Easier way to show different forms of Jacobi's formula for the derivative of the determinant

Question

Jacobi's formula for the derivative of the determinant of a matrix A is

$$ {\displaystyle {\frac {d}{dt}}\det A=\det A\;\mathrm {tr} \left(A^{-1}{\frac {dA}{dt}}\right)=\mathrm {tr} \left(\mathrm {adj} \ A\;{\frac {dA}{dt}}\right)} $$

A proof of this formula is given in the wikipedia and been asked before, link, but I was wondering is there a better way to prove it, probabily making use of the Leibniz formula for determinants ?

Probable Intuitive way $$ \frac{d}{dt}\det A(t)= \frac{d}{dt}\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \\ \end{vmatrix}\\ =\begin{vmatrix} \dot{a}_{11} & \dot{a}_{12} & ... & \dot{a}_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \\ \end{vmatrix}+\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n} \\ \dot{a}_{21} & \dot{a}_{22} & ... & \dot{a}_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \\ \end{vmatrix}+...+\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \dot{a}_{n1} & \dot{a}_{n2} & ... & \dot{a}_{nn} \\ \end{vmatrix}\\ $$ If $A(t)=I$ $$ \frac{d}{dt}\det A(t)=\begin{vmatrix} \dot{a}_{11} & \dot{a}_{12} & ... & \dot{a}_{1n} \\ 0 & 1 & ... & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & 1 \\ \end{vmatrix}+\begin{vmatrix} 1 & 0 & ... & 0 \\ \dot{a}_{21} & \dot{a}_{22} & ... & \dot{a}_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & 1 \\ \end{vmatrix}+...+\begin{vmatrix} 1 & 0 & ... & 0 \\ 0 & 1 & ... & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \dot{a}_{n1} & \dot{a}_{n2} & ... & \dot{a}_{nn} \\ \end{vmatrix}\\ =\dot{a}_{11}+\dot{a}_{22}+...+\dot{a}_{nn}=Tr(\dot{A}(t))\\ \implies \boxed{\frac{d}{dt}\det A(t)=Tr(\dot{A}(t))\quad \text{when }A(t)=I} $$

Let $B$ a constant and invertible matrix such that, $B.\psi(t)=I\implies\det\Big(B.\psi(t)\Big)=\det B.\det\psi(t)$ $$ \frac{d}{dt}\det B.\psi(t)=\det B.\frac{d}{dt}\det \psi(t)=Tr\;\big(B.\frac{d}{dt}{\psi(t)}\big)=Tr\;\Big(B.\dot{\psi}(t)\Big)\\ \boxed{ \det B.\frac{d}{dt}\det \psi(t)=Tr\;\Big(B.\dot{\psi}(t)\Big)\quad\text{when }B.\psi(t)=I\implies B=\psi^{-1}(t) }\\ \det B.\frac{d}{dt}\det\psi(t)=\frac{1}{\det{\psi(t)}}\frac{d}{dt}\det\psi(t)=Tr\;\Big(\psi^{-1}(t).\dot{\psi}(t)\Big)\\ \boxed{\frac{d}{dt}\det\psi(t)=\det\psi(t).Tr\;\Big(\psi^{-1}(t).\dot{\psi}(t)\Big)} $$ Is there atleast an intuitive way to prove that, $\det\psi(t).Tr\;\Big(\psi^{-1}(t).\dot{\psi}(t)\Big)=Tr\;\Big(adj \;\psi(t).\dot{\psi}(t)\Big)$ ?

Answer 1

Let $E_{ij}$ be the matrix with a $1$ at the $(i,j)$-th position and zeroes elsewhere. Denote the $(i,j)$-th minor of $A$ by $M_{ij}$. Jacobi's formula holds in the following special case: $$ \frac{d\det(A(t)+hE_{ij})}{dh} =\frac{d\left(\det(A)+h(-1)^{i+j}M_{ij}\right)}{dh} =(-1)^{i+j}M_{ij} =(\operatorname{adj}(A))_{ji} =\operatorname{tr}\left(\operatorname{adj}(A)E_{ij}\right). $$ Denote the determinant function by $F$. The general case now follows from the total derivative formula: \begin{aligned} &\frac{d\det(A(t))}{dt} =\frac{dF(A(t))}{dt}=\sum_{i,j}\frac{\partial F}{\partial a_{ij}}\frac{da_{ij}}{dt} =\sum_{i,j}\frac{d\det(A(t)+hE_{ij})}{dh}\frac{da_{ij}}{dt}\\ &=\sum_{i,j}\operatorname{tr}\left(\operatorname{adj}(A)E_{ij}\right)\frac{da_{ij}}{dt} =\operatorname{tr}\left(\operatorname{adj}(A)\sum_{i,j}\frac{da_{ij}}{dt}E_{ij}\right)=\operatorname{tr}\left(\operatorname{adj}(A)\frac{dA}{dt}\right). \end{aligned}

Answer 2

$\DeclareMathOperator{\sign}{sgn}$ $\DeclareMathOperator{\trace}{tr}$ $\DeclareMathOperator{\adjugate}{adj}$

