Relation between the determinant and the trace

See @arnold1992ordinary section 16.3. The determinant and the trace of a matrix has the following relation.

Let $A : R^{n} \to R^{n}$ be a linear operator, and let $ϵ \in R .$ Then, as $ϵ \to 0$ ,

\begin{matrix} (1) & det (I + ϵ A) = 1 + ϵ tr (A) + O (ϵ^{2}) . \end{matrix}

Also, from the point of view of the matrix exponential,

det e^{A} = e^{tr (A)} .

Conclusion: if the matrix $A$ represents an invertible linear map, the determinant is a kind of rate of volumes, see determinant. And if the matrix $A$ represents a direction of perturbation of the identity matrix, the trace is the derivative of the rate of volumes through this direction of perturbation, since from (1):

tr (A) \approx \frac{det (I + ϵ A) - det (I)}{ϵ}