Invertibility of $I + AB(x)$

Question

I am dealing with a matrix $$I + AB(x)$$ where $A, B(x)$ are square $n\times n$ real matrices and $x$ is a real variable. I want to find the values of $x$ for which this matrix is singular (and then an eigenvector). $A$ is invertible and $B(x)$ is invertible for the values of $x$ I am looking for. I looked into:

• Determinant: plainly awful
• Gaussian elimination: a bit better but leads to $n$ lengthy equations
• Neumann series: useless because it is not a necessary and sufficient condition
• Generalized eigenvalue: not very familiar with this theory but looks more suitable for numerical approximation

Any other ideas would be very appreciated! (For clarity I am hoping to find a somewhat palatable equation for x...) Thanks p

Answer 1

Presumably $B(x)$ is a nice analytic function of $x$, maybe polynomial. Determinant is not necessarily so awful: its computation (for a particular matrix) takes about the same time as Gaussian elimination. You might try Newton's method, using Jacobi's formula to compute the derivatives.

Answer 2

We can simplify the equation a bit more using the Singular Value Decomposition of $A$ as follows:$$A=UDV\implies |I+AB(x)|{=|I+UDVB(x)|\\=|U^H+DVB(x)|\\=|U^HU+DVB(x)U|\\=|I+DVB(x)U|\\\triangleq |I+DC(x)|}$$where $D$ is diagonal, $U$ and $V$ are unitary and $C(x)\triangleq VB(x)U$, but this time, the singularity of $I+AB(x)$ is equivalent to the singularity of $I+DC(x)$, with $D$ being diagonal. Any constraint on $C(x)$ then, can be converted to that for $B(x)$, and as @RobertIsrael said, if $B(x)$ is a nice function, then so is $C(x)$. I think this might reduce the computation time as $D$ is diagonal. Even you can use $|D^{-1}+C(x)|$ to decide.