Given $2\times 2$ matrices $\mathbf{U}_k$ for $k\in[L]$, what is the constraint on each of them such that the matrix $\mathbf{A}\in\mathbb{R}^{2^L \times 2}$
$$\mathbf{A} = \mathbf{U}_1 \odot \mathbf{U}_2 \odot \cdots \odot \mathbf{U_L}$$
has normalized rows?
The definition of Khatri-Rao product $\odot$ is as follow
$$\mathbf{A} \odot \mathbf{B} = [\mathbf{a}_1 \otimes \mathbf{b}_1 \quad\mathbf{a}_2 \otimes \mathbf{b}_2 \quad \ldots \quad \mathbf{a}_K \otimes \mathbf{b}_K] $$ where $\mathbf{a}_k$ and $\mathbf{b}_k$ are column vectors of $\mathbf{A}$ and $\mathbf{B}$.
Although the number of constraints (on row normalization) is $2^L$ and there are only $4 L$ variables, the requirement is not an impossibility. Consider
$$ \begin{pmatrix}a_0 & a_2 \\ a_1 & a_3 \end{pmatrix} \odot \begin{pmatrix}b_0 & b_2 \\ b_1 & b_3 \end{pmatrix} = \begin{pmatrix} a_0b_0 & a_2b_2 \\ a_0b_1 & a_2b_3\\ a_1b_0 & a_3b_2\\ a_1b_1 & a_3b_3 \end{pmatrix}$$
if we let $a_0=a_1$, $a_2=a_3$, $b_0=b_1$, $b_2=b_3$, then the normalization constraint becomes only
$$|a_0b_0|^2 + |a_2b_2|^2 = 1$$
and there would be plenty of degree of freedoms.
I'd like to know whether there is any local constraint (if and only if) on $\mathbf{U}_k$ such that the Khatri-Rao product of them will have normalized rows. Thanks very much.