[proofplan] The assertion is local on $X$, so we compute in an arbitrary holomorphic frame of $E$. In such a frame the Chern connection has connection matrix $A = H^{-1}\partial H$, where $H$ is the Hermitian metric matrix; in particular $A$ has type $(1,0)$. The curvature matrix is $dA + A \wedge A$, so the only non-immediate point is the vanishing of its $(2,0)$ component. This follows by differentiating the identity $H^{-1}H = I$ and using $\partial^2 = 0$, leaving only the $(1,1)$ component. [/proofplan]

[step:Compute the Chern connection in a holomorphic frame] Let $p \in X$. Choose an open neighbourhood $U \subset X$ of $p$ over which $E$ admits a holomorphic frame $e = (e_1,\dots,e_r)$, where $r = \operatorname{rank}E$. Let \begin{align*} H: U \to \operatorname{Herm}_r^{+}(\mathbb{C}) \end{align*} be the Hermitian metric matrix in this frame, defined by $H_{\alpha\beta} = h(e_\alpha,e_\beta)$ for $1 \leq \alpha,\beta \leq r$. In the frame $e$, a smooth section of $E|_U$ is represented by a smooth map \begin{align*} s: U \to \mathbb{C}^r. \end{align*} Here $\bar{\partial}_E:A^0(U;E|_U)\to A^{0,1}(U;E|_U)$ denotes the local Dolbeault operator induced by the holomorphic structure. Since $D^{0,1}=\bar{\partial}_E$ and the frame $e$ is holomorphic, the local connection matrix has no $(0,1)$ part. Thus the Chern connection has the local expression \begin{align*} D s = d s + A s, \end{align*} where the connection matrix satisfies \begin{align*} A \in A^{1,0}(U;\operatorname{End}\mathbb{C}^r). \end{align*} We use the column-vector convention: $s^*Ht$ denotes the Hermitian pairing $h(s,t)$ in the frame $e$, and the connection acts on local column vectors by $Ds=ds+As$. Metric compatibility says that for smooth sections $s,t:U\to\mathbb{C}^r$ represented in this frame, \begin{align*} d(s^*Ht) = (D s)^*Ht + s^*H(D t). \end{align*} Comparing the $(1,0)$ part under this convention and using arbitrariness of $s$ and $t$ gives \begin{align*} \partial H = H A. \end{align*} Multiplying on the left by $H^{-1}$ gives \begin{align*} A = H^{-1}\partial H. \end{align*} Here $\partial H$ denotes the matrix whose $(\alpha,\beta)$ entry is the $(1,0)$-form $\partial H_{\alpha\beta}$. Since $H^{-1}$ is a matrix of smooth functions and $\partial H$ has type $(1,0)$ entrywise, this formula is consistent with $A$ having type $(1,0)$. [/step]

[step:Write the curvature matrix and isolate its type components] In the same frame, the curvature $F_D = D^2$ is represented by the matrix-valued two-form \begin{align*} \Omega = dA + A \wedge A \in A^2(U;\operatorname{End}\mathbb{C}^r). \end{align*} Since $A$ has type $(1,0)$, its [exterior derivative](/theorems/1525) decomposes as \begin{align*} dA = \partial A + \bar{\partial}A, \end{align*} with $\partial A$ of type $(2,0)$ and $\bar{\partial}A$ of type $(1,1)$. Also $A \wedge A$ has type $(2,0)$ because it is the wedge product of two matrix-valued $(1,0)$-forms. Hence the type decomposition of $\Omega$ is \begin{align*} \Omega^{2,0} = \partial A + A \wedge A. \end{align*} The $(0,2)$ component is zero because the local connection matrix has no $(0,1)$ part in a holomorphic frame, equivalently because $D^{0,1}=\bar{\partial}_E$ and the holomorphic structure is integrable, so $\bar{\partial}_E^2=0$. [/step]

[step:Show that the $(2,0)$ component vanishes]We compute $\partial A + A \wedge A$ using $A = H^{-1}\partial H$. Since $H^{-1}H = I_r$, applying $\partial$ entrywise gives \begin{align*} \partial(H^{-1})H + H^{-1}\partial H = 0. \end{align*} Multiplying this identity on the right by $H^{-1}$ gives \begin{align*} \partial(H^{-1}) = -H^{-1}(\partial H)H^{-1}. \end{align*} Now apply the Leibniz rule for $\partial$ to $A = H^{-1}\partial H$. Since $\partial^2H=0$ entrywise, we obtain \begin{align*} \partial A = \partial(H^{-1}) \wedge \partial H. \end{align*} Substituting the formula for $\partial(H^{-1})$ gives \begin{align*} \partial A = -H^{-1}(\partial H)H^{-1}\wedge \partial H. \end{align*} By matrix multiplication of matrix-valued forms, the right-hand side is exactly $-A\wedge A$. Therefore \begin{align*} \partial A + A\wedge A = 0. \end{align*}[/step]

[guided]The only component that requires computation is the $(2,0)$ part. The connection matrix is \begin{align*} A = H^{-1}\partial H, \end{align*} where $H:U\to \operatorname{Herm}_r^{+}(\mathbb{C})$ is the Hermitian metric matrix. Since $A$ is a matrix-valued $(1,0)$-form, the possible $(2,0)$ curvature contribution is \begin{align*} \Omega^{2,0} = \partial A + A\wedge A. \end{align*} We now prove that this expression is zero. The key identity comes from differentiating the inverse relation \begin{align*} H^{-1}H = I_r. \end{align*} Applying $\partial$ to both sides and using the Leibniz rule gives \begin{align*} \partial(H^{-1})H + H^{-1}\partial H = 0. \end{align*} Multiplying on the right by $H^{-1}$ yields \begin{align*} \partial(H^{-1}) = -H^{-1}(\partial H)H^{-1}. \end{align*} Now differentiate $A = H^{-1}\partial H$. The operator $\partial$ satisfies the graded Leibniz rule, and $\partial^2H=0$ entrywise because $\partial^2=0$ on scalar-valued forms. Thus \begin{align*} \partial A = \partial(H^{-1})\wedge \partial H. \end{align*} Substituting the inverse-derivative formula gives \begin{align*} \partial A = -H^{-1}(\partial H)H^{-1}\wedge \partial H. \end{align*} But $A\wedge A$ is precisely the matrix product wedge \begin{align*} A\wedge A = H^{-1}(\partial H)H^{-1}\wedge \partial H. \end{align*} Therefore \begin{align*} \partial A = -A\wedge A. \end{align*} Hence \begin{align*} \partial A + A\wedge A = 0. \end{align*} This proves that the $(2,0)$ component of the curvature vanishes.[/guided]

[step:Conclude that the curvature has pure type $(1,1)$] On the neighbourhood $U$, the curvature matrix satisfies \begin{align*} \Omega = \bar{\partial}A. \end{align*} Thus $\Omega$ has type $(1,1)$ on $U$. Since $p \in X$ was arbitrary and type decomposition of $\operatorname{End}E$-valued forms is local, the global curvature form $F_D$ has no $(2,0)$ or $(0,2)$ component. Therefore \begin{align*} F_D \in A^{1,1}(X;\operatorname{End}E). \end{align*} This proves the theorem. [/step]