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: After obtaining a matrix representation of a $(0,2)$ (or $(2,0)$) tensor $g$ on a vector space $V$ using a basis , yes, you can of course consider the determinant/trace/eigenvalues/eigenvectors of the matrix. However, these notions are not basis-free, i.e if you consider a different basis, and look at the corresponding matrix, you'll get different results for these quantities. This is because if you take bases $\beta_1,\beta_2$ for $V$, then the matrix representations $[g]_{\beta_1}$ and $[g]_{\beta_2}$ are related by congruence: $[g]_{\beta_2}=Q^t[g]_{\beta_1}Q$. This is in sharp contrast to what we have for endomorphisms $T:V\to V$, where if you take two bases $\beta_1,\beta_2$, then the matrix representations are related by similarity: $[T]_{\beta_2}=P\cdot[T]_{\beta_1}\cdot P^{-1}$. The transpose vs inverse makes all the difference.
The fact that in the case of endomorphisms (linear maps $V\to V$) the matrix representations are related by similarity allows us to conclude that (using that determinant of product is product of determinants, and the cyclic property of trace) the determinant and trace are the same regardless of the choice of basis. This is why we can write abstractly $\det T$ and $\text{trace}(T)$ of the linear transformation itself. However, we cannot write $\det g$ and $\text{trace}(g)$; we have to always specify which basis we're referring to, eg $\det([g]_{\beta})$ or $\text{trace}([g]_{\beta})$.
A: 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.
