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.



Proving a lemma for a decomposition of orthogonal matrices
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}


