Could the linear algebra concepts regarding $n\times n$ matrices be used on any second order tensors? The metric tensor is actuality is a $(2,0)$ tensor define on a manifold, but in all practical use I've seen, it appears no different than a regular square matrix. In sense that it is just a collection of numbers of size $n\times n$.
It feels a bit perverse, but could one define thing such trace, determinant, eigen vector, rank etc on this object after viewing it in that way? In this interpretation would it obey all the laws of linear algebra such as determinant being independent of basis and such?
Related, also related

A metric tensor acts like an inner product (or bilinear form) at each point. We use it to compare vector fields and eventually to develop the connection operator which is important in the theory of curvature etc. 
The associated matrix simply gathers the effects it has on pairs of basis fields. For example:
$$g_{ij}|_p = g(\partial_i,\partial_j)\big|_p.$$
Studying the coefficients of the characteristic poly for this matrix can of course be done. However this matrix is not applied to one vector as in the eigenvalue problem. It appears as in $\langle v,w\rangle = v^tAw$ instead of as in $Av$. I've never seen a metric tensor with degeneracies (not full rank). This would be problematic, I imagine.  I recommend John Lee's Introduction to Smooth Manifolds and his follow up text on Curvature for more. Also Michael Artin's Algebra for discussion on Bilinear Forms.
