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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.

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    $\begingroup$ Do you know a formula for the determinant or trace, maybe in terms of eigenvalues? $\endgroup$
    – Stan Tendijck
    Commented Feb 28, 2019 at 12:13
  • $\begingroup$ yes, but I don't know how to proof this equation $\endgroup$
    – LinTIna
    Commented Feb 28, 2019 at 12:16
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    $\begingroup$ First you have to ensure that the expression $\log(A)$ makes sense (this is not the case for all hermitian matrices). $\endgroup$
    – Alpha001
    Commented Feb 28, 2019 at 12:17
  • $\begingroup$ Continuing the @Alpha001 comment: To answer this, you need a definition of $\log(A)$. So begin there. $\endgroup$
    – GEdgar
    Commented Feb 28, 2019 at 12:25
  • $\begingroup$ I just add an example, do you know what's wrong with it $\endgroup$
    – LinTIna
    Commented Feb 28, 2019 at 12:25

1 Answer 1

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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...

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