Jacobi's formula

We first prove a preliminary lemma:

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

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

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 A^T 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:

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

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

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

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

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

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/∂A_ij 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 ∂ / ∂A_ij:

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

Thus, by the product rule,

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

Now, if an element of a matrix A_ij and a cofactor adj^T(A)_ik of element A_ik lie on the same row (or column), then the cofactor will not be a function of A_ij, because the cofactor of A_ik is expressed in terms of elements not in its own row (nor column). Thus,

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

so

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

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

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

Therefore,

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

and applying the Lemma yields

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

Lemma 1. ${\displaystyle \det '(I)=\mathrm {tr} }$, where ${\displaystyle \det '}$ is the differential of ${\displaystyle \det }$.

This equation means that the differential of ${\displaystyle \det }$, evaluated at the identity matrix, is equal to the trace. The differential ${\displaystyle \det '(I)}$ is a linear operator that maps an n × n matrix to a real number.

Proof. Using the definition of a directional derivative together with one of its basic properties for differentiable functions, we have

${\displaystyle \det '(I)(T)=\nabla _{T}\det(I)=\lim _{\varepsilon \to 0}{\frac {\det(I+\varepsilon T)-\det I}{\varepsilon }}}$

${\displaystyle \det(I+\varepsilon T)}$ is a polynomial in ${\displaystyle \varepsilon }$ of order n. It is closely related to the characteristic polynomial of ${\displaystyle T}$. The constant term (${\displaystyle \varepsilon =0}$) is 1, while the linear term in ${\displaystyle \varepsilon }$ is ${\displaystyle \mathrm {tr} \ T}$.

Lemma 2. For an invertible matrix A, we have: ${\displaystyle \det '(A)(T)=\det A\;\mathrm {tr} (A^{-1}T)}$.

Proof. Consider the following function of X:

${\displaystyle \det X=\det(AA^{-1}X)=(\det A)\ \det(A^{-1}X)}$

We calculate the differential of ${\displaystyle \det X}$ and evaluate it at ${\displaystyle X=A}$ using Lemma 1, the equation above, and the chain rule:

${\displaystyle \det '(A)(T)=\det A\ \det '(I)(A^{-1}T)=\det A\ \mathrm {tr} (A^{-1}T)}$

Theorem. (Jacobi's formula) ${\displaystyle {\frac {d}{dt}}\det A=\mathrm {tr} \left(\mathrm {adj} \ A{\frac {dA}{dt}}\right)}$

Proof. If ${\displaystyle A}$ is invertible, by Lemma 2, with ${\displaystyle T=dA/dt}$

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

using the equation relating the adjugate of ${\displaystyle A}$ to ${\displaystyle A^{-1}}$. Now, the formula holds for all matrices, since the set of invertible linear matrices is dense in the space of matrices.