# For linear $A: V \to V$ strictly positive definite, does there exist linear $B: V \to V$ strictly positive definite such that $e^B = A$?

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.

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You're right, the statement is wrong. There is a self-adjoint logarithm of $A$, but it will strictly positive definite if and only if all eigenvalues of $A$ are greater than $1$.