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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.

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It can happen; take $A_1=I$, $A_2=-I$ and $t=\frac12$. –  Brian M. Scott Oct 14 '12 at 11:25
    
@dineshdileep You should add traceless... –  draks ... Oct 14 '12 at 11:31
    
@Brian M.Scott Yes I know it can happen. In fact, if $A_1$ and $A_2$ commute, it will definitely happen. I am looking for an answer in general. –  dineshdileep Oct 14 '12 at 11:51
    
It might be helpful to others to make that clearer in the question. –  Brian M. Scott Oct 14 '12 at 12:01
    
Consider when $A_1$ and $A_2$ are positive definite. –  chaohuang Oct 14 '12 at 14:39
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1 Answer

up vote 1 down vote accepted

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...

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Thanks!!, I came along this direction. Does it hold even if the matrices are singular or say real only?. I know specific examples where a solution doesn't exist when the matrices are real. Even then, the solution is helpful that you connected it to the generalized eigen value problem. –  dineshdileep Oct 15 '12 at 2:13
    
@dineshdileep yes there is quite a difference if the matrices are real only, see my edit, which addresses the singular matrices question as well. –  adam W Oct 22 '12 at 14:58
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