Projecting onto the space of Upper-Triangular-ish Matrices I want to solve the independent component analysis (ICA) problem for a single time domain channel of input. The problem is, in my formulation, the mixing matrix has a special structure. I want to know if I can find a least-squares type projection of my estimated mixing matrix onto this special matrix structure.
Background:
Let $t_{n}$ be the $n$-th sample point, our received signal is then:
\begin{equation}
    x(t_{n}) = \sum_{m=1}^{M} a_{m}(t_{n}) \cdot s_{m}(t_{n}) = \mathbf{a}^{T}(t_{n}) \mathbf{s}(t_{n})
\end{equation}
where $M$ is the number of sources, $\mathbf{a}(t_{n}) = [ a_{1}(t_{n}), \ldots, a_{M}(t_{n})  ]$ are the mixing coefficients, and $\mathbf{s}(t_{n}) = [s_{1}(t_{n}), \ldots, s_{M}(t_{n})]^{T}$ are the sources -- all at time $t_{n}$. Assuming our sampled signal is of length $L$, we can express this as a matrix-vector equation via:
\begin{equation}
    \mathbf{x} = \mathbf{A} \mathbf{s}
\end{equation}
where $\mathbf{x} = [ x(t_{1}), x(t_{2}), \ldots, x(t_{L})]^{T}$, $\mathbf{s} = [ \mathbf{s}(t_{1}), \mathbf{s}(t_{2}), \ldots, \mathbf{s}(t_{L})]^{T}$ and:
\begin{equation}
    \mathbf{A} = \begin{bmatrix} \mathbf{a}^{T}(t_{1}) & \mathbf{0}^{T} & \cdots & \mathbf{0}^{T} \\ \mathbf{0}^{T} & \mathbf{a}^{T}(t_{2}) & \cdots & \mathbf{0}^{T} \\ \vdots & \vdots & \ddots & \vdots \\  \mathbf{0}^{T} & \mathbf{0}^{T} & \cdots & \mathbf{a}^{T}(t_{L})   \end{bmatrix}
\end{equation}

We see that this problem has been formulated as a standard ICA problem, and so, we could potentially solve it using ICA techniques. However -- standard ICA assumes, generally, that the matrix $\mathbf{A}$ is dense (or at least imposes no special structure on it).
Suppose that I apply some standard ICA technique and get a dense $\mathbf{\hat{A}}$. Is there a way I could project this matrix onto the space of matrices that have the form:
\begin{equation}
    \mathbf{A} = \begin{bmatrix} \mathbf{a}^{T}(t_{1}) & \mathbf{0}^{T} & \cdots & \mathbf{0}^{T} \\ \mathbf{0}^{T} & \mathbf{a}^{T}(t_{2}) & \cdots & \mathbf{0}^{T} \\ \vdots & \vdots & \ddots & \vdots \\  \mathbf{0}^{T} & \mathbf{0}^{T} & \cdots & \mathbf{a}^{T}(t_{L})   \end{bmatrix}
\end{equation}
in such a way that the resulting projected matrix would be 'as close as possible' to the dense matrix in some way?
 A: Yes. As it turns out, the result of finding the projection of $\hat A$ onto the vector space of matrices with the required zero pattern relative to the Frobenius inner product coincides with the result of the "common sense" approach: take any coefficients that are required to be zero and set them equal to zero.
Here's a quick proof: let $\mathbf e_1,\dots,\mathbf e_m$ denote the standard basis vectors of $\Bbb R^m$. Let $\mathbf  E_{m,n}$ denote the matrix $\mathbf  A$
corresponding to the mixing coefficients $\mathbf a(t)$ such that $\mathbf a(n) = \mathbf e_m$ and $\mathbf a(t) = \mathbf 0$ for all $t \neq n$. The set $\{E_{m,n} : 1 \leq m \leq M, \ 1 \leq n \leq L\}$ forms an orthonormal basis for the vector subspace of interest (i.e. the set of matrices of the desired form). As such, the projection of $\hat{\mathbf A}$ onto this subspace can be computed as
$$
\mathbf A_{\text{proj}} = \sum_{m=1}^M \sum_{n = 1}^L \langle \hat{\mathbf A}, \mathbf E_{m,n} \rangle \mathbf E_{m,n}.
$$
Verify that this is the matrix obtained by setting any entries of $\hat{\mathbf A}$ outside of the "block-diagonal" equal to zero.
