Let $n,m,p,k_1,k_2$ be natural numbers.
Given two unitary matrices $U\in\mathbb{U}(n),V\in\mathbb{U}(m)$ and a decomposition of these as follows $$U=\bigg(\begin{matrix}A & C\\B& D\end{matrix}\bigg)~ ~ ~ ~ ~ ~V=\bigg(\begin{matrix}A' & C'\\B'& D'\end{matrix}\bigg)$$ where $A\in\mathcal{M}_{k_1,p}(\mathbb{C}),A'\in\mathcal{M}_{p,k_2}(\mathbb{C})$.
My question: is there a natural number $R$ and a unitary matrix $W\in\mathbb{U}(R)$ such that $$W=\bigg(\begin{matrix}AA' & E\\F& G\end{matrix}\bigg)$$ for some $E,F,G$.
Thoughts so far:
I think it should be true as the columns vectors that make $A$ and $A'$ are truncations of the column vectors of unitary matrices so have norm less that $1$. So I guess the column vectors that make $AA'$ should also have norm less that $1$ and it should be possible to add coordinates to these vectors to make them pairwise orthogonals and of norm $1$ (pairwise orthogonality is where I am stuck). Then one can conclude by completing the obtained orthonormal family into a basis and those vectors would form the columns of $W$.
The result is obviously true when $U,V$ are diagonals and I tried without success to extend this using the fact that a unitary is in the unitary conjugaison class of a diagonal matrix.
Any help would be hugely appreciated :)
$A$ and $A'$ each have spectral norm at most $1,$ so $AA'$ also has spectral norm at most $1.$ This means the matrix $I_p-(AA')^*(AA')$ is positive semidefinite, so has a $p\times p$ self-adjoint square root which we can take to be $F.$ This choice ensures $\begin{pmatrix}AA'\\ F\end{pmatrix}^*\begin{pmatrix}AA'\\ F\end{pmatrix}=I_p.$ Then the $2p\times p$ matrix $\begin{pmatrix}AA'\\ F\end{pmatrix}$ can be completed to a $2p\times 2p$ unitary matrix by taking an orthonormal basis of the orthogonal complement of the column space.