The other Coefficient of the Characteristic Polynomial of a 3x3 Matrix What's the geometric significance of $l_1l_2+l_1l_3+l_2l_3$ where $l_i,i=1,2,3$ are the eigenvalues of a 3x3 matrix? Is there a formula that depends only on the entries? It would be nice if there was something similar for higher dimensions and the other coefficients which aren't the determinant or trace.
(Note: Now that I think about it, I don't have a working intuition for the trace either; the trace is the divergence of the linear map viewed as a smooth vector field, but that doesn't mean anything to me, vector calculus courses handwave that it somehow measures, for infinitesimal spherical regions, the total "flow" that enters or leaves the region, but usually don't prove it or even state it properly.)
Thanks.
 A: Unfortunately this will all be covered by traces of different induced linear maps.
Let $A:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be a linear map. (You might aswell think of $A\in\text{Mat}_n(\mathbb{R})$.)
Let us define a polynomial
\begin{align} p_A(t):=\det(I+tA)=a_0+a_1t+\ldots+ a_nt^n\end{align}
which is closely related to the characteristic polynomial $\chi_A$ by the formula $\chi_A(t)=p_A(-t)$. Now we have to remember how the determinant was defined. It is the unique multilinear map $\det:\mathbb{R}^n\rightarrow \mathbb{R}$ which is alternating in its arguments and normalized for the standard basis. In that way it measures changes of volume enclosed by a set of $n$ linearly independent vectors - a paralleliped. The pull-back $A^* $ of $\det$ by $A$ defined by (and yields)
\begin{align}
(A^* \det)(v_1,\ldots,v_n):=\det(Av_1,\ldots,Av_n)=\det(A)\det(v_1,\ldots,v_n)
\end{align}
is a map that takes a map (like $\det$) measuring volume of a paralleliped and assigns it again to a map measuring volume, but after 'stretching' the paralleliped by $A$.
This can be extended to so called $k$-forms, that are $k-$multilinear, alternating maps. In other ways one can interprete these as volume forms like $\det$ but for parallelipeds consisting of $k$ linearly independent vectors. We will call these (induced) maps $A^*_k$.
One can show now that the coefficients $a_k$ defined at the top can be computed explicitly via
\begin{align}
a_k=\text{tr}(A^*_k),
\end{align}
that is the trace of the pull-backs on the corresponding space of $k$-forms. How does it fit into our 'classical' understanding? The space of $n$-forms on a vector space is one dimensional, so $\det$ is a basis of it. By the definition of pull-back we saw, that $\det$ is an eigenform of $A^* _n$ with eigenvalue $\det(A)$. Then tr$(A^* _n)=\det(A)$, which is the sum of all eigenvalues of $A^* _n$.
Returning to your initial question and the significance of $l_1l_2+l_1l_3+l_2l_3$: Let $v_1,v_2,v_3$ be the eigenvectors associated to these eigenvalues and let $V_i$ be the subspace of $\mathbb{R}^3$ spanned by $v_j$, $j\neq i$ (just to ease notation). Then associated to each $V_i$ there is a $2$-form $\omega_i$ acting on it 'like a two-dimensional $\det$'. (Formally: $\omega_1:=v^2\wedge v^3$, where $v^i$ is the co-vector dual to $v_i$ and $\wedge$ is the wedge-product.) Then
\begin{align}
(A^* _2 \omega_1)(v_2,v_3)=\omega_1(Av_2,Av_3)=l_2l_3\omega_1(v_2,v_3),
\end{align}
that is $\omega_1$ is an eigenform with eigenvalue $l_2l_3$. The same holds for $\omega_i$, $i=2,3$ which all together form a basis of the $2$-forms on $\mathbb{R}^3$ and hence the trace sums over all these eigenvalues and gives us tr$(A^*_2)=l_1l_2+l_1l_3+l_2l_3$. One may interpret the trace here as total change of volume of $2$-parallelipeds in $\mathbb{R}^3$, where 'total' refers to all directions.
Especially all $a_k$ will be sums over all (ordered) products of $k$ different eigenvalues $l_i$.
P.S.: From here one can return to the coefficients of the characteristic polynomial by the formula stated at the beginning. Thus $a_k$ changes its sign if $k$ is uneven.
