# Jacobi's formula

In matrix calculus, Jacobi's formula expresses the derivative of the determinant of a matrix A in terms of the adjugate of A and the derivative of A.[1] If A is a differentiable map from the real numbers to n × n matrices,

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

where tr(X) is the trace of the matrix X. Equivalently, if dA stands for the differential of A, the formula is

${\displaystyle d\det(A)=\mathrm {tr} (\mathrm {adj} (A)\,dA).}$

It is named after the mathematician Carl Gustav Jacob Jacobi.

## Derivation

We first prove a preliminary lemma:

Lemma. Let A and B be a pair of square matrices of the same dimension n. Then

${\displaystyle \sum _{i}\sum _{j}A_{ij}B_{ij}=\mathrm {tr} (A^{\rm {T}}B).}$

Proof. The product AB of the pair of matrices has components

${\displaystyle (AB)_{jk}=\sum _{i}A_{ji}B_{ik}.\,}$

Replacing the matrix A by its transpose AT is equivalent to permuting the indices of its components:

${\displaystyle (A^{\rm {T}}B)_{jk}=\sum _{i}A_{ij}B_{ik}.}$

The result follows by taking the trace of both sides:

${\displaystyle \mathrm {tr} (A^{\rm {T}}B)=\sum _{j}(A^{\rm {T}}B)_{jj}=\sum _{j}\sum _{i}A_{ij}B_{ij}=\sum _{i}\sum _{j}A_{ij}B_{ij}.\ \square }$

Theorem. (Jacobi's formula) For any differentiable map A from the real numbers to n × n matrices,

${\displaystyle d\det(A)=\mathrm {tr} (\mathrm {adj} (A)\,dA).}$

Proof. Laplace's formula for the determinant of a matrix A can be stated as

${\displaystyle \det(A)=\sum _{j}A_{ij}\mathrm {adj} ^{\rm {T}}(A)_{ij}.}$

Notice that the summation is performed over some arbitrary row i of the matrix.

The determinant of A can be considered to be a function of the elements of A:

${\displaystyle \det(A)=F\,(A_{11},A_{12},\ldots ,A_{21},A_{22},\ldots ,A_{nn})}$

so that, by the chain rule, its differential is

${\displaystyle d\det(A)=\sum _{i}\sum _{j}{\partial F \over \partial A_{ij}}\,dA_{ij}.}$

This summation is performed over all n×n elements of the matrix.

To find ∂F/∂Aij consider that on the right hand side of Laplace's formula, the index i can be chosen at will. (In order to optimize calculations: Any other choice would eventually yield the same result, but it could be much harder). In particular, it can be chosen to match the first index of ∂ / ∂Aij:

${\displaystyle {\partial \det(A) \over \partial A_{ij}}={\partial \sum _{k}A_{ik}\mathrm {adj} ^{\rm {T}}(A)_{ik} \over \partial A_{ij}}=\sum _{k}{\partial (A_{ik}\mathrm {adj} ^{\rm {T}}(A)_{ik}) \over \partial A_{ij}}}$

Thus, by the product rule,

${\displaystyle {\partial \det(A) \over \partial A_{ij}}=\sum _{k}{\partial A_{ik} \over \partial A_{ij}}\mathrm {adj} ^{\rm {T}}(A)_{ik}+\sum _{k}A_{ik}{\partial \,\mathrm {adj} ^{\rm {T}}(A)_{ik} \over \partial A_{ij}}.}$

Now, if an element of a matrix Aij and a cofactor adjT(A)ik of element Aik lie on the same row (or column), then the cofactor will not be a function of Aij, because the cofactor of Aik is expressed in terms of elements not in its own row (nor column). Thus,

${\displaystyle {\partial \,\mathrm {adj} ^{\rm {T}}(A)_{ik} \over \partial A_{ij}}=0,}$

so

${\displaystyle {\partial \det(A) \over \partial A_{ij}}=\sum _{k}\mathrm {adj} ^{\rm {T}}(A)_{ik}{\partial A_{ik} \over \partial A_{ij}}.}$

All the elements of A are independent of each other, i.e.

${\displaystyle {\partial A_{ik} \over \partial A_{ij}}=\delta _{jk},}$

where δ is the Kronecker delta, so

${\displaystyle {\partial \det(A) \over \partial A_{ij}}=\sum _{k}\mathrm {adj} ^{\rm {T}}(A)_{ik}\delta _{jk}=\mathrm {adj} ^{\rm {T}}(A)_{ij}.}$

Therefore,

${\displaystyle d(\det(A))=\sum _{i}\sum _{j}\mathrm {adj} ^{\rm {T}}(A)_{ij}\,dA_{ij},}$

and applying the Lemma yields

${\displaystyle d(\det(A))=\mathrm {tr} (\mathrm {adj} (A)\,dA).\ \square }$

## Corollary

The following is a useful relation connecting the trace to the determinant of the associated matrix exponential,

 ${\displaystyle \det e^{tB}=e^{\mathrm {tr} \left(tB\right)}}$.

This statement is clear for diagonal matrices, and a proof of the general claim follows.

For any invertible matrix A, the inverse A−1 is related to the adjugate by

A−1 = (det A)−1 adj A.

It follows that if A(t) is invertible for all t, then

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

which can be alternatively written as

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

Considering A(t) = exp(tB) in the first equation yields

${\displaystyle {\frac {d}{dt}}\det e^{tB}=\mathrm {tr} \left(B\right)\det e^{tB}}$.

The desired result follows as the solution to this ordinary differential equation.

## Notes

1. ^ Magnus & Neudecker (1999), Part Three, Section 8.3