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$\newcommand{\piff}[2]{\frac{\partial{#1}}{\partial{#2}}}\newcommand{\ppiff}[3]{\frac{\partial^2 {#1}}{\partial{#2}\partial{#3}}}\newcommand{\pa}[2]{\ppiff{|A|}{#1}{#2}}$ Suppose matrix $$A = \begin{pmatrix} x_{11} & x_{12} & \cdots & x_{1n}\\ x_{21} & x_{22} & \cdots & x_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ x_{n1} & x_{n2} & \cdots & x_{nn} \end{pmatrix}$$ Consider $\det{A}$, denoted $|A|$, as a $n^2$-variable function of variable $x_{ij}(1\le i,j \le n)$.

I've found that $$\piff{|A|}{A} = \begin{pmatrix} \piff{|A|}{x_{11}} & \piff{|A|}{x_{12}} & \cdots & \piff{|A|}{x_{1n}}\\ \piff{|A|}{x_{21}} & \piff{|A|}{x_{22}} & \cdots & \piff{|A|}{x_{2n}}\\ \vdots & \vdots & \ddots & \vdots\\ \piff{|A|}{x_{n1}} & \piff{|A|}{x_{n2}} & \cdots & \piff{|A|}{x_{nn}} \end{pmatrix} = C$$ where $C$ is the cofactor matrix.

Is there any matrix representation for follwing $n^2$ by $n^2$ matrix? $$ \begin{pmatrix} \pa{x_{11}}{x_{11}} & \pa{x_{11}}{x_{12}} & \cdots\cdots & \pa{x_{11}}{x_{nn}} \\ \pa{x_{12}}{x_{11}} & \pa{x_{12}}{x_{12}} & \cdots\cdots & \pa{x_{12}}{x_{nn}} \\ \vdots & \vdots & \ddots & \vdots \\ \pa{x_{nn}}{x_{11}} & \pa{x_{nn}}{x_{12}} & \cdots\cdots & \pa{x_{nn}}{x_{nn}} \end{pmatrix} $$

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For convenience, define the variables $$\alpha = \det(A),\quad a={\rm vec}(A),\quad c={\rm vec}(C)$$ From your previous result $$\frac{\partial\alpha}{\partial A} = \alpha A^{-T} = C$$ Continue on to find the differential and gradient of $C$. $$\eqalign{ dC &= A^{-T}d\alpha + \alpha\,dA^{-T} \cr &= C\alpha^{-1}\,d\alpha + \alpha\,dA^{-T} \cr &= C\alpha^{-1}(C:dA) - \alpha\,A^{-T}\,dA^T\,A^{-T} \cr }$$ Vectorize, i.e. flatten all the $n\times n$ matrices into $n^2\times 1$ vectors. $$\eqalign{ dc &= \alpha^{-1}c(c^Tda) - \alpha\,(A^{-1}\otimes A^{-T})K\,da \cr \frac{\partial c}{\partial a} &= \alpha^{-1}cc^T - \alpha\,(A^{-1}\otimes A^{-T})K \cr &= \alpha^{-1}\Big(cc^T - \big(C^T\otimes C\big)K\Big) \cr }$$ This result is the desired $n^2\times n^2$ matrix.
And $K$ is the commutation matrix associated with the vec-operation.

Update
The above result assumes that $\alpha\ne 0$, but it can be rearranged into a form valid for all $\alpha$, and with a more symmetric appearance. $$\eqalign{ \alpha\,\Big(\frac{\partial c}{\partial a}\Big) &= \Big(c^T\otimes c \,-\, \big(C^T\otimes C\big)K\Big) \, = \alpha\,\bigg(\frac{\partial^2 \alpha}{\partial a\,\partial a^T}\bigg) \cr }$$

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  • $\begingroup$ Thank you! But I have a question. Are we assuming that $\alpha$ never equals $0$? $\endgroup$
    – ronanhugo
    May 18, 2019 at 5:01
  • $\begingroup$ Yes, this derivation assumes that $\alpha\ne 0$ $\endgroup$
    – greg
    May 18, 2019 at 5:36

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