Jacobi's Formula

Theorem (Jacobi’s Formula)

In matrix calculus, Jacobi’s formula expresses the derivative of the determinant of a matrix $A$ in terms of its adjugate and the derivative of $A$.

Formally, if $A:\mathbb{R}\to {\mathbb{R}}^{n\times n}$ is a differentiable matrix-valued function, the derivative of its determinant is:

$$\frac{d}{dt}\det A(t)=\mathrm{tr}\left(\mathrm{adj}(A(t))\frac{dA(t)}{dt}\right)$$

where:

• $\mathrm{adj}(A(t))$ is the adjugate matrix (the transpose of the cofactor matrix) of $A(t)$, as discussed in Square Matrices.
• $\mathrm{tr}(\cdot )$ denotes the trace operator.

Invertible Case

If $A(t)$ is invertible (so $\det A(t)\ne 0$), we can use the identity $\mathrm{adj}(A(t))=\det A(t)A(t{)}^{-1}$ to write the formula in the more common form:

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

Matrix Derivative Form

In terms of matrix derivatives (often used in machine learning and optimization, see Stress-Strain Relation), Jacobi’s formula can be written as:

$$\frac{\partial \det A}{\partial A}=\mathrm{adj}(A{)}^{\top }$$

For invertible $A$, this simplifies to:

$$\frac{\partial \det A}{\partial A}=\det (A)A^{-\top }$$

where $A^{-\top }=(A^{-1}{)}^{\top }$.

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Lemma (Log-Determinant Identity)

For any invertible matrix $A\in {\mathbb{C}}^{n\times n}$, we have the identity:

$$\det A=e^{\mathrm{tr}(\log A)}$$

Case 1: Diagonalizable Matrices

Assume $A$ is diagonalizable. Then there exists an invertible matrix $P$ and a diagonal matrix $D$ such that:

$$A=PD{P}^{-1}$$

where $D=\mathrm{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$ contains the eigenvalues of $A$. The determinant of $A$ is the product of its eigenvalues:

$$\det A=\det (PD{P}^{-1})=\det P\cdot \det D\cdot \det (P^{-1})=\det D=\prod _{i=1}^n{\lambda }_i$$

The matrix logarithm of a diagonalizable matrix is defined by applying the scalar logarithm to its eigenvalues:

$$\log A=P(\log D)P^{-1}=P\cdot \mathrm{diag}(\log {\lambda }_1,\log {\lambda }_2,\dots ,\log {\lambda }_n)\cdot P^{-1}$$

Taking the trace of both sides and utilizing the cyclic property of the trace ($\mathrm{tr}(BC)=\mathrm{tr}(CB)$):

$$\mathrm{tr}(\log A)=\mathrm{tr}\left(P(\log D)P^{-1}\right)=\mathrm{tr}\left((\log D)P^{-1}P\right)=\mathrm{tr}(\log D)$$

The trace of a diagonal matrix is the sum of its diagonal entries:

$$\mathrm{tr}(\log D)=\sum _{i=1}^n\log {\lambda }_i$$

The exponential of the trace:

$$e^{\mathrm{tr}(\log A)}=e^{{\sum }_{i=1}^n\log {\lambda }_i}=\prod _{i=1}^n{e}^{\log {\lambda }_i}=\prod _{i=1}^n{\lambda }_i$$

Since both sides equal ${\prod }_{i=1}^n{\lambda }_i$, the identity holds for all diagonalizable matrices:

$$\det A=e^{\mathrm{tr}(\log A)}$$

Case 2: General Invertible Matrices (Continuity Argument)

