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In matrix calculus, Jacobi's formula expresses the differential of the determinant of a matrix A in terms of the adjugate of A and the differential of A. The formula is

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

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

Derivation

We first prove a preliminary lemma:

Lemma. Given a pair of square matrices A and B of the same dimension n, then

$\sum_i \sum_j A_{ij} B_{ij} = \mbox{tr} (A^\top B).$

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 A^T is equivalent to permuting the indices of its components:

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

The result follows by taking the trace of both sides:

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

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

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

$\mbox{det}(A) = \sum_j A_{ij} \mbox{adj}^\top (A)_{ij}.$

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,..., A 21, A 22,..., A n n)

so that, by the chain rule its differential is

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

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

To find ∂F / ∂A_ij consider that in the right side of Laplace's formula, 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:

${\partial \, \mbox{det}(A) \over \partial A_{ij}} = {\partial \sum_k A_{ik} \mbox{adj}^\top(A)_{ik} \over \partial A_{ij}} = \sum_k {\partial A_{ik} \mbox{adj}^\top(A)_{ik} \over \partial A_{ij}}.$

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,

${\partial \, \mbox{adj}^\top(A)_{ik} \over \partial A_{ij}} = 0,$

${\partial \, \mbox{det}(A) \over \partial A_{ij}} = \sum_k \mbox{adj}^\top(A)_{ik} {\partial A_{ik} \over \partial A_{ij}}.$

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

${\partial \, \mbox{det}(A) \over \partial A_{ij}} = \sum_k \mbox{adj}^\top(A)_{ik} \delta_{jk} = \mbox{adj}^\top(A)_{ij}.$

Therefore,

$d(\mbox{det}(A)) = \sum_i \sum_j \mbox{adj}^\top(A)_{ij} \,d A_{ij},$

and applying the Lemma yields

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

References

Magnus, Jan R.; Neudecker, Heinz (1999), Matrix Differential Calculus with Applications in Statistics and Econometrics, Wiley, ISBN 0-471-98633-X

Part Three, Section 8.3

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