Jacobi's formula

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,

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

where tr(X) is the trace of the matrix X. 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 Carl Gustav Jacob 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 A^T 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/∂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:

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

{\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}\,dA_{ij},

and applying the Lemma yields

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

Corollary

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

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

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

{\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)

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

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

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

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 (1997). Introduction to Matrix Analysis. SIAM. ISBN 0-89871-399-4.

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