If $A$ is invertible but not diagonalizable, we can use the property that diagonalizable matrices are dense in ${\mathbb{C}}^{n\times n}$. There exists a sequence of diagonalizable matrices $\{A_k{\}}_{k=1}^{\infty }$ that converges to $A$:

$$\underset{k\to \infty }{\lim }{A}_k=A$$

Since the functions $\det (\cdot )$, $\log (\cdot )$, and $\mathrm{tr}(\cdot )$ are continuous on the domain of invertible matrices, we can take the limit of the identity:

$$\det A=\det \left(\underset{k\to \infty }{\lim }{A}_k\right)=\underset{k\to \infty }{\lim }\det A_k=\underset{k\to \infty }{\lim }{e}^{\mathrm{tr}(\log A_k)}=e^{\mathrm{tr}\left(\log \left({\lim }_{k\to \infty }{A}_k\right)\right)}=e^{\mathrm{tr}(\log A)}$$

This completes the proof for any invertible matrix $A$.

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Proof (Jacobi’s Formula)

Below are three different proofs of Jacobi’s formula, each highlighting different mathematical techniques.

Proof 1: Via the Log-Determinant Identity

We differentiate the log-determinant identity:

$$\det A=e^{\mathrm{tr}(\log A)}$$

Differentiating both sides with respect to $t$ using the chain rule:

$$\frac{d}{dt}(\det A)=\frac{d}{dt}\left(e^{\mathrm{tr}(\log A)}\right)=e^{\mathrm{tr}(\log A)}\cdot \frac{d}{dt}\left(\mathrm{tr}(\log A)\right)=\det A\cdot \mathrm{tr}\left(\frac{d}{dt}\log A\right)$$

To evaluate $\mathrm{tr}\left(\frac{d}{dt}\log A\right)$, we use the formula for the derivative of the matrix exponential. If $A=e^X$, then:

$$\frac{dA}{dt}={\int }_0^1{e}^{sX}\frac{dX}{dt}{e}^{(1-s)X}ds$$

Multiplying by $A^{-1}=e^{-X}$:

$$A^{-1}\frac{dA}{dt}=e^{-X}{\int }_0^1{e}^{sX}\frac{dX}{dt}{e}^{(1-s)X}ds={\int }_0^1{e}^{(s-1)X}\frac{dX}{dt}{e}^{(1-s)X}ds$$

Taking the trace and using the cyclic property of trace:

$$\mathrm{tr}\left(A^{-1}\frac{dA}{dt}\right)={\int }_0^1\mathrm{tr}\left(e^{(s-1)X}\frac{dX}{dt}{e}^{(1-s)X}\right)ds={\int }_0^1\mathrm{tr}\left(\frac{dX}{dt}\right)ds=\mathrm{tr}\left(\frac{dX}{dt}\right)$$

Since $X=\log A$, we have $\mathrm{tr}\left(\frac{d}{dt}\log A\right)=\mathrm{tr}\left(A^{-1}\frac{dA}{dt}\right)$. Substituting this back gives:

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

Proof 2: Via Differentiation at the Identity

We first establish the formula when $A(t)=I$ at a given point $t=t_0$. Let $B=\dot{A}(t_0)$. Using the Leibniz formula for the determinant:

$$\det (I+ϵB)=1+ϵ\mathrm{tr}(B)+O({ϵ}^2)$$

This is because in the expansion of $\det (I+ϵB)$, the only term that does not contain any $ϵ$ is $1$, and the terms linear in $ϵ$ must come from the product of the diagonal elements $(1+ϵB_{ii})$, which yields $ϵ{\sum }_{i=1}^n{B}_{ii}=ϵ\mathrm{tr}(B)$. Therefore, the derivative of the determinant at the identity is:

$${\frac{d}{dϵ}\det (I+ϵB)|}_{ϵ=0}=\mathrm{tr}(B)$$

For a general invertible matrix $A(t)$, we can factor out $A(t)$:

$$\begin{array}{rl}\frac{d}{dt}\det A(t)& =\underset{ϵ\to 0}{\lim }\frac{\det A(t+ϵ)-\det A(t)}{ϵ}\\ & =\underset{ϵ\to 0}{\lim }\frac{\det (A(t)+ϵ\dot{A}(t)+O({ϵ}^2))-\det A(t)}{ϵ}\\ & =\det A(t)\underset{ϵ\to 0}{\lim }\frac{\det (I+ϵA(t{)}^{-1}\dot{A}(t)+O({ϵ}^2))-1}{ϵ}\\ & =\det A(t)\mathrm{tr}\left(A(t{)}^{-1}\dot{A}(t)\right)\end{array}$$

