Can the metric tensor be treated as a linear transformation? In general relativity, the metric tensor $g$ is a covariant, second rank, symmetric tensor that can be written down as a 4x4 matrix. The metric tensor generalizes the notion of distance between points and angle between vectors in Cartesian coordinates to arbitrary coordinates whose axes may not be orthonormal.
Each row of the metric tensor describes how a coordinate axis transforms. The identity metric tensor reconstructs the standard Cartesian coordinates. Can the metric tensor then be treated as a linear, not necessarily affine, transformation of a set of coordinate axes? It basically describes how the Cartesian coordinate axes are modified to form a new basis?
Note: I do not come from a physics background. If my wording is not exact, please correct me.
 A: The metric can be thought of as a linear map between two vector spaces, but they're not the same vector space.  Rather, we can think of the metric as a map from $V \to V^*$, the dual space of the original vector space $V$. If you've never encountered a dual space $V^*$ before, it is defined as the set of all linear maps from a vector space $V \to \mathbb{R}$.  (Or to $\mathbb{C}$ if we're talking about complex vector spaces, as in quantum mechanics.)  It is not too hard to see that the space of all maps $V \to \mathbb{R}$ is itself a vector space:  there is a zero map, the sum of two maps is also a map, we have a notion of scalar multiplication, etc.  So $V^*$, so defined, is a vector space.
The metric $g$ is defined as a bilinear function from $V \times V \to \mathbb{R}$.  But the presence of a metric also defines a correspondence between $V$ and $V^*$, as follows:  for any element $v \in V$, define the corresponding element $v^* \in V^*$ as the map $v^*(w) \equiv g(v,w)$.  It is not too hard to see that $v^* \in V^*$;  since $g$ is bilinear, it is a linear map from $V \to \mathbb{R}$.
What's more, this correspondence map is in fact a linear isomorphism between $V$ and $V^*$:

*

*The correspondence map is linear, since the map $(\alpha v)^*$ maps $w \to g(\alpha v, w) = \alpha g(v,w)$, and so it is equal to the map $\alpha (v^*)$.  Similarly, it is not hard to show that $(u + v)^* = u^* + v^*$.

*The correspondence map is one-to-one, since if we have two elements $u,v \in V$ such that $u^* = v^*$ then we must have $g(u, w) = g(v,w)$ for all $w \in V$, meaning that $g(u - v, w) = 0$ for all $w$.  This then implies that $u = v$ because of the non-degeneracy of the metric.

*The correspondence map is onto.  This proof is a little trickier, but the easiest way to do it is to define a basis of $n$ vectors $\{v_\mu\}$ for $V$ (where $n = \dim V$), define a set of $n$ dual vectors $\{{v^*}^\nu\}$ by the relation ${v^*}^\nu(v_\mu) = \delta^\nu {}_\mu$, and then show that these vectors form a basis for $V^*$, establishing that $\dim V = \dim V^*$.

So you can think about the metric as a linear isomorphism from one basis to another, but the catch is that the bases in question belong to two different vector spaces.  In particular, it maps a coordinate basis for $V$) to a corresponding coordinate basis for $V^*$.
