Orthonormal columns of block matrices expanded with Kronecker products Let $W_{i,1},W_{i,2},W_{i,3} \in \mathbb{R}^{n \times n}$, $i \in {1,2}$ be such that
$$
\eqalign{ 
    \Big[\matrix{W_{i,1}^T & W_{i,2}^T & W_{i,3}^T}\Big] 
    \left[\matrix{W_{i,1}\\W_{i,2}\\W_{i,3}}\right]
    = I
}
$$
where $I$ is the identity matrix. Now, let
$$\eqalign{
    W = \left[\matrix{
        W_{2,1}W_{1,1} \\
        W_{2,2}W_{1,1} \\
        W_{2,3}W_{1,1} \\
        W_{1,2} \\
        W_{1,3}} \right]
}$$
Then, $W^{T}W = I$ but I don't see why. How can this idea be expanded towards the Kronecker products, i.e., $W_{i,j} \otimes W_{i,k}$ for $i \in \{1,2\}$ and $j,k \in \{1,2,3\}$?
 A: Let's start with showing $W^\top W = I$.
We can write the assumption about the $W_{i,j}$ as
$$
\begin{align}
    \label{eq:condition}\tag{\(*\)}
    W_{i,1}^\top W_{i,1} + W_{i,2}^\top W_{i,2} + W_{i,3}^\top W_{i,3}
    = I,
    \qquad i = 1,2.
\end{align}
$$
Now let's expand $W^\top W$ in a similar way.
$$
\begin{align*}
    W^\top W
    &= \Big[\begin{array}{ccccc}
        (W_{2,1}W_{1,1})^\top &
        (W_{2,2}W_{1,1})^\top &
        (W_{2,3}W_{1,1})^\top &
        W_{1,2}^\top &
        W_{1,3}^\top
    \end{array}\Big]
    \left[\begin{array}{c}
        W_{2,1}W_{1,1} \\
        W_{2,2}W_{1,1} \\
        W_{2,3}W_{1,1} \\
        W_{1,2} \\
        W_{1,3}
    \end{array}\right]
    \\
    &= W_{1,1}^\top W_{2,1}^\top W_{2,1} W_{1,1}
     + W_{1,1}^\top W_{2,2}^\top W_{2,2} W_{1,1}
     + W_{1,1}^\top W_{2,3}^\top W_{2,3} W_{1,1}
     + W_{1,2}^\top W_{1,2}
     + W_{1,3}^\top W_{1,3}
    \\
    &= W_{1,1}^\top \left(W_{2,1}^\top W_{2,1} 
     + W_{2,2}^\top W_{2,2} 
     + W_{2,3}^\top W_{2,3} \right)W_{1,1}
     + W_{1,2}^\top W_{1,2}
     + W_{1,3}^\top W_{1,3}
\end{align*}
$$
Now recognize that the sum in the parentheses is \eqref{eq:condition} with $i = 2$.
Then we have a simplification:
$$
\begin{align*}
    W^\top W
    &= W_{1,1}^\top (I) W_{1,1}
     + W_{1,2}^\top W_{1,2}
     + W_{1,3}^\top W_{1,3}
    \\
    &= W_{1,1}^\top W_{1,1}
     + W_{1,2}^\top W_{1,2}
     + W_{1,3}^\top W_{1,3}
    \\
    &= I,
\end{align*}
$$
using \eqref{eq:condition} again but with $i=1$.
Not sure how this relates to the Kronecker product though, especially since $W$ has matrix-multiplied blocks instead of element-wise multiplications. Note though that $W$ has orthonormal columns, since $W^\top W = I$, but the $W_{i,j}$ don't necessarily, because $W_{i,j}^\top W_{i,j} = I$ for all $i,j$ would contradict \eqref{eq:condition}.
