Jacobi's Formula

Lemma (Log-Determinant Identity)

The identity relates the determinant of an invertible matrix $A\in {\mathbb{C}}^{n\times n}$ to the trace of its matrix logarithm:

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

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 $D=\text{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$ 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:

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

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

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

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

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

The exponential of the trace:

$$\begin{array}{rl}{e}^{\text{tr}(\log A)}& =e^{{\sum }_{i=1}^n\log {\lambda }_i}\\ & =\prod _{i=1}^n{e}^{\log {\lambda }_i}\\ & =\prod _{i=1}^n{\lambda }_i\end{array}$$

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

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

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 $\text{tr}(\cdot )$ are continuous on the domain of invertible matrices, we can take the limit of the identity:

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

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

Jacobi’s Formula

The derivative of the determinant of a matrix $A$ (if $A$ is invertible) is given by:

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

Proof:

First, we use the identity:

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

Differentiating both sides with respect to $t$:

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

To evaluate $\text{tr}\left(\frac{d}{dt}\log A\right)$, we recall 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 ($\text{tr}(BC)=\text{tr}(CB)$):

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

Since $X=\log A$, we have $\text{tr}\left(\frac{d}{dt}\log A\right)=\text{tr}\left(A^{-1}\frac{dA}{dt}\right)$. Substituting this back into our expression for the derivative of the determinant:

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

Note: This derivation assumes $A$ is invertible so that $\log A$ is well-defined. The formula holds more generally (even if $A$ is singular) as $\frac{d}{dt}\det A=\text{tr}(\text{adj}(A)\dot{A})$.