Homogeneous vector bundles with zero chern classes

We know that a line bundle $$L$$ on the complex flag variety $$G/P$$ is trivial iff $$c_1(E) = 0$$. But if we have a homogeneous vector bundle $$E$$ of higher rank, then is it true that $$c_i(E) = 0$$ $$\forall i$$ implies that $$E$$ is trivial? If not, is there anything analogous we can say about such vector bundles? I apologize if this question is trivial as my Representation theory background isn't that great.

Take for instance $$G/P = Gr(k,n)$$, and let $$0 \to U \to V \otimes \mathcal{O} \to Q \to 0$$ be the tautological exact sequence. Then $$c_\bullet(U \oplus Q) = c_\bullet(U) \cdot c_\bullet(Q) = c_\bullet(V \otimes \mathcal{O}) = 1,$$ but $$U \oplus Q$$ is a nontrivial vector bundle.
• Thanks for your answer. I realized after asking this question that it's actually really easy to produce counterexamples: \mathcal{O}(-n) \bigoplus \mathcal{O}(n) on \mathbf{P}^n being easy counterexamples. In fact more generally, any homogeneous bundle all of whose weights are non zero linear combinations of the weights corresponding to the parabolic $P$ will have zero chern classes, but not be trivial. I think this question should be taken off this site: it's too silly. – user69183 Oct 20 '19 at 19:18
• Not quite, in fact $c_2(\mathcal{O}(-n) \oplus \mathcal{O}(n)) \ne 0$. – Sasha Oct 21 '19 at 4:32
• Yes, you're right, it should be $-n^2$. But whatever, the Chern class formula for homogeneous bundles sees only the weights which aren't in $P$. So there exist lots of counterexamples. – user69183 Oct 21 '19 at 15:33