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 $M$ 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, i.e., the family of matrices $I + ϵ A$ , the trace is the derivative of the rate of volumes generated by these matrices, since from (1):

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

It is not as strange as it can seem. For example, in the typical Youtube videos they don't use a single transformation matrix M, but a perturbation of the identity, to illustrate how the transformation is done. If you keep track of the resulting volumes, and you derive with respect to time, what you obtain is the trace, not of M, but of the matrix that let you go from the identity towards M