Jacobi's formula - Alchetron, The Free Social Encyclopedia

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,

Derivation

We first prove a preliminary lemma:

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

∑ i ∑ j A i j B i j = tr ⁡ ( A T B ) .

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

( A B ) j k = ∑ i A j i B i k .

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

( A T B ) j k = ∑ i A i j B i k .

The result follows by taking the trace of both sides:

tr ⁡ ( A T B ) = ∑ j ( A T B ) j j = ∑ j ∑ i A i j B i j = ∑ i ∑ j A i j B i j . ◻

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

d det ( A ) = tr ⁡ ( adj ⁡ ( A ) d A ) .

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

det ( A ) = ∑ j A i j adj T ⁡ ( A ) i j .

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 det ( A ) = ∑ i ∑ j ∂ F ∂ A i j d A i j .

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:

∂ det ( A ) ∂ A i j = ∂ ∑ k A i k adj T ⁡ ( A ) i k ∂ A i j = ∑ k ∂ ( A i k adj T ⁡ ( A ) i k ) ∂ A i j

Thus, by the product rule,

∂ det ( A ) ∂ A i j = ∑ k ∂ A i k ∂ A i j adj T ⁡ ( A ) i k + ∑ k A i k ∂ adj T ⁡ ( A ) i k ∂ A i j .

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,

∂ adj T ⁡ ( A ) i k ∂ A i j = 0 ,

so

∂ det ( A ) ∂ A i j = ∑ k adj T ⁡ ( A ) i k ∂ A i k ∂ A i j .

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

∂ A i k ∂ A i j = δ j k ,

where δ is the Kronecker delta, so

∂ det ( A ) ∂ A i j = ∑ k adj T ⁡ ( A ) i k δ j k = adj T ⁡ ( A ) i j .

Therefore,

d ( det ( A ) ) = ∑ i ∑ j adj T ⁡ ( A ) i j d A i j ,

and applying the Lemma yields

d ( det ( A ) ) = tr ⁡ ( adj ⁡ ( A ) d A ) . ◻

Corollary

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

d d t det A ( t ) = ( det A ( t ) ) tr ⁡ ( A ( t ) − 1 d d t A ( t ) ) ,

which can be alternatively written as

d d t log ⁡ det A ( t ) = tr ⁡ ( A ( t ) − 1 d d t A ( t ) ) .

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

d d t det e t B = tr ⁡ ( B ) det e t B .

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

References

Jacobi's formula Wikipedia

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Contents

Derivation

Corollary

References