# Is there any matrix representation for the second order derivation of a determinant?

$$\newcommand{\piff}[2]{\frac{\partial{#1}}{\partial{#2}}}\newcommand{\ppiff}[3]{\frac{\partial^2 {#1}}{\partial{#2}\partial{#3}}}\newcommand{\pa}[2]{\ppiff{|A|}{#1}{#2}}$$ 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? $$\begin{pmatrix} \pa{x_{11}}{x_{11}} & \pa{x_{11}}{x_{12}} & \cdots\cdots & \pa{x_{11}}{x_{nn}} \\ \pa{x_{12}}{x_{11}} & \pa{x_{12}}{x_{12}} & \cdots\cdots & \pa{x_{12}}{x_{nn}} \\ \vdots & \vdots & \ddots & \vdots \\ \pa{x_{nn}}{x_{11}} & \pa{x_{nn}}{x_{12}} & \cdots\cdots & \pa{x_{nn}}{x_{nn}} \end{pmatrix}$$

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