Rank of eigenvector projections Consider the following two $n\times n$ matrices:
$$A_1:=H_1\Sigma_1H_1' $$
$$A_2:=H_2\Sigma_2H_2'$$
where $\Sigma_1$,$\Sigma_2$ are positive definite $k\times k$ matrices ($k<n$), and $H_1$,$H_2$ are $n\times k$ matrices such that
$$H_1'H_1=H_2'H_2=I_k \hspace{1cm}\text{and} \hspace{1cm} H_1'H_2=\begin{bmatrix}
I_r & 0 \\
0 & 0 
\end{bmatrix} .$$
Now I perform the eigendecomposition of $A_1$ and $A_2$ (with eigenvalues in decreasing order and orthonormal eigenvectors), but keep only the first $k$ eigenvectors  of $A_1$ and the first $s$ eigenvectors $(s>k)$ of $A_2$. Let $U_1$ and $U_2$ be the $n\times k$ and $n\times s$ matrices of eigenvectors for $A_1$ and $A_2$ respectively. I want to show that the matrix
$$U_1-U_2U_2'U_1=MU_1$$
has rank $n-r$, where $M:=I_n-U_2U'_2$. This is similar to here, but the columns of $H_1$ and $H_2$ are not fully independent.
Any help is greatly appreciated.
 A: The columns of $U_1$ form an orthonormal basis for the column space of $A_1$, which is in turn the column space of $H_1$. Let $S_1$ denote this column space.
The columns of $U_2$ form an orthonormal basis for the sum of the column space of $A_2$ and an arbitrary $(s-k)$-dimensional subspace of $\ker(A_2)$. Let $S_2$ denote the column space of $U_2$.
From the construction of $A_1$ and $A_2$, we know that the intersection of the column spaces of $A_1$ and $A_2$ is $r$-dimensional (in particular, it is equal to the column space of the matrix $J_0$ defined in the comments on your question). Let $S_0$ denote this intersection.
The rank of $(I - U_2U_2')U_1$ is equal to $\dim(S_1) - \dim(S_1 \cap S_2)$. Because $S_0 \subseteq S_2$, the dimension of this intersection is at most equal to
$$
\dim(S_1) - \dim(S_0) = k - r.
$$
However, if $S_0 \subsetneq S_2$, then this rank will be strictly smaller. In other words, if the arbtirary $(s-k)$-dimensional subspace of $\ker(A_2)$ has a non-trivial intersection with the column space of $A_1$, then the rank will be less than $k-r$.
If this $(s-k)$-dimensional subspace is "randomly" selected, then the probability of a non-trivial intersection occurring by chance is extremely small, which would explain your empirical results.

Illustrative counterexample:
$$
H_1 = \pmatrix{1&0\\0&1\\0&0\\0&0}, \quad H_2 = \pmatrix{1&0\\0&0\\0&1\\0&0}, \Sigma_1 = \Sigma_2 = \pmatrix{2&0\\0&1}.
$$
Note that the following is possible:
$$
U_1 = H_1, \quad U_2 = \pmatrix{1&0&0\\0&0&1\\0&1&0\\0&0&0}.
$$
We compute
$$
(I - U_2U_2')U_1 = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1} \pmatrix{1&0\\0&1\\0&0\\0&0} = 0,
$$
which has rank smaller than $k - r = 2 - 1 = 1$. On the other hand, if you take the final column of $U_2$ to be a "random" unit vector from the set $\{(0,x_2,0,x_4): x_2,x_4 \in \Bbb C\}$, you should find that the matrix $(I - U_2U_2')U_1$ has rank $1$ in practically all cases.