Hi Sooraj. Your proof starts fine. Basically you are using the definition $$ \det(A(t)) = \sum_{\sigma \in S_n} \sign(\sigma) \prod_{i=1}^{n} a_{i,\sigma(i)}(t), $$ and then you are taking a derivative to obtain $$ \begin{align} \frac{d}{dt}\det(A(t)) &= \sum_{\sigma \in S_n} \sign(\sigma) \sum_{k=1}^{n}\dot{a}_{k,\sigma(k)}\prod_{i\neq k}^{n} a_{i,\sigma(i)}(t) \\ &= \sum_{k=1}^{n} \sum_{\sigma \in S_n} \sign(\sigma) \dot{a}_{k,\sigma(k)}\prod_{i\neq k}^{n} a_{i,\sigma(i)}(t) \\ &= \sum_{k=1}^{n} \sum_{\sigma \in S_n} \sign(\sigma) \prod_{i=1}^{n} b_{i,\sigma(i)}^{(k)}(t), \end{align} $$ where $$ b^{(k)}_{i,j}= \left\{ \begin{array}{rl} \dot{a}_{i,j}, & i=k \\ {a}_{i,j}, & \text{else}. \end{array} \right. $$ This proves your formula $$ \frac{d}{dt}\det A(t)= \frac{d}{dt}\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \\ \end{vmatrix}\\ =\begin{vmatrix} \dot{a}_{11} & \dot{a}_{12} & ... & \dot{a}_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \\ \end{vmatrix}+\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n} \\ \dot{a}_{21} & \dot{a}_{22} & ... & \dot{a}_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & ... & a_{nn} \\ \end{vmatrix}+\cdots+\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \dot{a}_{n1} & \dot{a}_{n2} & ... & \dot{a}_{nn} \\ \end{vmatrix}. $$ However, as soon as you say $A(t)=I$, all your matrices become zero and at the end you are basically proving that $0=0$. Let me suggest an alternative approach which combines your idea with was already mentioned in the link that you provided by Jean Van Schaftingen.

Since the determinant is a function from the space of $n\times n$ matrices to $\mathbb{R}$, i.e., $\det:M_n(\mathbb{R})\to\mathbb{R}$, we can ask the question: how does $\det$ change as we move in the direction of another matrix $U$, when we are located at the identity matrix, i.e., at the position $X=I$? This is a directional derivative $$ \left.\frac{\partial \det}{\partial U}\right|_{X=I} = \lim_{t\to 0} \frac{\det(I+tU)-\det(I)}{t} $$ Define $$g(t)=\det(I+tU).$$ Then $$ \left.\frac{dg}{dt}\right|_{t=0} = \left.\frac{\partial \det}{\partial U}\right|_{X=I} $$ So we can use your formula to obtain $$ \left.\frac{dg}{dt}\right|_{t=0}=\begin{vmatrix} {u}_{11} & {u}_{12} & ... & {u}_{1n} \\ 0 & 1 & ... & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & 1 \\ \end{vmatrix}+\begin{vmatrix} 1 & 0 & ... & 0 \\ {u}_{21} & {u}_{22} & ... & {u}_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & ... & 1 \\ \end{vmatrix}+\cdots+\begin{vmatrix} 1 & 0 & ... & 0 \\ 0 & 1 & ... & 0 \\ \vdots & \vdots & \ddots & \vdots \\ {u}_{n1} & {u}_{n2} & ... & {u}_{nn} \\ \end{vmatrix}\\ ={u}_{11}+{u}_{22}+...+{u}_{nn}=\trace(U), $$ and conclude that $$ \left.\frac{dg}{dt}\right|_{t=0} %=\left.\frac{\partial \det}{\partial U}\right|_{X=I} =\left.\frac{d\det(I+tU)}{dt}\right|_{t=0} = \trace(U). $$ Now let $A(t) \in M_n(\mathbb{R})$, fix a time $s$, and assume that $A(t)$ is invertible and differentiable in a neighborhood of $t=s$. Furthermore, consider the function $$ H(X(t)) = \det(A(s))\cdot\det(A^{-1}(s)X(t)). $$ Note that $H(A(s)) = \det(A(s))$. Basically, what we want to do is observe how $H$ changes as we move from $X=A(s)$ to $X=A(s+t)$. Formally, we are interested in the limit $$ \begin{align} \left.\frac{dH(A(t))}{dt}\right|_{t=s} = \lim_{t\to 0} \frac{H(A(s+t))-H(A(s))}{t} \end{align} $$ Since $A(t)$ is differentiable, write its Taylor expansion about $t=s$ to obtain $$ A(t+s)=A(s)+t\dot{A}(s)+\mathcal{O}(t^2) $$ and define the trajectory $$ X(t) = A(s)+t\dot{A}(s) = A(t+s)+\mathcal{O}(t^2), $$ which describes the behavior $A$ near $A(s)$. Furthermore, note that $$ \begin{align} H(X(t)) &= \det(A(s))\cdot\det(A^{-1}(s)(A(s)+t\dot{A}(s)))\\ &= \det(A(s))\cdot\det(I+tA^{-1}(s)\dot{A}(s))),\\ H(X(0)) &= \det(A(s)). \end{align} $$ Hence, $$ \begin{align} \left.\frac{d H}{dt}\right|_{t=0} &= \lim_{t\to 0} \frac{H(X(t))-H(X(0))}{t} \\ &= \lim_{t\to 0} \frac{ \det(A(s))\cdot\det(I+tA^{-1}(s)\dot{A}(s))) - \det(A(s)) }{t} \\ &= \det(A(s)) \lim_{t\to 0} \frac{\det(I+tA^{-1}(s)\dot{A}(s))) - \det(I) }{t} \\ &= \det(A(s)) \trace(A^{-1}(s)\dot{A}(s)). \end{align} $$ Lastly, using the fact that the trace is a linear operator, we can bring $\det(A(s))$ into the argument and use the identity $$ \det(A) A^{-1} = \adjugate(A), $$ to obtain $$ \left.\frac{d\det(A(t))}{dt}\right|_{t=s} =\trace(\adjugate(A)(s)\dot{A}(s)). $$