The following is from a set of lecture notes I'm following and I have had some difficulties understanding it.
Let us discuss a few equivalent formulations of the Hermite-Einstein condition ($\Lambda_\omega F_\nabla = \lambda \text{id}_E$). Firstly, one can always write the curvature of the Chern connection $\nabla$ on any bundle $E$ as $$ F_{\nabla}=\frac{\operatorname{tr}\left(F_{\nabla}\right)}{\operatorname{rk}(E)} \cdot \mathrm{id}+F_{\nabla}^{\circ} $$ where $F_{\nabla}^{\circ}$ is the trace free part of the curvature. Let us now assume that $g$ is a Kähler metric, i.e. that $\omega$ is closed (and thus, automatically, harmonic). Then the connection is Hermite-Einstein if and only if $\operatorname{tr}\left(F_{\nabla}\right)$ is an harmonic $(1,1)$-form and $F_{\nabla}^{\circ}$ is (locally) a matrix of primitive $(1,1)$-forms. Indeed, if $\nabla$ is Hermite-Einstein, then $i \cdot \operatorname{tr}\left(F_{\nabla}\right)=(\operatorname{rk}(E) \cdot \lambda) \cdot \omega+\alpha$ with $\alpha$ a primitive $(1,1)$-form. Since the trace is closed, the form $\alpha$ is closed and hence harmonic. The other assertion and the converse are proved analogously. Secondly, the Hermite-Einstein condition for the curvature of a Chern connection is equivalent to writing $$ i \cdot F_{\nabla}=(\lambda / n) \cdot \omega \cdot \mathrm{id}_{E}+F_{\nabla}^{\prime} $$ where $F_{\nabla}^{\prime}$ is locally a matrix of $\omega$-primitive $(1,1)$-forms. Here, $n=\operatorname{dim}_{\mathbb{C}} X$. The factor $(1 / n)$ is explained by the commutator relation $[L, \Lambda]=H$, which yields $\Lambda L(1)=n$. One easily finds that $h$ is Hermite-Einstein if and only if $$ i \cdot F_{\nabla} \wedge \omega^{n-1}=(\lambda / n) \cdot \omega^{n} \cdot \operatorname{id}_{E} $$
The first equation is clear. Locally it's the same thing as writing a matrix as a sum of the trace-free part. The equation
$$ i \cdot \operatorname{tr}\left(F_{\nabla}\right)=(\operatorname{rk}(E) \cdot \lambda) \cdot \omega+\alpha $$
is the first one that stumps me. As I followed the text I thought that this would have been derived from the sum formula for $F_\nabla$ above, but I have come to the conclusion that this is not the case. To my current understanding this seems to be derived from the primitive decomposition (for $k=2$), which for vector spaces is
$$ \bigwedge^2 V^*= \omega \Bbb R \oplus P^2, $$
where $P^k$ denotes the subspace of all primitive elements, i.e. elements $\alpha$ for which $\Lambda \alpha = 0$. Here $\Lambda$ is the dual of the Lefschetz operator.
Q1: If this indeed is due to the primitive decomposition, how does one derive the coefficient $\operatorname{rk}(E)\lambda$ for $\omega$? Supposedly this comes somehow from the Hermite-Einstein condition, but how?
Q2: Following this, how does one conclude that the Hermite-Einstein condition is equivalent to $$i \cdot F_{\nabla}=(\lambda / n) \cdot \omega \cdot \mathrm{id}_{E}+F_{\nabla}^{\prime}$$ and furthermore to $$i \cdot F_{\nabla} \wedge \omega^{n-1}=(\lambda / n) \cdot \omega^{n} \cdot \operatorname{id}_{E}?$$
