If $\det{e^A}$ is maximized, also is $\det{A}$?

Question

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 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 $\det{A} = \prod_i \lambda_i$.

Answer 1

Render

$$ A = \pmatrix{-1 & 0\cr 0 & b}$$

Then $\det(e^A)=e^{b-1}$ is maximized as $b\to+\infty$ but $\det(A)=-b$ is maximized as $b\to-\infty$. The claim is as wrong as could be in this case.

The determinant of $e^A$ is maximized when the trace of $A$ is.

Answer 2

By Jacobi’s formula, $$\det(e^A) = e^{\mathrm{tr}(A)}$$

So if you want to maximise $\det(e^A)$, you want to be maximising the trace of A (and vice-versa), not the determinant of A. These are not the same.

So in your case, you have a maximised trace, $\mathrm{tr}(A) = \sum_i \lambda_i$.

Answer 3

No. For example, try

$$ A = \pmatrix{2 & 0\cr 0 & 8}, B = \pmatrix{1 & 0 \cr 0 & 10}$$

$\det(A) = 16 > \det(B) = 10$, but $\det(\exp(A) = e^{10}) < \det(\exp(B)) = e^{11}$.