I want to know that if the equation in the title always holds? I have generated a random Hermitian matrix $A$, and then compute $\log(\det A)$ and $ tr(\log(A))$ in matlab, it is not equal. So I'm really puzzled.
Can you give me the proof of this equation if it is true. Thanks in advance
I just show an example:
$$A=\begin{bmatrix}30.9186 + 0.0000i & -0.5120 - 0.0197i \\-0.5120 + 0.0197i & 10.3822 + 0.0000i \end{bmatrix}$$
$K>> log2(det(A))$
$ans = 8.3253 + 0.0000i$
$K>> trace(log2(A))$
$ans =8.3264 + 0.0000i$
what's wrong about it?
it is essentially different from these problem such as
How to prove $\det(e^A) = e^{\operatorname{tr}(A)}$?
because I think this equation is totally wrong, so I don't know why it can be proved.
If $A$ is hermitian $>0$, then your formula is valid is we choose the principal logarithm $(\log(re^{i\theta})=\log(r)+i\theta$ where $\theta\in (-\pi,\pi)$).
If $A$ is invertible hermitian, your formula is -in general- not valid; example
$A=-I_2$. The required equality is (*) $\log(1)=2\log(-1)$.
If we choose the following $\log$: $\log(re^{i\theta})=\log(r)+i\theta$ where $\theta\in (-\pi+1/10,\pi+1/10)$, then $\log(1)=0,\log(-1)=i\pi$ and (*) is not satisfied.
EDIT. About your edit, you are not serious.
If $spectrum(A)=(\lambda_i)$, then $spectrum(e^A)=(e^{\lambda_i})$. Therefore, $\det(e^A)=\Pi_i e^{\lambda_i}$ and $e^{tr(A)}=e^{\Sigma_i \lambda_i}$. I hope that you are convinced...