I need to solve for $B$ in the equation $A B = C$ where: $A$ and $C$ are known $1 \times 6$ vectors and $C$ is an unknown $ 6 \times 6$ transition probability matrix (i.e. rows sum = to $1$).
As far as I understand that gives us $12$ equations for $36$ unknowns meaning there could be infinitely many solutions. I know that $11$ of the transition probabilities are usually zero or very close so we could set those to zero to inform our solution a bit more.
I want to choose a solution that is reasonable in the context of the problem, in the context these matrices typically have much higher values in the diagonals (in the order of $0.7$) and adjacent cells (in the order of $0.1$).
What is the best way of incorporating this loose criterion?
I thought of maybe finding a way to generate many solutions and then picking one that is closest to what we could expect in reality.
Would that be possible or does anyone have a better idea?
I work in insurance and use R for my matrix calculations.
You could call this as the " Inverse Outer Product " problem. [ Let us choose the vector $B$ arbitrarily as $[b_1 , b_2 , ... , b_n] $. Denote $s= b_1+b_2+...+b_n$ .Then choose the vector $A$ as $[1/s , 1/s,...,1/s]$. If you take the outer product now as $A^t.B$ we will get an $n$ x $n$ matrix, whose row sum is one.] But, in our case we have $A,C$ to be known. We'll derive one condition taking a simple case $n=2$. Let $A=[a,b],B=[x_1,x_2]$. Then the matrix $A^t.B$ has row vectors $[ax_1,ax_2]$ and $[bx_1,bx_2]$.If the given matrix $C$ has row vectors $[p,1-p]$ and $[r,1-r]$. For the solution to be unique, we need $p/x_1 = (1-p)/x_2$ and $r/x_1=(1-r)/x_2$. Also, $x_1+x_2=1/a=1/b$. At the end we get two solutions like this $x_1=p/a, x_2=(1-p)/a$ and $x_1=r/b,x_2=(1-r)/b$ . $ \begin{bmatrix} a \\ b\\ c\\ d\\ e\\ f\\ \end{bmatrix} \begin{bmatrix} x_1 & x_2 &x_3 &x_4 &x_5& x_6 \end{bmatrix}= \begin{bmatrix} c_{11} & c_{12}&c_{13} &c_{14} &c_{15}& c_{16}\\ c_{21} & c_{22}&c_{23} &c_{24} &c_{25}& c_{26}\\ c_{31} & c_{32}&c_{33} &c_{34} &c_{35}& c_{36}\\ c_{41} & c_{42}&c_{43} &c_{44} &c_{45}& c_{46}\\ c_{51} & c_{52}&c_{53} &c_{54} &c_{55}& c_{56}\\ c_{61} & c_{62}&c_{63} &c_{64} &c_{65}& c_{66}\\ \end{bmatrix}=\begin{bmatrix} ax_1 & ax_2 &ax_3 &ax_4 &ax_5& ax_6\\ bx_1 &b x_2 &bx_3 &bx_4 &bx_5& bx_6\\ cx_1 & cx_2 &cx_3 &cx_4 &cx_5&c x_6\\ dx_1 & dx_2 &dx_3 &dx_4 &dx_5& dx_6\\ ex_1 & ex_2 &ex_3 &ex_4 &ex_5& ex_6\\ fx_1 & fx_2 &fx_3 &fx_4 &fx_5& fx_6\\ \end{bmatrix} $
So, for this system directly we can get 6 candidates, which may or maynot contain a solution. Once we come to know which 11 entries can be made zero, we could take this further.