Firstly, I'm but a mere physicist, so please be gentle :-) I want to explicitly show that the derivative of the (natural) logaritm of a general $n \times n$ (diagonalizable) matrix $X(x)$ w.r.t. $x$ is

$$\frac{\text{d}}{\text{d}x}\Big(\ln{\left[X(x)\right]}\Big) = X'(x)X^{-1}$$

where $X'(x)$ is the derivative of $X$ w.r.t. $x$.

I'm going about this in a similar way to how I would prove it for $X$ being just a scalar function of $x$, meaning I start from the definition of the derivative

$$ \newcommand{\D}[2]{\frac{\text{d}#1}{\text{d}#2}} \D{}{x}\Big(\ln{[X(x)]}\Big) = \lim_{\Delta x\rightarrow 0}{\frac{\ln{[X+\Delta X]}-\ln{X}}{\Delta x}} $$

where I rewrite $\Delta X = X'\Delta x$:

$$ \newcommand{\D}[2]{\frac{\text{d}#1}{\text{d}#2}} \D{}{x}\Big(\ln{[X(x)]}\Big) = \lim_{\Delta x\rightarrow 0}{\frac{\ln{[X+X'\Delta x]}-\ln{X}}{\Delta x}} $$

The idea is then to use some logarithm properties to get $e$ out of it$^1$:

$$\newcommand{\D}[2]{\frac{\text{d}#1}{\text{d}#2}} \D{}{x}\Big(\ln{[X(x)]}\Big) = \lim_{\Delta x\rightarrow 0}{\frac{1}{\Delta x}\Big(\ln{[XX^{-1}+X'X^{-1}\Delta x]}\Big)} \\ \D{}{x}\Big(\ln{[X(x)]}\Big) = \lim_{\Delta x\rightarrow 0}{\frac{1}{\Delta x}\Big(\ln{[\mathbb{I}+X'X^{-1}\Delta x]}\Big)} \\ \D{}{x}\Big(\ln{[X(x)]}\Big) = \lim_{\Delta x\rightarrow 0}{\ln{\left[\left(\mathbb{I}+X'X^{-1}\Delta x\right)^{\frac{1}{\Delta x}}\right]}} \\ \D{}{x}\Big(\ln{[X(x)]}\Big) = \lim_{U\rightarrow 0}{\ln{\left[\left(\mathbb{I}+U\right)^{X'X^{-1}U^{-1}}\right]}} \\ \D{}{x}\Big(\ln{[X(x)]}\Big) = X'X^{-1}\lim_{U\rightarrow 0}{\ln{\left[\left(\mathbb{I}+U\right)^{U^{-1}}\right]}} \\ \D{}{x}\Big(\ln{[X(x)]}\Big) = X'X^{-1}\lim_{U\rightarrow 0}{\ln{e}} \\ \D{}{x}\Big(\ln{[X(x)]}\Big) = X'X^{-1} $$

But I'm not at all convinced about all my steps there. Furthermore, I used the logarithm property $\ln{A}-\ln{B} = \ln{AB^{-1}}$ which only holds if $A$ and $B$ commute. I suppose in the limit of $\Delta x$ approaching zero, $\Delta X=X'\Delta x$ and $X^{-1}$ would commute (and $X$ and $X^{-1}$ always do), but I'd like to find out what a mathematician thinks of this.

Lastly I want to add that if I just assume the definition of the matrix logarithm as a power series$^2$,

$$\ln{X} = -\sum_{k=1}^{\infty}{\frac{1}{k}(\mathbb{I}-X)^k},$$

and then differentiate this series, I exactly find $X^{-1}X'$. Again the assumption has to be made, however, that $X$ and $\Delta X$ commute inside a limit.

So my question is: am I right to feel a bit sketchy about my attempt at an explicit proof for the derivative of the matrix logarithm? And can we generally assume $X$ and $\Delta X$ commute when the limit of small $\Delta X$ is to be taken?

