Is there a simple identity for the derivative of a matrix logarithm w.r.t. a real parameter?

Question

Let $A(t)$ be an invertible square matrix that depends (differentiably) on a real parameter $t$. It is well known that for example $$ \frac{d}{dt} A(t)^{-1}=-A(t)^{-1}\ \dot{A}(t)\ A(t)^{-1} $$ and $$ \frac{d}{dt} \text{det}\ A(t)=\text{det}( A(t))\ \text{tr}(A(t)^{-1}\dot{A}(t)). $$ Is there a similarly simple identity for $$ \frac{d}{dt}\log A(t)=? $$

Answer 1

First, note that the chain rule for matrix functions (i.e. functions which map matrices to matrices) results in a rank-4 tensor: $$ \frac{d}{dt}F(A(t))_{ab} = \sum_{cd} F'(A(t))_{ab;cd} \frac{dA(t)_{cd}}{dt} $$ where $F'(A(t))$ is a rank-4 tensor which encodes the derivative of $F$ and $a$, $b$, $c$, and $d$ are indices of the above matrices and tensors. For example, if $F(A) = A^{-1}$, then $$ F'(A(t))_{ab;cd} = - (A(t)^{-1})_{ac} (A(t)^{-1})_{db} $$ which reproduces the expression for $\frac{d}{dt}A(t)^{-1}$ given in the question.

For the case $F = \log$ and if $A(t)$ is diagonalizable with no eigenvalues that are zero or on the negative real axis (i.e. the principal branch cut of $\log$), then the answer is given on page 146 (see 2nd to last equation) of Jog, C.S. J Elasticity (2008) 93: 141. doi:10.1007/s10659-008-9169-x and can be expressed as $$ \log'(A(t))_{ab;cd} = \sum_{ij} P^{(i)}_{ac} P^{(j)}_{db} \begin{cases} \lambda_i^{-1} & \lambda_i = \lambda_j \\ \frac{\log\lambda_i - \log\lambda_j}{\lambda_i - \lambda_j} & \lambda_i \neq \lambda_j \end{cases} $$ where $i$ and $j$ index the eigenvalues $\lambda$ of $A(t)$, and $P^{(i)}_{ab} \equiv Q_{ai} (Q^{-1})_{ib}$ projects onto the $i$-th eigenvector where $Q$ is the matrix of eigenvectors of $A(t)$ given by the eigendecomposition $A(t) = Q \Lambda Q^{-1}$. Therefore $$ \frac{d}{dt}\log A(t) = \sum_{ij} P^{(i)} \cdot \frac{dA(t)}{dt} \cdot P^{(j)} \begin{cases} \lambda_i^{-1} & \lambda_i = \lambda_j \\ \frac{\log\lambda_i - \log\lambda_j}{\lambda_i - \lambda_j} & \lambda_i \neq \lambda_j \end{cases} $$ (I checked this equation in a Mathematica notebook.)

Answer 2

A common definition of the logarithm for (finite dimensional) matrices is via the Dunford-Taylor integral:

$$ \ln(T) := \frac{1}{2\pi i} \oint_\Gamma \ln(z) (z-T)^{-1} dz \, , $$

Where $\Gamma$ is a simple closed smooth curve that contains all the eigenvalues of $T$ in anticlockwise direction. Note that the above is a seemingly natural generalization of Cauchy's integral formula.

Let the matrix $T$ depend on some parameter $x$. As you noted yourself one has

$$ \frac{\partial }{\partial x} (z - T)^{-1}= (z - T(x))^{-1} T' (z- T(x))^{-1} \, , $$

where prime denotes differentiation with respect to $x$.

Combining the two equations we get

$$ \frac{\partial \ln(T)}{\partial x} = \frac{1}{2\pi i} \oint_\Gamma \ln(z) (z - T(x))^{-1} T' (z- T(x))^{-1} dz \, . $$

a good reference is the book of Kato Perturbation theory for linear operators.

Answer 3

First, let us do not care about convergence issues, then we know \begin{equation} \exp(\log(A(t)))=A(t) \end{equation} Take the derivative with respect to both sides we have \begin{equation} \exp(\log(A(t))) (\log(A(t)))'=A'(t) \end{equation} So we have \begin{equation} (\log(A(t)))'=A^{-1}(t)~A'(t) \end{equation} modulo convergence issues.