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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 $A_2$ (both matrices are guaranteed to have at least one zero eigen value), 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.

( This is my first question in mathoverflow. I am not a mathematician, but from an engineering back ground. In my application, this kind of problem arises. My level of mathematical maturity is not enough to solve it. I hope some one here can.)

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Let $X$ be a subset of a vector space defined by polynomial equations, here, $\det = 0$. If $X$ has your convexity property, then $X$ is a translate of a linear subspace, i.e. the equations can be taken to be linear. But $\det A =0$ is not equivalent to linear conditions on $A$. –  Allen Knutson Oct 15 '12 at 12:33
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3 Answers

up vote 1 down vote accepted

Consider the characteristic polynomial $$ \lambda^N -a_1(t)\lambda^{N-1} + \dots +(-1)^{N} a_N(t) = 0 $$ of your Hermitian matrix $tA_1 +(1-t)A_2$ which has all roots real and whose coefficients are real analytic as functions of $t$. By a theorem of Rellich, the roots can be arranged as real analytic functions of $t$ also. Moreover, the eigenvectors of $tA_1 + (1-t)A_2$ can also be arranged real analytic in $t$. See [Dmitri Alekseevky, Andreas Kriegl, Mark Losik, Peter W. Michor: Choosing roots of polynomials smoothly, Israel J. Math 105 (1998)] [(pdf)]1. (2.5 is wrong in this paper.)

By your assumptions you have at both $t=0$ and $t=1$: At least one positive root, one negative root, and one root zero.
From this follows nothing, and it is easy to come up with simple examples with any outcome.

But if you can ascertain that a positive eigenvector for $A_1$ turns into a negative one for $A_2$, then the corresponding eigenvalue must go through 0 in between.

Maybe this can help you playing with your assumptions.

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I suspect that you mean $0 < t < 1$ in your question.

Then the answer in still no under the added condition that both matrices are not of full rank. Consider $$ A_1 = \begin{pmatrix}1 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & -1\end{pmatrix}\quad A_2 = \begin{pmatrix}0 & 0 & 0\\\\ 0 & 1 & 0\\\\ 0 & 0 & -1\end{pmatrix}. $$

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The OP actually specifies that both matrices are indefinite, but your example is easily modified to meet that. –  Igor Rivin Oct 15 '12 at 12:36
    
Oh yes - why not... –  Dirk Oct 15 '12 at 15:06
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Trivially no: consider the case $A_1 = A_2$.

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Thanks for that!!. But for me, both of them are not full rank (I have updated the question.), so a non-zero vector does exist in the example you pointed out. –  dineshdileep Oct 15 '12 at 6:31
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