Proof 3: Via Cofactor Expansion

Recall the cofactor expansion of the determinant along any row $i$:

$$\det A=\sum _{j=1}^n{A}_{ij}{C}_{ij}$$

where $C_{ij}$ is the cofactor of entry $A_{ij}$. Since $C_{ij}$ is the determinant of the submatrix obtained by deleting row $i$ and column $j$, it does not depend on any entry in row $i$. Thus:

$$\frac{\partial \det A}{\partial A_{ij}}=C_{ij}$$

By definition, the adjugate matrix $\mathrm{adj}(A)$ is the transpose of the cofactor matrix, meaning $\mathrm{adj}(A{)}_{ji}=C_{ij}$. Therefore:

$$\frac{\partial \det A}{\partial A_{ij}}=\mathrm{adj}(A{)}_{ji}$$

Using the multivariable chain rule, we compute the total derivative with respect to $t$:

$$\frac{d}{dt}\det A(t)=\sum _{i=1}^n\sum _{j=1}^n\frac{\partial \det A}{\partial A_{ij}}\frac{d{A}_{ij}}{dt}=\sum _{i=1}^n\sum _{j=1}^n\mathrm{adj}(A{)}_{ji}\frac{d{A}_{ij}}{dt}=\mathrm{tr}\left(\mathrm{adj}(A)\frac{dA}{dt}\right)$$

This proof is completely general and does not assume $A$ is invertible.

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Application (Ordinary Differential Equations and Continuum Mechanics)

Theorem (Liouville’s Formula)

Jacobi’s formula is the key tool used to prove Liouville’s formula (or Abel’s identity) for linear systems of Ordinary Differential Equations. Consider a homogeneous linear system:

$$\dot{x}(t)=P(t)x(t)$$

If $\Phi (t)$ is a fundamental matrix solution (so $\dot{\Phi }(t)=P(t)\Phi (t)$), then the derivative of its determinant (called the Wronskian $W(t)=\det \Phi (t)$) is given by:

$$\frac{d}{dt}W(t)=W(t)\mathrm{tr}\left(\Phi (t{)}^{-1}\dot{\Phi }(t)\right)=W(t)\mathrm{tr}\left(\Phi (t{)}^{-1}P(t)\Phi (t)\right)$$

Using the cyclic property of trace, $\mathrm{tr}({\Phi }^{-1}P\Phi )=\mathrm{tr}(P\Phi {\Phi }^{-1})=\mathrm{tr}(P(t))$. Therefore:

$$\frac{d}{dt}W(t)=\mathrm{tr}(P(t))W(t)$$

Integrating this scalar ODE yields Liouville’s formula:

$$W(t)=W(t_0)\exp \left({\int }_{t_0}^t\mathrm{tr}(P(s))ds\right)$$

Theorem (Mass Conservation Equation)

In Continuum Mechanics, Jacobi’s formula relates the change in volume of a deforming body to the divergence of its velocity field. If $\varphi (X,t)$ is the flow map and $F=d\varphi$ is the deformation gradient, the local volume ratio is $J=\det F$. The rate of change of $J$ is:

$$\dot{J}=J\mathrm{tr}(F^{-1}\dot{F})$$

Since $\dot{F}{F}^{-1}=L$ is the spatial velocity gradient $\nabla \mathbf{u}$, we have:

$$\dot{J}=J\mathrm{tr}(\nabla \mathbf{u})=J(\nabla \cdot \mathbf{u})=J\mathrm{div}(\mathbf{u})$$

This identity is central to deriving the continuity equation for mass conservation.