Linear combination of invertible matrices which is singular says that if $\det A < 0$ and $\det B > 0$, then there's a real number $t$ with $tA + (1-t)B$ singular.
Assuming that $A$ and $B$ are real matrices, the natural proof is to look at $$ f(t) = \det (tA + (1-t)B) $$ and apply the intermediate value theorem to find a $t$ for which $f(t) = 0$, and you're done.
Here's my question: What if we don't assume we're working over the reals? In particular:
Let $F$ be a subfield of the reals, and $A$ and $B$ be matrices with entries in $F$, $\det A < 0, \det B > 0$. Must there be an element $t \in F$ with $\det (tA + (1-t)B) = 0$ ?
I have a feeling that the answer's either "no" or "obviously, yes," but at the moment I have no idea which.
Take subfield $\mathbb Q$ and $$ A = \left( \begin{array}{cc} 3 & 1 \\ 1 & -2 \end{array} \right) $$ $$ B = \left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right) $$
I get the linear combination as $$ \left( \begin{array}{cc} 2 + t & 1 \\ 1 & 2 - 4t \end{array} \right) $$ and singular for some irrational $t \; , \;$ when $$ 4 t^2 + 6t - 3 = 0 $$