Consider $n$ matrices $A_k \in \mathbb R^{n \times n}$. The equation $\det\left(\sum_{k=1}^n x_k A_k\right) = 0$ describes (in general) a hypersurface when $x \in \mathbb R^n$.
Below is an example of such surface in $\mathbb R^3$ given by $\det \begin{pmatrix} x_1 + x_2 & x_3 & 0\\ x_2 & x_2 + x_3 & x_1\\ x_2 & 0 & x_1 + x_3 \end{pmatrix} = 0$
Since $\det\left(\sum_{k=1}^n x_k A_k\right) = 0$ implies $\det\left(\sum_{k=1}^n (\lambda x_k) A_k\right) = 0$ it is enough to consider this equation on the unit sphere $x^2 = 1$.
I'm wondering if these surfaces are known and studied. Particularly I'm interested in testing whether for given $A_k$ and starting and terminating points $A, B$ there exists a continuous path $x(t)$ such that $x(0) = A, \quad x(1) = B$ which never crosses the surface: $\det (A_k x_k(t)) \neq 0, \forall t \in [0, 1]$.
My observations:
- If all matrices are diagonal: $A_k = \operatorname{diag}_i(d_i^{(k)})$ the determinant is easy to compute: $$ 0 = \det (\sum_{k=1}^n x_k A_k) = \prod_{i=1}^n \sum_{k=1}^n x_k d_i^{(k)} $$ The solution is a union of $n$ planes given by $\sum_{k=1}^n x_k d_i^{(k)} = 0$.
- For nonsingular $S, T \in \mathbb R^{n \times n}$ the substitution $A_k \to S A_k T$ does not change the equation. Combined with the first observation this gives a solution in case when all $A_k$ can be simultaneously diagonalized by a pair of matrices.
- If matrices are symmetric and positive definite then $\mathbb R_+^n$ does not intersect with the surface.
This problem arose when I tried to solve numerically a quadratic system of equations given by $x^\top B_k x = c_k$ and observed that the numerical method "hits a wall" at a point where jacobian of the system vanishes. In my particular case matrices $B_k$ are indefinite.