I want to see how
$$\text{trace} [\log (A^{\otimes n})]$$
scales with $n$, where $A$ is a $d\times d$ positive definite matrix and
$$A^{\otimes n}=\underbrace{A\otimes A \otimes \cdots \otimes A}_{n\; \text{times}} $$
My feeling is that $\text{trace} [\log (A^{\otimes n})]$ should scale polynomial in $n$ because of the logarithm, but if I write it in terms of eigenvalues of $\log (A^{\otimes n})$, since $\log (A^{\otimes n})$ is a $d^n\times d^n$ matrix, we will have $d^n$ terms, i.e.
$$\text{trace} [\log (A^{\otimes n})]=\sum_{i=1}^{d^n}\lambda_i$$
$\lambda_i$s are eigenvalues of $\log (A^{\otimes n})$. Thus, it scales exponential in $n$.
Am I making a mistake here?
You have $$ \text{trace}\,(\log(A^{\otimes n}))=\log(\det(A^{\otimes n}))=\log((\det A)^n)=n\log\det A=n\,\text{trace}\,(\log A). $$