\top}C_{2}\underbrace{C_{2}^{\top}D_{1}v}_{=0_{r}}=0_{r} \end{align}