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.


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