Jacobi's formula

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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,

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

It is named after the mathematician Carl Gustav Jacob Jacobi.


We first prove a preliminary lemma:

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

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

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

The result follows by taking the trace of both sides:

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

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

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:

so that, by the chain rule, its differential is

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:

Thus, by the product rule,

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,


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

where δ is the Kronecker delta, so


and applying the Lemma yields


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


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


which can be alternatively written as


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


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


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