Let $A ∈ M(n,\mathbb{C}$).
(i) Let $f(t) = det(e^{tA})$ and $g(t) = e^{t \cdot tr(A)}$. Show that $f(t)$ and $g(t)$ both solve the first order DEQ $u'= tr(A)u$.
(ii) Using (i), show that $\det(exp(A)) = exp(tr(A))$.
Hint: $\det(1 + εA)= 1 + εtr(A) + O(ε^2)$
Solution: to show that $g(t)$ is indeed a solution is straightforward. But how can someone show that $f(t)$ is also a solution by using the linear algebra result from the hint? Since $f(t)$ and $g(t)$ satisfy the same initial conditions the answer to (ii) is obvious.
Use the definition $$ e^{tA}=\lim_{N\to\infty}\left(I+\frac{t}{N}A\right)^N. $$ Generally this limit has bad convergence, is not practically useful. But here it works nicely together with the determinant. Then apply the hint for $ε=\frac{t}{N}$ small enough $$ \det(I+εA)=\prod_{i=1}^n(1+εA_{ii})+O(ε^2)=1+ε\sum_{i=1}^n A_{ii}+O(ε^2). $$ The determinant expansion has the diagonal product, and any other product term contains at least 2 off-diagonal elements, which are proportional to $ε$.