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This arises as a part of my work. Given a positive number $t$, two hermitian matrices $P_1$ and $P_2$, I am interested in knowing if a unit norm vector $z$ exists such that \begin{align} z^{H}P_1z\geq t \\ z^{H}P_2z\geq t \end{align}

A possible reformulation

The vector being unit norm, one can reformulate the above problem as to check whether a unit norm solution exists for \begin{align} z^{H}(P_1-tI)z\geq 0 \\ z^{H}P_2-tI)z\geq 0 \end{align} Note that $t$ is a given positive constant.

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up vote 0 down vote accepted

HINT: expand $z$ in terms of the eigenvectors of $P_1$ or $P_2$, it's easy to show that the necessary condition is $\min(\lambda_1,\lambda_2) \ge t$, where $\lambda_1$ and $\lambda_2$ are the largest eigenvalue of $P_1$ and $P_2$, respectively.

And if the eigenvector corresponding to $\lambda_1$ is also the eigenvector associated with $\lambda_2$, then this condition is also sufficient.

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It is not a sufficient condition. If you take $P_1 = \pmatrix{ 1 & 0 \\ 0 & -1 }$ and $P_2 = -P_1$, then both the largest eigenvalues are $\geq 1$, but $z^*P_1 z = - z^* P_2 z$, so they can never be simultaneously greater than $0$. – copper.hat Nov 20 '12 at 16:02
@copper.hat But the OP asked for $z^HP_iz \ge 0, \text{not} z^HP_iz > 0 \; (i=1,2)$ – chaohuang Nov 20 '12 at 18:49
The OP asked for $\geq t > 0$. – copper.hat Nov 20 '12 at 19:04
Thanks @copper.hat answer is edited. – chaohuang Nov 20 '12 at 20:33
@chaohuang thanks for the answer, but that doesn't help in general. In my case, I will be generating this matrices randomly, and the probability for the situation you mentioned happening is almost always zero. – dineshdileep Nov 21 '12 at 5:14

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