There exist only finitely many λ ∈ C such that det(λA + (1 − λ)B) = 0

Question

I found this question in the Lie Groups book, this is the start of the proof of connectedness of GL(n, C).

A, B ∈ GL(n, C), show that there exist only finitely many λ ∈ C such that det(λA + (1 − λ)B) = 0

Answer 1

The key observation is that $f(t):= \det (tA+(1-t)B)$ is a polynomial with coefficients in $\mathbb{C}$ with degree $\leq n$.

By the Fundametal Theorem of Algebra, $f$ has at most $n$ distinct roots, unless it's identically zero.

But obviously $f(1)=\det(A)\neq 0$, so $f\not\equiv 0$.

Answer 2

Note that $$ \lambda A+(1-\lambda)B=B-\lambda(B-A)=(I-\lambda(B-A)B^{-1})B $$ so we are reduced to prove that $f(\lambda)=\det(I-\lambda C)=0$ for only finitely many values of $\lambda$, where $C=(B-A)B^{-1}$, because $\det B\ne0$.

Suppose $f(\lambda)=0$. Then there exists a nonzero vector $v$ such that $$ (I-\lambda C)v=0 $$ which means $\lambda Cv=v$. In particular, $\lambda\ne0$ and $\lambda^{-1}$ is an eigenvalue of $C$. Since $C$ has only finitely many eigenvalues, we're done.