[proofplan] The proof is a local computation in the chosen holomorphic frame. The condition $D^{0,1}=\bar{\partial}_E$ forces the connection matrix of the Chern connection to have only a $(1,0)$-part in a holomorphic frame. Metric compatibility then gives a matrix identity for the differential of the Hermitian metric matrix $H$; taking the $(1,0)$-part eliminates the conjugate connection term and yields $\partial H = H A$. Since $H$ is pointwise positive definite, it is invertible, and left multiplication by $H^{-1}$ gives the formula. [/proofplan] [step:Write the Chern connection in the holomorphic frame] Let $s_U: U \to \mathbb{C}^r$ denote the column-vector representative of a smooth local section $s \in C^\infty(U;E)$ in the frame $e$, so that \begin{align*} s = \sum_{j=1}^r e_j (s_U)_j. \end{align*} The local connection matrix of the Chern connection $D$ in this frame is the matrix-valued one-form $A \in A^1(U;\operatorname{End}(\mathbb{C}^r))$, where $A^1(U;\operatorname{End}(\mathbb{C}^r))$ denotes smooth one-forms on $U$ with values in $\operatorname{End}(\mathbb{C}^r)$, defined by \begin{align*} (Ds)_U = d s_U + A s_U. \end{align*} Equivalently, for each frame vector $e_j$, \begin{align*} D e_j = \sum_{i=1}^r e_i A_{ij}, \end{align*} where $A_{ij} \in A^1(U;\mathbb{C})$ is the $(i,j)$-entry of $A$. [/step] [step:Use holomorphicity of the frame to identify the type of the connection matrix] Because $D$ is the Chern connection, its $(0,1)$-part equals the holomorphic structure operator $\bar{\partial}_E: C^\infty(U;E) \to A^{0,1}(U;E)$: \begin{align*} D^{0,1} = \bar{\partial}_E. \end{align*} Since the frame $e_1,\dots,e_r$ is holomorphic, each frame section satisfies \begin{align*} \bar{\partial}_E e_j = 0. \end{align*} Taking the $(0,1)$-part of \begin{align*} D e_j = \sum_{i=1}^r e_i A_{ij} \end{align*} therefore gives \begin{align*} 0 = D^{0,1}e_j = \sum_{i=1}^r e_i A_{ij}^{0,1}. \end{align*} The frame vectors are pointwise linearly independent, so $A_{ij}^{0,1}=0$ for all $i,j$. Hence \begin{align*} A = A^{1,0}. \end{align*} [guided] The reason a holomorphic frame is useful is that it removes the antiholomorphic part of the connection matrix. The Chern connection is characterized by the identity \begin{align*} D^{0,1} = \bar{\partial}_E, \end{align*} where $\bar{\partial}_E: C^\infty(U;E) \to A^{0,1}(U;E)$ is the holomorphic structure operator. Now fix an index $j \in \{1,\dots,r\}$. Since $e_j$ is a holomorphic local section, the definition of a holomorphic frame gives \begin{align*} \bar{\partial}_E e_j = 0. \end{align*} On the other hand, by the definition of the connection matrix in the column-vector convention, \begin{align*} D e_j = \sum_{i=1}^r e_i A_{ij}. \end{align*} Taking the $(0,1)$-component of both sides gives \begin{align*} D^{0,1}e_j = \sum_{i=1}^r e_i A_{ij}^{0,1}. \end{align*} The left-hand side equals $\bar{\partial}_E e_j$, so it is zero. Therefore \begin{align*} \sum_{i=1}^r e_i A_{ij}^{0,1} = 0. \end{align*} Because $e_1,\dots,e_r$ is a frame, the representation of an $E$-valued form in this basis is unique. Thus every coefficient $A_{ij}^{0,1}$ vanishes. Since this holds for every pair of indices, the full matrix $A$ has no $(0,1)$-part: \begin{align*} A = A^{1,0}. \end{align*} This is the step where holomorphicity of the frame is used. [/guided] [/step] [step:Apply metric compatibility to the frame vectors] For $1 \leq i,j \leq r$, metric compatibility of $D$ with $h$ gives \begin{align*} d(H_{ij}) = d(h(e_i,e_j)) = h(D e_i,e_j) + h(e_i,D e_j). \end{align*} Using \begin{align*} D e_i = \sum_{k=1}^r e_k A_{ki} \end{align*} and the convention that $h$ is conjugate-linear in the first argument and linear in the second argument, we obtain \begin{align*} h(D e_i,e_j) = \sum_{k=1}^r \overline{A_{ki}} H_{kj} \end{align*} and \begin{align*} h(e_i,D e_j) = \sum_{k=1}^r H_{ik} A_{kj}. \end{align*} Thus the matrix identity is \begin{align*} dH = A^*H + HA, \end{align*} where $A^*$ denotes the conjugate transpose of the matrix of one-forms, namely $(A^*)_{ij}=\overline{A_{ji}}$ with complex conjugation applied to the scalar form coefficients. [/step] [step:Take the $(1,0)$-part of the metric identity] Since $A$ has type $(1,0)$, conjugating the scalar form coefficients in $A^*$ changes each $(1,0)$-form coefficient into a $(0,1)$-form coefficient, so the matrix $A^*$ has type $(0,1)$. Taking the $(1,0)$-part of \begin{align*} dH = A^*H + HA \end{align*} therefore gives \begin{align*} \partial H = HA. \end{align*} Here $H$ is a matrix of smooth functions, while $A$ is a matrix of $(1,0)$-forms, so the product $HA$ is the usual matrix product with wedge multiplication by functions. [/step] [step:Invert the Hermitian metric matrix to obtain the connection matrix] For every point $x \in U$, the matrix $H(x)$ is positive definite because $h$ is a Hermitian metric and $e_1(x),\dots,e_r(x)$ is a basis of $E_x$. Hence $H: U \to \operatorname{Herm}_r^+$ takes values in invertible matrices, and the smooth inverse matrix $H^{-1}: U \to GL(r,\mathbb{C})$ is defined pointwise. Multiplying \begin{align*} \partial H = HA \end{align*} on the left by $H^{-1}$ gives \begin{align*} A = H^{-1}\partial H. \end{align*} Substituting this expression into the local formula $(Ds)_U = d s_U + A s_U$ gives \begin{align*} D = d + H^{-1}\partial H \end{align*} in the holomorphic frame $e$, as claimed. [/step]