This question appeared in some work I'm doing at the moment. I don't need the answer, it just interested me.
Let $A$ be an invertible $n\times n$ matrix, and let $B$ be another $n\times n$ matrix, both over some field $k$. I guess $k$ should be algebraically closed, just so I can split polynomials. Let $\lambda\in k$. We are interested in the matrix
$$ A+\lambda B,$$
and in particular whether it is invertible.
Treating $\lambda$ as a variable, it's clear that $f(\lambda)=\det(A+\lambda B)$ is a polynomial of degree at most $n$. Since $f(0)\neq 0$, $f$ is not the zero polynomial, and therefore there are at most $n$ values of $\lambda$ for which $A+\lambda B$ is non-invertible. In fact, it is easy to see that there are at most $\mathrm{rank}(B)$ such values by change of basis. It is not hard to give constructions that show that all values between $0$ and $\mathrm{rank}(B)$ can occur.
My interaction with this problem comes from bilinear forms, where an alternating form on a vector space was a linear combination of two, one of rank $56$ and the other of rank $12$. Because my form came from points on a $1$-dimensional torus, I knew that there was exactly one value of $\lambda$ which yielded a form with a radical, i.e., a non-invertible matrix.
I came to the conclusion that $B$ in this case must be very special. In particular, I guessed that the collection of all $B$ such that $f(\lambda)$ does not have $n$ distinct roots is small, probably Zariski closed.
So my questions are:
Is this true?
and
Is this obviously true or false?
My third, softer, question, is
Is this sort of thing well-known/previously looked at?
Of course one obvious case where it is looked at is where $B=-I_n$, as this is the characteristic polynomial! If it is looked at more generally, my guesses would be in analysis, if $k=\mathbb{R}$, or perturbation theory for small $\lambda$.
(For $A=I_n$ the answer is clear, as it is all matrices $B$ with distinct eigenvalues.)
Best Answer
The question of whether $A + \lambda B$ is invertible is known as the generalized eigenvalue problem. It is indeed well studied.
To your first two questions: the answer is yes. Note that in the "generic case", $B$ is invertible, which means that $$ \det(A + \lambda B) = \det(B)\det(AB^{-1} + \lambda I). $$ In this case, we find that for a "generic" choice of $A$, $AB^{-1}$ will have $n$ distinct eigenvalues.