Given a holomorphic vector bundle $\mathcal{V}$ over a compact complex manifold $M$, it seems that even if $\mathcal{V}$ is nontrivial, then it can still have trivial Euler characteristic, that it, $$ \sum_{k=0}^{\mathrm{dim}(M)} (1)^k H^{(0,k)}(\mathcal{V}) = 0. $$ Is it true that for a positive line bundle $\mathcal{L}$, one can always find a large enough $n$, such that the twisted vector bundle $\mathcal{V} \otimes \mathcal{L}^{\otimes n}$ will have nontrivial Euler characteristic?

3$\begingroup$ If $M$ is projective, then for $n$ large enough one has that $\mathcal V\otimes \mathcal L^{\otimes n}$ is globally generated and that $H^q(M, V\otimes \mathcal L^{\otimes n})=0$ for $q>0$. Therefore, the holomorphic Euler characteristic of $V\otimes \mathcal L^{\otimes n}$ is positive. $\endgroup$– HenriOct 10, 2019 at 12:57

1$\begingroup$ ... and since $M$ carries a positive line bundle (by assumption), it is in fact projective. $\endgroup$– BenOct 15, 2019 at 17:49