I recently learned about the symmetric logarithmic derivative $L_\rho(A)$, defined implicitly by $$i\left[\rho,A\right]=\frac 1 2\left\{\rho,L_\rho(A)\right\}.$$ Here $\rho$ is a Hermitian, positive semi-definite matrix and $A$ is also Hermitian. This quantity is important in quantum metrology.
On the wikipedia page it is mentioned that this operator is linear: \begin{align} L_\rho(\mu A)&=\mu L_\rho(A)\\ L_\rho(A+B)&=L_\rho(A)+L_\rho(B) \end{align}
This last fact confuses me. As a generalized version of the logarithmic derivative (is it?), I should expect it to behave at least somewhat like the ordinary logarithmic derivative, which has scale invariance and which does not distribute: \begin{align} \left[\frac{d}{dx}\ln\right](af(x))&=\frac{f'(x)}{f(x)}\\ \left[\frac{d}{dx}\ln\right](f(x)+g(x))&=\frac{f'(x)+g'(x)}{f(x)+g(x)}\neq\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)} \end{align} So why should we expect linearity to hold the symmetric logarithmic derivative instead of scale invariance and the product Le
The way I've seen the symmetric logarithmic derivative defined is as follows.
Let $\rho(\lambda)$ be a continuously parameterized family of states. Then the logarithmic derivative of $\rho$ is defined as the operator $L_\rho$ that satisfies $$ \frac{L_\rho\rho+\rho L_\rho}{2} = \frac{d\rho}{d\lambda}. $$ If $\rho$ is just a one-by-one matrix, then $$ L_\rho = \frac{1}{\rho}\frac{d\rho}{d\lambda}=\frac{\rho'}{\rho}, $$ which is the regular logarithmic derivative. That's where the name comes from.
If specifically $\rho(\lambda)=e^{-i\lambda A}\rho e^{i\lambda A}$, we get the definition you have. It no longer has the same properties that the classical logarithmic derivative has.