Jacobi's formula

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

:$sum\_i\; sum\_j\; A\_\{ij\}\; B\_\{ij\}\; =\; mbox\{tr\}\; (A^\; op\; B).$

"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:

:$(A^\; op\; B)\_\{jk\}\; =\; sum\_i\; A\_\{ij\}\; B\_\{ik\}.$

The result follows by taking the trace of both sides:

:$mbox\{tr\}\; (A^\; op\; B)\; =\; sum\_j\; (A^\; op\; B)\_\{jj\}\; =\; sum\_j\; sum\_i\; A\_\{ij\}\; B\_\{ij\}\; =\; sum\_i\; sum\_j\; A\_\{ij\}\; B\_\{ij\}.\; square$

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

:$mbox\{det\}(A)\; =\; sum\_j\; A\_\{ij\}\; mbox\{adj\}^\; op\; (A)\_\{ij\}.$

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":

:$mbox\{det\}(A)\; =\; F(A\_\{11\},\; A\_\{12\},\; ...\; ,\; A\_\{21\},\; A\_\{22\},\; ...\; ,\; A\_\{nn\})$

so that 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":

:$\{partial\; ,\; mbox\{det\}(A)\; over\; partial\; A\_\{ij\; =\; \{partial\; sum\_k\; A\_\{ik\}\; mbox\{adj\}^\; op(A)\_\{ik\}\; over\; partial\; A\_\{ij\; =\; sum\_k\; \{partial\; A\_\{ik\}\; mbox\{adj\}^\; op(A)\_\{ik\}\; over\; partial\; A\_\{ij.$

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,

:$\{partial\; ,\; mbox\{adj\}^\; op(A)\_\{ik\}\; over\; partial\; A\_\{ij\; =\; 0,$

so

:$\{partial\; ,\; mbox\{det\}(A)\; over\; partial\; A\_\{ij\; =\; sum\_k\; mbox\{adj\}^\; op(A)\_\{ik\}\; \{partial\; A\_\{ik\}\; over\; partial\; A\_\{ij.$

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

:$\{partial\; ,\; mbox\{det\}(A)\; over\; partial\; A\_\{ij\; =\; sum\_k\; mbox\{adj\}^\; op(A)\_\{ik\}\; delta\_\{jk\}\; =\; mbox\{adj\}^\; op(A)\_\{ij\}.$

Therefore,

:$d(mbox\{det\}(A))\; =\; sum\_i\; sum\_j\; mbox\{adj\}^\; op(A)\_\{ij\}\; ,d\; A\_\{ij\},$

and applying the Lemma yields

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