Logarithm of a positive-definite matrix If $z$ is a nonzero complex number, then we can write it in the form $z=e^w$ where $w$ is another complex number which is not unique. For example we can do $z=re^{i\theta}=e^{\ln(r)+i(\theta+2k\pi)}$, where $k\in \mathbb{Z}$. So in this case we have multiple complex logarithms $w_k=\ln(r)+i(\theta+2k\pi)$. Now if $z$ is a positive real number $z=r>0$, it still has multiple complex logarithms given by $\ln (r)+i(2k\pi)$, but only one of these complex logarithms is real (which corresponds to $k=0$). Negative numbers do not have a real logarithm, for example the complex logarithms of $-1$ are $i(\pi+2k\pi)$, none of them is real.
If $A$ is an invertible complex matrix, then there exists a matrix $B$ such that $A=e^B$, where $e^B$ is the matrix exponential. As in complex numbers, the matrix $B$ is of course not unique. Now if we assume that $A$ is Hermitian  positive definite, then  we can write it in the form $A=P^*DP$ where $D=diag(\lambda_1,\dots\lambda_n)$ and $P$ is unitary. Since $A$ is Hermitian  positive definite, its eigenvalues $\lambda_1,\dots\lambda_n$ are positive and hence $D=e^{D'}$ where $D'=diag(\ln\lambda_1,\dots, \ln\lambda_n)$. So this way we can write
$$A=P^*DP=P^*e^{D'}P=e^{P^*{D'}P}$$
So $P^*{D'}P$ is a Hermitian logarithm of the positive definite matrix $A$. Is this the only Hermitian logarithm?
 A: To prove unicity, it is simpler to reformulate the problem with Hermitian endomorphisms of some Hermitian space $E$. This enables to give an intrinsic definition of the logarithm of a definite positive Hermitian endomorphism.
Is $u \in H^{++}(E)$, $v \in H(E)$ and $e^v=u$, then $u$ and $v$ commute. For every eigenvalue $\lambda$ of $u$, the eigensubspace $E_\lambda = \mathrm{Ker}(u-\lambda)$ is stable by $u$ and the endomorphism $v_{E_\lambda}$ induced by $v$ on $E_\lambda$ is diagonalisable since $v$ is. Given any  eigenvalue $\mu$ of $v_{E_\lambda}$, $e^\mu$ is an eigenvalue of $\exp(v_{E_\lambda}) = u_{E_\lambda} = \lambda \mathrm{Id}_{E_\lambda}$, so $\mu =. \ln \lambda$. Hence $v_{E_\lambda} = \ln \lambda \mathrm{Id}_{E_\lambda}$
As a result,
$$v = \sum_{\lambda \in \mathrm{Sp}(u)} \ln \lambda \times p_\lambda,$$
where $p_\lambda$ denotes the orthogonal projection on $p_{E_\lambda}$ (remind that the eigensubspaces are pairwise orthogonal).
Conversely, if one defines $v$ by the last equality above,
then $e^v = u$ since these two endomorphisms coincide on each eigensubspace $E_\lambda$.
