Ngô Quốc Anh

November 2, 2010

Jacobi’s formula for the differential of the determinant of matrices

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 15:34

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

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

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

We first prove a preliminary lemma.

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

\displaystyle\sum_i \sum_j A_{ij} B_{ij} = \mbox{tr} (A^\top 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^\top is equivalent to permuting the indices of its components

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

The result follows by taking the trace of both sides

\displaystyle \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}.

Theorem. It holds

\displaystyle 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

\displaystyle\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

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

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

To find \frac{\partial F}{\partial 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 \frac{\partial}{\partial A_{ij}}

\displaystyle {\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}}.

Thus by product rule,

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

Now, if an element of a matrix A_{ij} and a cofactor \mbox{adj}^\top(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 \, \mbox{adj}^\top(A)_{ik} \over \partial A_{ij}} = 0,


\displaystyle {\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.

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

where \delta is the Kronecker delta, so

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


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

and applying the Lemma yields

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


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