Let $C_1$,$C_2$,...$C_N$ be $M \times M$ indefinite hermitian matrices. What can we say about the following quadratic constriants \begin{align} w^{H}C_1w>0 \\\ w^{H}C_2w>0 \\\ ...~~~~~~~~~~ \\\ ...~~~~~~~~~~ \\\ w^{H}C_Nw>0 \end{align} where $w$ is a nonzero $M\times 1$ complex vector. Can we come up with conditions for existence of such a $w$? I tried using the CVX package to determine the same (after using semidefinite relaxation). It always gives the zero vector as the answer.

As mentioned in its user's guide, CVX does not distinguish between constraints in the form of $f(x) < 0$ and $f(x) \leq 0 $, which is very common in numerical optimization. Instead, you should use $w^{H} C_i w \geq 1, \quad i = 1,2,\cdots,N$ to avoid the zero vector solution. Of course, you can replace $1$ by any positive constant. This is because we can always find $w$ via scaling so that the above conditions hold provided that the conditions in your original post are satisfied, i.e., there exists $\hat{w}$ such that $\hat{w}^{H} C_i \hat{w} = \epsilon_i > 0, \quad i = 1,2,\cdots,N$. 


One option for a convex relaxation is to search for a positive semidefinite hermitian $W$ with $\mathrm{Trace}(WC_i)\geq 1$ for all $i$. These conditions are equivalent to the ones you have written when $W$ is also rank one ($W = ww^H$), i.e., they are necessary but not sufficient for existence of the $w$ you seek. If you like you could also try minimizing the nuclear norm of $W$ subject to these constraints because that is a heuristic convex surrogate for minimizing rank. If the resulting optimal $W$ is not rank one you could also try for example taking the SVD of the optimal $W$ and see if any of the first few singular vectors has the property you seek. 

