Can eigenvalues be expressed in terms of geometric relations? Let $A$ be a $3$ by 3 matrix with each row being a unit vector in
the unit sphere of $\mathbb{R}^{3}$, then can the eigenvalues of $AA^{t}$ be expressed
in terms of some geometric relations or geometric quantities among the row vectors of $A$
in the unit sphere? Thanks a lot.
 A: The term  "geometric'' usually refers to something that is independent of the Euclidean coordinates.  The rows of $A$ are not geometric  in this sense.
Here is a  dynamical interpretation of the eigenvalues. This is geometric in the sense that it is independent of coordinates. I'll assume $A$ is generic. Then the surface $E=(AA^tx,x)=1$ $\newcommand{\bR}{\mathbb{R}}$ is an ellipsoid in $\bR^3$.
Think of the  sphere $\Sigma_t=\{\Vert x\Vert=t\}$ as  a moving wave front generated by a ping! at the origin, at time $t=0$. This  wave front will become tangent to $E$ at three moments of time $t_1<t_2< t_3$. The eigenvalues of $AA^t$ are $t_1^{-2}, t_2^{-2}, t_3^{-2}$.
A: In addition to the case in the comment (orthogonal rows $\Leftrightarrow$ unit eigenvalues), another simple case is where the three row vectors lie in a plane. The angle between the $i$th-row and the $j$-th row is $\theta_{ij}$. One eigenvalue is zero, the other two are
$$\tfrac{3}{2} \pm\sqrt{ \cos^2 \theta_{12}+ \cos^2  \theta_{13}+ \cos^2  \theta_{23}-\tfrac{3}{4}}.$$
The fully general case has expressions that are very much more complicated.
