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3Blue1Brown has great visual explanations of what some commonly used properties of matrices mean, such as determinant, rank, and kernel, but I have absolutely no idea what the trace of a matrix has in linear algebra and what its visual representation (if any) is.

Rodrigo de Azevedo
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    For real matrices, determinants represent volumes of parallelepipeds, and the trace is roughly speaking, the derivative at the identity of the determinant. So in other words, one geometric interpretation of the trace is it is closely related to the rate of change of volume of parallelepipeds. – peek-a-boo Jun 03 '23 at 16:33
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    Cf. https://mathoverflow.net/questions/13526/geometric-interpretation-of-trace – Travis Willse Jun 03 '23 at 17:29
  • It may not be a "visual" explanation, but the trace of $A$ is exactly the sum of the eigenvalues of $A$. This is important, see here. – Dietrich Burde Jun 03 '23 at 17:54
  • @peek-a-boo do you refer to $\frac{\mathrm{d}}{\mathrm{d}M}\det(M)$ at the identity? I assume a Fréchet derivative. So, the linear operator that stands for the derivative would just be the trace, meaning $\det(M)-1-\operatorname{tr}(M-I)$ is $o(|M-I|)$ as $M\to I$, which is not clear to me. I know you said ‘roughly speaking’, I’m just curious what the precise formulation is – FShrike Jun 03 '23 at 22:42
  • @FShrike I mean the Frechet derivative of $\det$ at the identity equals the trace functional: $D(\det){I}(\cdot)=\text{trace}(\cdot)$. If you know how to differentiate multilinear functionals (google for the proof of smoothness) you’ll easily see that $D(\det)_I(\xi)=\sum{i=1}^n\det(e_1,\dots, e_{i-1},\xi_i,e_{i+1},\dots, e_n)=\sum_{i=1}^n(\xi_i)_i=\text{trace}(\xi)$, where I used column operations to evaluate the determinant. A more linear-algebraic proof (reference) can be found in the links above, or in the second half of my answer here. – peek-a-boo Jun 04 '23 at 04:16

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The trace of the infinitesimal strain tensor, $\boldsymbol{\varepsilon},$ in the theory of elasticity represents a fractional volume change in the elastic material when it is deformed, so that $$\dfrac{V_{\rm deformed}-V_{\rm initial}}{V_{\rm initial}}= \varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}, $$ and minus one third of the trace of the Cauchy stress tensor, $\boldsymbol{\sigma},$ represents the isotropic compressive pressure in the material, $$p=-\dfrac{\sigma_{11}+\sigma_{22}+\sigma_{33}}{3}.$$

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An example is the Frobenius norm of an operator in a square matrix representation. The trace does not depend on the base of the matrix representation (is invariant). If $A$ is the matrix (or the linear operator), then $\|A\| = \sqrt{\mathrm{tr}(A^* A)}$. And The "operator norm of a linear operator $T:V->W$ is the largest value by which $T$ stretches an element of $V$". There are also, of course, other examples or interpretations of the trace of a matrix.

Pablo
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