I'm not a mathematician but the engineering problem I'm considering is more of a mathematical question, that's why I post it here:
Consider the matrices $M$ ($n \times 1$), $T$ ($n \times m$) and $F (m \times 1)$, and:
$$
M = TF
$$
I solved for $F$ using the Moore-Penrose pseudo inverse. For this special case (where $T$ has full rank), $F$ is simply
$$
F = T^+M=T^T (TT^T)^{-1}M
$$
with $T^+$ the pseudo-inverse.
However, I want to find only $F \geq 0$. Is there a similar (simple) way to find $F$? Is it a classical problem? There is an infinite amount of solutions $F$, but I want to find $F$ that e.g. minimises the norm of $F$ (like the pseudo-inverse), or e.g. the sum of $F$.
Please let me know if the question is clear or if I need to be more precise
Best Answer
For this case, the target matrix $\mathbf{T}$ has full column rank. Therefore the null space $\mathcal{N}\left( \mathbf{A} \right)$ is trivial. Therefore the least squares solution is unique and given by the point $$ F = \mathbf{T}^{+}M = \left( \mathbf{T}^{*} \, \mathbf{T} \right)^{-1} \mathbf{T}^{*} M. $$
There is but one solution $F$. Options to compute $\mathbf{T}^{+}$ via the SVD, or $\left( \mathbf{T}^{*} \, \mathbf{T} \right)^{-1} \mathbf{T}^{*}$ through conventional means.