How can one visualize a real, symmetric $3 \times 3$ tensor with zero trace? I am looking for the simplest way to visualize a real, symmetric 3x3 tensor than has vanishing trace. (All entries are real numbers.)
It cannot be an ellipsoid, because an ellipsoid has three positive eigenvalues, and thus its trace is positive, and not vanishing.
Is there something equally simple for the case of zero trace?
(Zero trace means that the sum of all diagonal elements is zero.)
P.S. Also asked, without success, in https://math.stackexchange.com/q/4577043
 A: Here is 1 point of view. The Lie group $G=O(3)$ of orthogonal transformations in 3D acts via similarity transformation/conjugation on the 5-dimensional subspace $V$ of traceless symmetric $3\times 3$ real matrices.
This may be identified with a change of the tensor components under a change of orthonormal basis.
In fact $V$ furnishes an irreducible spin-2 representation. E.g. $V$ can be decomposed into eigenspaces for the angular momentum $L_z$.
A: I can only come up with a visualization for a 3×3 skew-symmetric matrix.
$$ \begin{bmatrix}
0 & -c & b \\ c & 0 & -a \\ -b & a & 0 
\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$
Corresponds to a line through the origin along the $(a,b,c)$ vector since
$$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = t\, \begin{bmatrix} a \\ b \\ c \end{bmatrix}$$
trivially solves the above equation.
To make the line go though a specific point $(u,v,w)$ not at the origin, then use the equation
$$ \begin{bmatrix}
0 & -c & b \\ c & 0 & -a \\ -b & a & 0 
\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} b w - c v \\ c u - a w \\ a v - b u \end{bmatrix} $$

So I tried to find a nifty solution to the 3×3 symmetric matrix
$$ \begin{bmatrix}
0 & c & b \\ c & 0 & a \\ b & a & 0 
\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$
for $(x,y,z)$ with a geometric interpretation and did find anything. Only $(x,y,z)=0$ is a solution which isn't helpful.

Moving into homogeneous coordinates with $(x,y,z,1)$ you can define a surface with
$$ \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix}^\intercal \begin{bmatrix}
0 & c & b & 0 \\ c & 0 & a & 0  \\ b & a & 0 & 0 \\ 0 & 0 & 0 & -1 
\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1 \end{bmatrix} = 0$$
produces a surface with equation
$$ a (y z) + b ( x z ) + c ( x y ) = \tfrac{1}{2} $$
this is a 3D hyperbolic surface since the $z=0$ slice is the curve $x y = \tfrac{1}{2 c}$
