# Partial derivative of matrix

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

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.

• What is $H$ in the line $Diag(1^TH)$? Or is it meant to be $A$? – user1936752 Mar 20 '19 at 13:11
• 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). – OliveR Mar 20 '19 at 19:51

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 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 }