I understand the proof of the identity in the title for $A$ Hermitian. One uses that any Hermitian matrix can be diagonalized as $A = X \Lambda X^{-1}$, such that $$ \det{A} = \prod_i \lambda_i, $$ and we have $$ \exp(Tr(\log(A)) = \exp(Tr(X\log\Lambda X^ {-1}) = \exp(\sum_i\log(\lambda_i)) = \prod_i \lambda_i. $$

However, is it possible to show the identity for $A$ not Hermitian? My motivation for this question is that in physics the identity is often used without it being clear that $A$ is Hermitian.

  • $\begingroup$ You do need to assume $A$ is nonsingular, else $\ln(A)$ doesn't make sense. $\endgroup$ – Robert Israel Oct 19 '15 at 17:24

If $\ln(A) = B$, the identity says

$$ \det(\exp(B)) = \exp(\text{Tr}(B)) $$

which is more usual form for this identity, true for all $n \times n$ matrices $B$ over $\mathbb C$ (avoiding questions about whether $\ln(A)$ is defined, and which of the possible logarithms to use).

One way to do this is to show it first for diagonalizable matrices $B$, then use the fact that diagonalizable matrices are dense and both sides of the equation are continuous functions of $B$.

A second way is to use Jordan canonical form.

A third way is to note that both $\det(\exp(tB))$ and $\exp(t \text{Tr}(B))$ satisfy the differential equation $y' = \text{Tr}(B) y$ with initial value $y(0) = 1$.


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