linear algebra - Geometric Interpretation of Trace

What seems really odd to me is this limitation set by the original question.

The divergence application of trace is somewhat interesting, but again, not really what we are looking for.

Maybe that is rejected because it involves a metric tensor in most textbooks about differential geometry, but the divergence requires only an affine connection, even in differential geometry. In flat Cartesian space (without a norm or inner product), it's even simpler.

First consider that matrices have two main applications, as the components of linear maps and as the components of bilinear forms. Let's ignore the bilinear forms. Linear maps are really where matrices come from because matrix multiplication corresponds to composition of linear maps.

We know that the determinant is the coefficient of the characteristic polynomial at one end of the polynomial, and the trace is at the other end, as the coefficient of the linear term. So we should think in terms of linearization and volume, or some combination of these two concepts. We know that the determinant can be interpreted as the relative volume expansion of the map $xmapsto Ax$. So we should think in terms of maybe linearizing this in some way.

Define a velocity vector field $V(x)=Ax$ on $mathbb{R}^n$ and integrate the flow for a short time. What happens to the volume of any region? The rate of increase of volume equals $mathrm{Tr}(A)$. This is because the integral curves have the form $x(t)=exp(At)x(0)$. (See Jacobi's formula.)

Thus the determinant tells you the volume multiplier for a map with coefficient matrix $A$, whereas the trace tells you the multiplier for a map whose rate of expansion has component matrix $A$.

That sounds very neat and simple to me, but only if you avoid the formulas in the DG literature which try to interpret divergence in terms of absolute volume by referring to a metric tensor or inner product.

PS. To avoid analysis, to keep it completely algebraic apart from the geometric meaning of the determinant, consider the family of transformations $x(t)=x(0)+tAx(0)$ for $tinmathbb{R}$ for all $x(0)inmathbb{R}^n$. Then the volume of a figure (such as a cube) is a polynomial function of $t$. The linear coefficient of this polynomial with respect to $t$ is $mathrm{Tr}(A)$. There are no derivatives, integrals or exponentials here. The trace also happens to be the linear component of the characteristic polynomial. I think this is a pretty close tie-up.

PS 2. I forgot to mention that the divergence of the field $V(x)=Ax$ is $textrm{div} V=mathrm{Tr}(A)$. Therefore trace equals divergence. That's the geometrical significance of the trace. The function $V$ is the linear map with coefficient matrix $A$. And the trace equals its divergence if it is thought of as a vector field rather than just a linear map. You could even write $mathrm{Tr}(A)=mathrm{div}(A)$ if you identify the matrix with the corresponding linear map.