If anyone feels particularly inclined, I was also wondering if the power series I've taken as the definition of the matrix logarithm above is indeed the definition and if so, why that one is chosen. Is it purely in analogy to the Taylor expansion of $\ln{x}$? If this would be better asked as a separate question, I'll go ahead and do that.

There's a fair amount of related questions on here already, but they haven't allowed me to figure out the answers to my questions in a way that I'm 100% sure I understand.

$^1$ By the way, can anyone tell me why the align-environment doesn't work on here? It works just fine for me on Physics.SE .

$^2$ Can anyone confirm that this series converges if $\max_{i}{|1-\lambda_i|} < 1$ ?

  • 1
    $\begingroup$ Hi, fellow mere physicist here - in fact, last did physics a long time ago. I just wanted to recommend two books that I made frequent use of in my career. They deal with issues like those you are considering and are really valuable. By chance they are available online, but I believe one should pay for such things - this is just a convenient hyper-reference: math.bme.hu/~balint/oktatas/fun/notes/Reed_Simon_Vol1.pdf , poncelet.sciences.univ-metz.fr/~gnc/bibliographie/… $\endgroup$ Mar 23, 2014 at 14:10
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    $\begingroup$ (1) is $X(x)$ an $n\times n$ matrix? (2) Is $X(x)$ Hermitian, or normal? $\endgroup$ Mar 23, 2014 at 14:13
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    $\begingroup$ @JasonZimba Thanks for the references! In my particular case $X(x)$ is a general (square) diagonalizable matrix. $\endgroup$
    – Wouter
    Mar 23, 2014 at 14:20

3 Answers 3


A simple expression can be derived by manipulating the Taylor series $\ln X = \sum_{n=1}^\infty -\frac{(-1)^n}{n}(X-1)^n$ with the result $$\frac{d}{ds}\ln X(s) = \int_0^1 \frac{1}{1-t\,(1-X(s))} X'(s) \frac{1}{1-t\,(1-X(s))}\, dt\ .$$ While not in closed form, this formula can be easily computed numerically, for example. In the above expressions, 1 is the unit matrix.

To derive: $$\frac{d}{ds}\ln X(s) = -\sum_{n=1}^\infty \frac{(-1)^n}{n}\sum_{a=0}^{n-1}(X-1)^a X' (X-1)^{n-1-a}\\ =-\sum_{a=0}^\infty \sum_{n=a+1}^\infty \frac{(-1)^n}{n}(X-1)^a X' (X-1)^{n-1-a}\\ = -\sum_{a=0}^\infty\sum_{b=0}^\infty\frac{(-1)^{a+b+1}}{a+b+1}(X-1)^a X' (X-1)^{b}\\ = \sum_{a=0}^\infty\sum_{b=0}^\infty \int_0^1 dt\, t^{a+b}(1-X)^a X' (1-X)^{b}\ . $$ On performing the sums over $a$ and $b$ one gets the formula stated above.

  • $\begingroup$ This is really cool! Since the derivative of the exponential has a similar expression, do you know of any standard reference for this kind of manipulations? $\endgroup$ Sep 7, 2020 at 0:15
  • $\begingroup$ No, sorry, I don't have a reference; I just derived this one myself, inspired by the one for exponentials. $\endgroup$
    – dbkaplan
    Sep 8, 2020 at 1:42

You can write $d\log X = dX\,X^{-1}$ if and only if $X$ and $dX$ commute. In that case, of course: $$ dX\,X^{-1} = X^{-1}dX. $$

In the general case they do not commute, and there is no simple rule for the derivative of the logarithm. Even though the expressions $dX\,X^{-1} $ and $X^{-1}dX$ are called "logarithmic derivatives", as they share some properties with the actual derivatives of the logarithm, they are not.

