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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. If A is a differentiable map from the real numbers to n × n matrices,

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

Equivalently, if dA stands for the differential of A, the formula is

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

It is named after the mathematician C.G.J. 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

$$\sum_i \sum_j A_{ij} B_{ij} = \mathrm{tr} (A^{\rm T} B).$$

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

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

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

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

The result follows by taking the trace of both sides:

$$\mathrm{tr} (A^{\rm T} B) = \sum_j (A^{\rm T} B)_{jj} = \sum_j \sum_i A_{ij} B_{ij} = \sum_i \sum_j A_{ij} B_{ij}.\ \square$$

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

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

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

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

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:

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

so that, by the chain rule, its differential is

$$d \det(A) = \sum_i \sum_j {\partial F \over \partial A_{ij}} \,dA_{ij}.$$

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:

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

Thus, by the product rule,

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

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,

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

so

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

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

$${\partial A_{ik} \over \partial A_{ij}} = \delta_{jk},$$

where δ is the Kronecker delta, so

$${\partial \det(A) \over \partial A_{ij}} = \sum_k \mathrm{adj}^{\rm T}(A)_{ik} \delta_{jk} = \mathrm{adj}^{\rm T}(A)_{ij}.$$

Therefore,

$$d(\det(A)) = \sum_i \sum_j \mathrm{adj}^{\rm T}(A)_{ij} \,d A_{ij},$$

and applying the Lemma yields

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

Corollary

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

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

which can be alternatively written as

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

Furthermore, taking A(t ) = exp(tB ) in the first equation, we obtain

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

solved by

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

a useful relation connecting the trace to the determinant of the associated matrix exponential.
Alternative Derivation

A quicker proof of Jacobi's formula is as follows. By the chain rule, we have

$$\frac{d}{dt}\mbox{det}\left(A\left(t\right)\right)=\left(\nabla\mbox{det}\left(A\left(t\right)\right)\right):\left(\frac{d}{dt}A\left(t\right)\right)=\mbox{tr}\left(\mbox{adj}\left(A\left(t\right)\right)\frac{d}{dt}A\left(t\right)\right)$$

where (:) denotes tensor double-contraction.
Notes

Magnus & Neudecker (1999), Part Three, Section 8.3

References

Magnus, Jan R.; Neudecker, Heinz (1999), Matrix Differential Calculus with Applications in Statistics and Econometrics, Wiley, ISBN 0-471-98633-X
Bellmann, Richard (1987), Introduction to Matrix Analysis, SIAM, ISBN 0898713994

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