In an exam today about mathematical methods of physics, we were asked to prove the following statement:
Let $V$ be a complex finite dimensional inner product space. Let $A: V \to V$ be a strictly positive definite linear map. Show that there exists a unique strictly positive definite linear map $B: V \to V$ such that $e^B = A$.
(I'm not 100% sure about the exact statement, but this should be it)
Note that in our course, positive definite or strictly positive definite also means that the map must be self-adjoint (since non-real eigenvalues can't be "positive" or "not positive"). I wanted to prove the statement using the spectral theorem, however, when I got to the part to show that $B$ is strictly positive definite too, I ran into a problem.
Let $A = \mathrm{Id}_V$ denote the identity. $A$ is self-adjoint and strictly positive definite, however, I think that the only map $B$ satisfying $e^B = A$ would be the zero map, which clearly is not a strictly positive definite map. For $V = \mathbb{C}$, the statement claims that there exists a strictly positive real number $x$ such that $e^x = 1$, which cannot be true.
Question: Did I go wrong with my reasoning at some point or is the claim that $B$ is strictly positive definite too strong? I tried proving this taking the logarithm of the eigenvalues of $A$ which can also be a negative number.
Thanks in advance for any clarification.
