In the wikipedia article on Jacobi's formula
$$\frac{d}{dt}\det A(t) = \textrm{tr}\left(\textrm{adj}(A(t))\frac{dA(t)}{dt}\right)$$ they offer a proof via the chain rule that i am trying to understand.
The lemma states that $\det'(I) = \textrm{tr}$, where $I$ is the identity matrix and $\det'$ is the differential of $\det$. This is where I'm confused. What is the differential of $\det$? I figured it meant the derivative of the determinant of a matrix with respect to the matrix's indices, but then how does it make sense to apply it to the specific matrix $I$?
Best Answer
The differential of a function $\mathrm{d}f$ is its best linear approximation at some point $x_0$. In this case $\det$ is a function $\det:M_{n\times n} \rightarrow\mathbb{R}$ where $M_{n\times n}$ is the space of $n\times n$ square matrices. Therefore, a matrix is the equivalent of a point for real functions. The best linear approximation to $\det$ near the identity is given by:
$$ \det(\mathbf{I}+\mathbf{M})=\det(\mathbf{I})+\mathrm{d}(\det(\mathbf{I}))\mathbf{M}+R(\mathbf{I},\mathbf{M}),\qquad \lim_{\|\mathbf{M}\|\rightarrow 0} \frac{R(\mathbf{I},\mathbf{M})}{\|\mathbf{M}\|}= 0 \tag{1}\label{eq1}.$$
The analogous expression for $f: \mathbb{R} \rightarrow\mathbb{R}$ is
$$ f(x_0+\epsilon)=f(x_0)+\mathrm{d}(f(x_0))\epsilon+R(x_0,\epsilon),\qquad \lim_{|\epsilon|\rightarrow 0} \frac{R(x_0,\epsilon)}{|\epsilon|}= 0 \tag{2}\label{eq2}.$$
Regarding the original question, $\det'(\mathrm{I})=\mathrm{tr}$ is equivalent to the following:
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \det(\mathbf{I}+t\mathbf{B})\right] \right|_{t=0} = \mathrm{tr}(\mathbf{B}) \tag{3} \label{eq3}$$
You can use induction to prove $\eqref{eq3}$. The first step is to prove it for $n=2$: $$\frac{\mathrm{d}}{\mathrm{d}t} \left . \begin{bmatrix} 1+tB_{11} & tB_{12} \\ tB_{21} & 1+tB_{22} \end{bmatrix} \right|_{t=0} = \frac{\mathrm{d}}{\mathrm{d}t} \left. \left[ (1+tB_{11})(1+tB_{22}) - t^2B_{12}B_{21} \right] \right|_{t=0}=B_{11}+B_{22}. \tag{4} \label{eq4}$$ Assuming that $\eqref{eq3}$ is valid for $n=k-1$, we will prove it for $n=k$.
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \det(\mathbf{I}+t\mathbf{B})\right] \right|_{t=0} = \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \sum_{i=1}^k (\delta_{i,k}+tB_{i,k})M_{i,k}(t) \right] \right|_{t=0} \tag{5} \label{eq5}$$
here $M_{i,k}(t)$ represent the $(i,k)$ minor of $\mathbf{I}+t\mathbf{B}$.
$$ \frac{\mathrm{d}}{\mathrm{d}t} \left. \left [ \sum_{i=1}^k (\delta_{i,k}+tB_{i,k})M_{i,k}(t) \right] \right|_{t=0}=\sum_{i=1}^k\left. \left [ B_{i,k}M_{i,k}(t)+(\delta_{i,k}+tB_{i,k})\frac{\mathrm{d}}{\mathrm{d}t} M_{i,k}(t) \right] \right|_{t=0} \tag{6} \label{eq6}$$
evaluating the first term in the RHS of $\eqref{eq6}$ at $t=0$ gives zero except when $i=k$ since each minor has lower order proportional to $t$. On the other hand, $M_{k,k}(0)=\det(\mathbf{I})=1$. Therefore,
$$ \sum_{i=1}^k\left. \left [ B_{i,k}M_{i,k}(t)+(\delta_{i,k}+tB_{i,k})\frac{\mathrm{d}}{\mathrm{d}t} M_{i,k}(t) \right] \right|_{t=0}=B_{k,k}+\frac{\mathrm{d}}{\mathrm{d}t} M_{k,k}(0)= \sum_{j=1}^k B_{j,j}\tag{7} \label{eq7}$$ In the last step we use the hypothesis $n=k-1$ since $M_{k,k}$ is $\det(\mathbf{I^{(k-1)\times(k-1)}}+t\mathbf{B^{(k-1)\times(k-1)}})$.