expansion of logarithmic matrix Let $A,B$ be Hermitian matrices ($A$ has positive or zero eigenvalues and $Tr A=Tr[A+\lambda B]=1$), and $\lambda$ is infinitesimal constant. How to expand
\begin{equation}  
\ln(A+\lambda B)
\end{equation}
and 
\begin{equation}  
Tr[(A+\lambda B)\ln(A+\lambda B)]
\end{equation}
into powers of $\lambda$ as usually do in Taylor expansion of functions?
 A: First note the derivative of the scalar function 
$$\eqalign{
 f(x)&=x\log x\cr
f^\prime=\frac{df}{dx}&=1+\log x
}$$
Next, define a new matrix variable 
$$\eqalign{
 M(\lambda) &= A+B\lambda \cr
dM &= B\,d\lambda\cr
}$$
Then use this result for the differential of the trace of a matrix function
$$\eqalign{
 d\operatorname{tr}f(M) &= f^\prime(M^T):dM
}$$where colon denotes the Frobenius product, i.e. $\,\,A:B=\operatorname{tr}(A^TB)$

Now to adress your second question, let $$T(\lambda)=\operatorname{tr}(M\log M)=\operatorname{tr}(f(M))$$ and use the preceeding to find its gradient $g(\lambda)$
$$\eqalign{
 dT &= f^\prime(M^T):dM \cr
  &=(I+\log M)^T:B\,d\lambda \cr
  &= \operatorname{tr}(B(I+\log M))\,d\lambda \cr\cr
g(\lambda) &= \frac{\partial T}{\partial\lambda} = \operatorname{tr}(B+B\log M) \cr\cr
}$$
At $\lambda=0$, we can evaluate each of these quantities 
$$\eqalign{
M(0) &= A \cr
T(0) &= \operatorname{tr}(A\log A) \cr
g(0) &= \operatorname{tr}(B+B\log A) \cr\cr
}$$
Finally, we can expand $T(\lambda)$ about $\lambda=0$
$$\eqalign{
T(\lambda) &= T(0) + \lambda g(0) \cr
  &\approx \operatorname{tr}(A\log A) + \lambda\operatorname{tr}(B+B\log A) \cr\cr\cr
}$$
As for your first question, you could apply the block-triangular method of Kenney and Laub 
$$\eqalign{
L = \log\Bigg(\begin{bmatrix}A&B\\0&A\end{bmatrix}\Bigg) =  \begin{bmatrix}\log A&\frac{d\log M}{d\lambda}|_{\lambda=0}\\0&\log A\end{bmatrix} \cr\cr
}$$
and then 
$$\eqalign{
 \log(M) &= \log(A+\lambda B) \cr
 &\approx \log(A) + \,\lambda\,\,\begin{bmatrix}I&0\end{bmatrix}L\begin{bmatrix}0\\I\end{bmatrix} \cr
}$$
A: Stephen L Adler wrote a nice set of notes, entitled Taylor Expansion and Derivative Formulas for Matrix Logarithms.
He shows (at a physicist level of rigor) that
$$\ln (A + t B) = \ln(A) + t \int_0^\infty \frac{1}{A+z}B\frac{1}{A+z}dz+\mathcal O(t^2).$$
The starting point is the matrix expansion:
$$\frac{1}{A+tB}=\frac{1}{A}-t\frac{1}{A}B\frac{1}{A}+t^2\frac{1}{A}B\frac{1}{A}B\frac{1}{A}-\dots$$
which can be verified by multiplying both sides by $A+tB$ and expanding.
A second important identity is
$$\int_0^u \frac{1}{X+z}dz=\ln(X+u)-\ln(X).$$
For us, matrix $X$ could be either $A$ or $A+tB$. Expanding these inverse matrices under the integral, then manipulating the resulting logarithms, then finally taking the limit $u \rightarrow \infty$, gives the desired result for $\ln (A+tB)$ to all orders in $t$.
