What is the interpretation of the eigenvectors of the jacobian matrix? I'm trying to think about the jacobian matrix as a abstract linear map.
What is the interpretation of the eigenvalues and eigenvectors of the jacobian?
 A: Suppose that $F:M\rightarrow N$ is a smooth mapping of Riemannian manifolds, with $F(p)=q$.  Let $<\ , \ >_p$ and $<\ , \ >_q$ denote the metrics in $T_pM$ and $T_qN$.
We can define a bilinear pairing on $T_pM$ by $B(\vec{v},\vec{w})=<DF_p(\vec{v},DF_p(\vec{w})>_q$.  Assuming we used orthonormal bases at $T_pM$ and $T_qN$ to express the matrix for $DF_p$, this is the same as $<DF_p^tDF_p(\vec{v}),\vec{w}>q$.  The matrix of the pairing with respect to the basis at $T_pM$ will be the matrix $DF_p^tDF_p$.  By Lagrange's theorem we can find an orthogonal matrix $P$ so that $P^t(DF_p^tDF)P$ is diagonal. The diagonal entries of that matrix are the eigenvalues of $DF_p^tDF_p$.  They tell you how much the mapping $F$ is distorting distances near $p$, as you map into $N$.
It isn't exactly the eigenvalues of $DF_p$, but you could say things about the eigenvalues of $DF_p^tDF_p$ in terms of the eigenvalues of $DF_p$. 
Admittedly it's not pithy and aha.  But it is an understanding of how the eigenvalues of the matrix $DF_p$ are reflected in the local geometry.
A: What Frank Science has said in the question comment above is right. I'm simply expanding on his comment here:
Since the Jacobian has eigenvectors, it is square i.e. the input and output space have same dimensions. If there is some kind of natural interpretation to the input and output basis and they can be mapped to each other, the eigenvectors of the Jacobian represent those directions in the input space, where if you move locally (small amount), you move in the same direction in the output space. This interpretation is no different from that of the eigenvectors of any matrix. The only addition is that of local (small) movement, because the Jacobian approximates the original function locally.
