Convex Combination of Hermitian Matrices Assume all the matrices I discuss about are $N \times N$. Consider any two hermitian matrices $A_1$ and $A_2$ which are indefinite. The question is, In general, for any $A_1$ and $A_2$, does there exist any positive number $t$ such that equation
\begin{align}
(tA_1+(1-t)A_2)x=0
\end{align}
has a non-zero vector $x$ as a solution.
 A: If divide through by $t$ and set $s=\frac{1}{t}$ it is the generalized eigenvector problem with matrices $M = A_1 -A_2$ and $A_2$
\begin{align}
(tA_1+(1-t)A_2)x &=0 \\
(tA_1-tA_2 + A_2)x &=0 \\
(t(A_1-A_2) + A_2)x &=0 \\
(A_1-A_2 + sA_2)x &=0 \\
(M + sA_2)x &=0 \\
\end{align}
The generalized eigenvector problem does have solutions in general.
Could also write it as
$$(A_2 + tM)\mathbf{x}=\mathbf{0}$$
EDIT
I would like to point out that the equation you have is the basis for an argument using continuity (the eigenvalues of a matrix vary continuously with the entries). Look at $t$ at $0$ and $1$. 

For $t=0$, the equation is $A_2 x =0$ and the question is if $A_2$ is singular ($x \ne 0$ solves it only for singular $A_2$).

I realize that you are interested in positive $t$, so either consider $t$ very small (and thus consider an approximate solution), or look now at $t=1$. Does $A_1 x = 0$ have a solution?

Now look at $t \to \infty$, in that case $(A_1-A_2)x=0$ is the question.

I would consider Sylvester's law of inertia if the matrices were real symmetric (the theorem is for real symmetric only). It is much easier to find a diagonalized congruence rather than a similarity, and the law states that the inertia is constant under congruency (inertia is the triplet number, number of positive eigenvalues, the number of $0$ eigenvalues, and number of negative eigenvalues). If congruency were preserving the inertia with hermitian (not a theorem as far as I know) then you would look at the inertia of the different matrices from $t=0$, $t=1$, and $t \to \infty$. If their inertia differs, then there is some point at which a positive crosses to be negative, and thus some point of singularity, giving the $t$ at which the equation has a solution.

But since you asked about hermitian indefinite, that is the difference...