The reason behind this is that, for general matrices: $$ e^A\,dA\ne d(e^A) \ne dA\,e^A, $$

unless $A$ and $dA$ commute. This can be seen from the definition by the Taylor series: $$ d(e^A) = d \left( 1 + A + \frac{1}{2}A^2 +\dots \right) = 0 + dA + \frac{1}{2}A\,dA + \frac{1}{2}dA\,A +... $$

which is not equal to: $$ dA + dA\,A +...= dA (1+A+...) = dA\,e^A, $$

because $\frac{1}{2}(dA\,A+A\,dA)\ne dA\,A$ in general.

You might feel that if $dA$ is "small", then the commutator is "small". That is a dangerous assumption, the truth is that the commutator is the same order as $dA$, so it matters.

  • $\begingroup$ Interesting, would $\text{d}\log{X} = \text{d}X X^{-1}$ hold if $X$ were a diagonal matrix? If not, is there any other particular property that $X$ must have for this to hold? And would I be right to say that the definition in terms of a Taylor series is the fundamental one for the matrix exponential and the matrix logarithm? $\endgroup$
    – Wouter
    Mar 23, 2014 at 15:04
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    $\begingroup$ Well it depends on what you mean by "diagonal". Intuitively, if $A$ and $dA$ do not commute, what happens is that $A(x)$ does not commute with $A+dA=A(x+dx)$. So if $A$ is diagonal at $x$, it is not diagonal at $x+dx$, because non-commuting matrices are not simultaneously diagonalizable. Viceversa, if $A$ is everywhere diagonal, then of course it commutes with itself at different points, and it commutes with $dA$. For many, the Taylor series is the fundamental definition. It depends on the point of view. For sure, it is very important. $\endgroup$
    – geodude
    Mar 23, 2014 at 15:12
  • $\begingroup$ Hmm, in that case I'll probably have to ask another question because I'm trying to prove $\delta \det{X} = (\det{X}) \mathrm{Tr}\,(\delta M M^{-1})$. A friend asked me about this and I told him I had proved it in the context of a course on general relativity. But when I went back and looked at that proof, I noticed some of these subtleties that I seem to have brushed over when I originally wrote down the proof. $\endgroup$
    – Wouter
    Mar 23, 2014 at 15:28
  • $\begingroup$ @Wouter I'm trying to prove the exact same thing. Have you made any progress as to what assumptions are required about the matrix $M$? $\endgroup$
    – balu
    Nov 7, 2016 at 12:21
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    $\begingroup$ Wouter, @balu you probably know the proof by know, but for reference this is known as Jacobi's formula, which holds for any matrix. The proof follows essentially from the definition of the determinant, and the computation of the matrix inverse from the adjugate (see for example Wikipedia) $\endgroup$ Jul 4, 2018 at 3:51

Since $X$ is diagonalizable $$\eqalign{ \def\b{\beta} \def\M{M^{-1}} \def\BR#1{\Big[#1\Big]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\diag#1{\op{diag}\LR{#1}} \def\Diag#1{\op{Diag}\LR{#1}} \def\qiq{\quad\implies\quad} X &= MB\M \qiq B = \Diag{\b_k} \\ }$$ you can use the $\sf Daleckii$-$\sf Krein$ Theorem $$\eqalign{ F &= f(X) \\ dF &= M\,\BR{R\odot\LR{\M\,dX\:M}}\,\M \\ R_{jk} &= \begin{cases} {\large\frac{f(\b_j)\,-\,f(\b_k)}{\b_j\,-\,\b_k}}\qquad{\rm if}\;\b_j\ne\b_k \\ \\ \quad{\small f'(\b_k)}\qquad\qquad{\rm otherwise} \\ \end{cases} \\ }$$ where the derivatives wrt $x$ have been abbreviated as $$\eqalign{ dF \equiv \frac{dF}{dx}\qquad\quad dX \equiv \frac{dX}{dx} }$$ $\odot$ denotes the Hadamard product, and $f(x)$ can be any function such that
$f(\b_k)\,$ and $f'(\b_k)$ are defined.

In this particular case, $f(x)=\log(x)$ and $f'(x)=x^{-1}\,$ so none of the eigenvalues
can be zero (i.e. the matrix must be invertible).


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