The logarithm of a symmetric positive definite matrix as a function Can the logarithm of a symmetric,positive definite matrix always be expressed locally as a power series? That is: Is the logarithm function analytic on the space of symmetric, positive definite matrices?
 A: Yes. If you fix the usual branch (i.e., remove the segment $(-\infty,0]$, the function $z\longmapsto \log z$ is analytic. That doesn't mean that there is a power series expression that expresses $\log z$ globally, because a power series is analytic on disks, and $\log z$ cannot be analytic on disks around $0$. Locally, however (on disks that do not contain $0$), $\log z$ can be expressed as a power series, as any analytic function can be. 
A: Proposition. Let $S^+$ be the set of symmetric $>0$ $n\times n$ matrices. Then $A\in S^+\rightarrow \log(A)$ is real anlytic.
Using the user1952009's post, we can do the job. Yet, we can also give the following simpler proof. Note that an essential tool is the fact that $(*)$ the spectrum of $A=[a_{i,j}]$ is a continuous function of the $(a_{i,j})$.
Proof. Note that if $A=Pdiag((\lambda_i)_i)P^{-1}$ where $\lambda_1\geq \lambda_2\geq \cdots\geq \lambda_n>0$, then $\log(A)=Pdiag((\log(\lambda_i))_i)P^{-1}$ and that this definition does not depend on the chosen invertible matrix $P$. 
We consider $a>0$ and the open set (cf. $(*)$): $Z=\{X\in S^+| spectrum(X)\subset (0,2a)\}$. We show that there exists a power series for $\log(X)$ that is valid for any $X\in Z$. When $X\in Z$, $X=a(I+U)$ where $||U||_2=\rho(U)<1$. It is easy to see that $\log(X)=\log(a)I_n+\log(I+U)=\log(a)I_n+U-\dfrac{1}{2}U^2+\cdots$; since this series converges for $||U||_2<1$, we are done.
Considering the hermitian $>0$ matrices, the $\log(.)$ is again a real analytic function. Note that it is not a complex analytic function.
