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$\renewcommand{\v}[1]{\mathrm{vec}\left(#1\right)} \renewcommand{\m}[1]{\mathbf{#1}} \renewcommand{\trace}[1]{\mathrm{trace}\left(#1\right)} \renewcommand{\diag}[1]{\mathrm{diag}\left(#1\right)}$

Suppose we have the diagonal matrix $\mathbf{D} = Diag(\mathbf{1}^{T}\mathbf{H})$, $1$ is a column vector with ones.

How can we calculate the partial derivative of the following wrt matrix calculus? ($A$ is known matrix.) $$\frac{\partial ({ \mathbf D^{-1} \mathbf A)}}{\partial\m H}$$

Since now, I have reached the following using matrix cookbook:

$$\frac{\partial ({ \mathbf D^{-1} \mathbf A)}}{\partial\m H} = -\mathbf{A} \mathbf{D}^{-1} \frac{\partial { \mathbf D}}{\partial\m H} \mathbf{D}^{-1} = -\mathbf{A} \mathbf{D}^{-1} \mathbf{1}^{T} \mathbf{J} \mathbf{D}^{-1} $$

I think I am missing something with 1 and J.

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  • $\begingroup$ What is $H$ in the line $Diag(1^TH)$? Or is it meant to be $A$? $\endgroup$ – user1936752 Mar 20 at 13:11
  • $\begingroup$ H is the uknown matrix. That's why the partial derivative is wrt H. And D is basically a diagonal matrix with its diagonal entries be the sum of each column of H. (that's what $\mathbf{1}^{T}$ does). $\endgroup$ – OliveR Mar 20 at 19:51
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Specify the dimensions of all the vectors and matrices involved. $$\eqalign{ A & &\in {\mathbb R}^{n\times p} \cr H & &\in {\mathbb R}^{m\times n} \cr h &= {\rm vec}(H) &\in {\mathbb R}^{mn\times 1} \cr v &= H^T1_m &\in {\mathbb R}^{n\times 1} \cr L &= (I_n\otimes 1_n)\odot(1_n\otimes I_n) &\in {\mathbb R}^{n^2\times n} \cr B &= {\rm Diag}(v) &\in {\mathbb R}^{n\times n} \cr b &= {\rm vec}(B) = Lv &\in {\mathbb R}^{n^2\times 1} \cr &= LH^T1_m \cr &= {\rm vec}(1_m^THL^T) \cr &= (L\otimes 1_m^T)\,h \cr }$$ where $I_n$ is the $n\times n$ identity matrix, $1_n$ is the all-ones vector of length $n$, $\odot$ is the Hadamard product, and $\otimes$ is the Kronecker product.

The function of interest is matrix-valued, so it must be flattened/vectorized in order to express the resulting derivative as a matrix (instead of a fourth-order tensor).

The steps are to calculate the function differential, vectorize it, and formulate the matrix derivative. $$\eqalign{ F &= B^{-1}A\cr dF &= -B^{-1}\,dB\,B^{-1}A \cr &= -B^{-1}\,dB\,F \cr {\rm vec}(dF) &= -(F^T\otimes B^{-1})\,{\rm vec}(dB) \cr df &= -(F^T\otimes B^{-1})\,db \cr &= -(F^T\otimes B^{-1})\,(L\otimes 1_m^T)\,dh \cr \frac{\partial f}{\partial h} &= -(F^T\otimes B^{-1})\,(L\otimes 1_m^T) &= \frac{\partial \,{\rm vec}(F)}{\partial \,{\rm vec}(H)} \cr }$$

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