for a matrix we know that $$\det(aA)=a^n \det(A) $$
but what happens for an INFINITE dimensional operator ?? should we have
$$\det(aA)=a^{Z(0)}\det(A) $$
$$ Z(s)= \sum_{n=0}^\infty \lambda_n^{-s} $$
since $ \sum_{n=0}^\infty 1 =Z(0) $ and the zeta function is taken over all the eigenvalues of "$ A $
You have the right intuition, though one does need to be a bit careful to make things rigorous. In the infinite dimensional context, when possible, one defines $$ \det(A) = e^{-\zeta_A^\prime(0)}, $$ where $$ \zeta_A(s) = \operatorname{tr}A^{-s} $$ has been extended to $0$ by analytic continuation. In particular, then, $$ \zeta_{aA}(s) = \operatorname{tr}(aA)^{-s} = a^{-s}\operatorname{tr}A^{-s} = a^{-s}\zeta_A(s), $$ so that $$ \zeta_{aA}^\prime(s) = -\log(a)a^{-s}\zeta_A(s) + a^{-s}\zeta_A^\prime(s), $$ and hence $$ \zeta_{aA}^\prime(0) = -\log(a)\zeta_A(0) + \zeta_A^\prime(0), $$ yielding $$ \det(aA) = e^{-\zeta_{aA}^\prime(0)} = e^{\log(a)\zeta_A(0)-\zeta_A^\prime(0)} = a^{\zeta_A(0)}\det(A). $$ Note, however, that one does need $\zeta_A$ to admit analytic continuation to $0$ and $a >0$.
