# Are there methods of solving a set of homogeneous linear equations where some coefficients are unknown?

I am aware that a standard formulation for a set of homogeneous linear equations is:

$$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$

I have a problem which can be modelled as a set of homogeneous linear equations, except that some of the values in the matrix $$A$$ are also unknown. So, I may have equations such as:

$$5 x_1 + 3 x_2 + a_{13} x_3 = 0$$

$$3 x_1 + 2 x_2 + a_{23} x_3 = 0$$

$$7 x_1 + a_{32} x_2 + 10 x_3 = 0$$

Note that these exact equations are made-up examples, and I am aware that even in normal homogeneous systems a trivial solution may be the only solution, so I do not know for sure if I will even get an interesting result. I suspect an iterative solution might be possible but I don't know if there is already work on this problem - so far my searching has not been successful.

You might do the following. Consider the matrix $$A(x,y,z) = \begin{pmatrix} 5 & 3 & x \\ 3 & 2 & y \\ 7 & z & 10 \end{pmatrix}.$$ Then the determinant of this matrix is a polynomial $$f$$ in the variables $$x,y,z$$. This polynomial has a set $$S_2$$ of zeros outside of which the linear system of equations $$A(x,y,z)v = 0$$ only has the trivial solution $$v = 0$$. Inside $$S_2$$ there are, in principle, three possible values for $$\text{rank }A(x,y,z),$$ namely $$0, 1,$$ and $$2$$. In other words, $$S_2$$ includes all values of $$(x, y, z)$$ such that the rank of $$A(x,y,z)$$ is at most $$2$$.
We can construct a similar set $$S_1$$ where the rank of $$A(x,y,z)$$ is at most $$1$$ by considering the common zero sets of polynomials corresponding to the $$2\times 2$$-minors. In your example, this is empty because the top left $$2\times 2$$-minor is $$\det \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix} = 1.$$ The set $$S_0$$ will be the common set of zeros of the $$1 \times 1$$-minors, which will be empty since the top left $$1 \times 1$$-minor is $$5$$, which is not zero.
Now you need to consider four cases separately: First, if $$(x,y,z) \in \mathbb R^3 \setminus S_2$$, then there is only the trivial solution $$v= 0$$. Second, if $$(x,y,z) \in S_2 \setminus S_1$$, there will be a $$1$$-dimensional linear subspace of solutions. Third, if $$(x,y,z) \in S_1 \setminus S_0$$, there will be a two-dimensional subspace of solutions. Fourth and finally, if $$(x,y,z) \in S_0$$, then $$A(x,y,z) = 0$$ so all vectors get mapped to zero, in which case every $$v \in \mathbb R^3$$ is a solution.