Ngô Quốc Anh

March 13, 2011

Jacobi’s formula for the derivative of a determinant revisited

Filed under: Uncategorized — Ngô Quốc Anh @ 12:23

Last time, we discussed [here] 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).

A more useful formula is the following

\displaystyle \frac{d}{dt}\mbox{det} (A+tB) = \mbox{det}(A+tB) \mbox{tr}\big((A+tB)^{-1}B\big).

Let us firstly reprove the Jacobi formula. Assuming (A^{ij}) is the cofactor matrix with respect to A=(a_{ij}). It then holds



\displaystyle\frac{d}{{d{a_{ij}}}}(\det (A)) = \frac{d}{{d{a_{ij}}}}\left( {\sum\limits_k {{a_{ik}}{A^{ik}}} } \right) = {A^{ij}}.

Now the fact that

\displaystyle d \text{det}(A) = \sum\limits_{i,j} {\frac{d}{{d{a_{ij}}}}} (\det (A))d{a_{ij}}

will help us to conclude

\displaystyle d {\text{det}}(A) = \sum\limits_{i,j} {{A^{ij}}d{a_{ij}}} = \mbox{tr} (\mbox{adj}(A) \, dA).

Let us now consider the following second derivative

\displaystyle d^2\mbox{det} (A).

To this purpose, it is well-known that AA^{-1}=I, then


In other words,


This and the fact that \mbox{adj}(A)=\mbox{det}(A) A^{-1} give us

\displaystyle d\mbox{adj} (A)=\mathcal S(A,dA)

where \mathcal S is given by

\displaystyle\mathcal{S}(B,Z) = \frac{1}{{{\text{det(B)}}}}({\text{tr}}({\text{adj}}(B)Z){\text{adj}}(B) - {\text{adj}}(B)Z{\text{adj}}(B))

provided \mbox{det}(B)\ne 0. Thus

\displaystyle\begin{gathered} {d^2}{\text{det(A) = }}d{\mkern 1mu} {\text{tr}}({\text{adj}}(A)dA) \hfill \\ \qquad= {\mkern 1mu} {\text{tr}}(d({\text{adj}}(A)dA)) \hfill \\ \qquad= {\mkern 1mu} {\text{tr}}(d({\text{adj}}(A))dA + {\text{adj}}(A){d^2}A) \hfill \\ \qquad= {\mkern 1mu} {\text{tr}}(\mathcal{S}(A,dA)dA + {\text{adj}}(A){d^2}A) .\hfill \\ \end{gathered}

We also have the following useful identity

\displaystyle\frac{{{d^2}}}{{d{t^2}}}{\text{det}}(A + tB) = {\text{det}}(A + tB)\left[ {{\text{tr}}{{\big((A + tB)}^{ - 1}}B{\big)^2} + \frac{d}{{dt}}{\text{tr}}{{\big((A + tB)}^{ - 1}}B\big)} \right].

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