If i have a colection of matrices $\{A\_1, A\_2,...,A\_n\}$, where $A\_k$ is an positive defite matrix.

Choosing an matrix $A$ that maximizes $\det{e^{A}}$, also maximizes $\det{A}$?

Refered to [this question](https://math.stackexchange.com/questions/118579/eigenvalues-of-a-matrix-a-and-ea)
If we rewrite $A$ in its Jordan normal form, we have: $A=PJP^{-1}$.

Then $e^A=Pe^JP^{-1}$. Thus the eigenvalues of $A$ are $\{\lambda\_i\}$ and the eigenvalues of $e^A$ are $\{e^{\lambda\_i}\}$.

Since $\det{e^{A}}=\prod\_i e^{\lambda\_i}$, can i say that maximizing this is the same of maximizing its eigenvalues?

If so, by maximizing $\{e^{\lambda\_i}\}$ also maximizes $\{\lambda\_i\}$ and consequently $\det{A} = \prod\_i \lambda\_i$.