Convex combination of matrices This question popped up in my head while doing a similar question.
Let $A$ and $B$ be two invertible $n \times n$ matrices. Define $C(t) \colon[0,1]\rightarrow M_{n \times n}(\mathbb{R})$ as
$$C(t) = (1-t)A+tB$$
Certainly $C(t)$ may not be invertible. For example if $A=-B$. Then $C \left(\frac{1}{2} \right)$ is not invertible.
I would like to know under what conditions can we guarantee invertibilty of $C(t)$ for all $t$ or at least infinite values of $t$? 
Edit: I'm looking for general conditions on something like determinant, trace, eigenvalues, etc. The condition that $A$ is a scalar multiple of $B$ is an un-interesing case.
 A: You will always have an infinite (indeed, uncountable) number of invertible matrices along the segment between $A$ and $B$.  Determinant is a continuous function of the entries of a matrix.  So in some neighborhood of both $A$ and $B$ you are guaranteed invertibility, since the determinant of $\epsilon A + (1-\epsilon)B$ will vary from det A by only a small amount, and therefore not be zero.
One case where you're guaranteed invertiblity is when $A$ and $B$ have a complete set of equal eigenvectors, where the corresponding eigenvalue pairs all have the same sign (and are nonzero).  In this case, the eigenvalues are given by $\lambda_i=t\lambda_{iA}+(1-t)\lambda_{iA}$, and since the eigenvalue pairs have the same sign, none of the eigenvalues of $C$ will be zero.
A: $\det C(t)$ is a polynomial of degree $\le n$. Indeed it is closely
related to the characteristic polynomial of $B^{-1}A$. As it's
a polynomial, and $\det C(0)\ne0$ and $\det C(1)\ne0$ it only vanishes at at most $n$ values of $t$. Unless $A=B$ the polynomial won't be constant, and over $\Bbb C$ at least, there will be some solution of $\det C(t)=0$.
