Determining whether a space is really three or two dimensional? A space purports to be three dimensional with the metric 
$$dl^2=dx^2+dy^2+dz^2-\left(\frac{3}{13}dx+\frac{4}{13}dy+\frac{12}{13}dz\right)^2$$
How can I show that it actually represents a two dimensional space?
Comments:
I tried diagonalizing it to see if it had a zero eigenvalue, for then it would imply that there exists a basis in which the representation of the metric tensor is really a 2X2 matrix i.e to show that there exists such a coordinate transformation which makes it a 2X2 matrix.
 A: It is not a space-time because it is not Lorentzian. It is actually Riemannian. This exercise may be from a general relativity book, but is in fact a geometry question. So I take it that the question is to show that it represents a two dimensional space. But since it is in the general relativity tag one can be smart and guess the following.
Consider the vector $n^i=\langle \frac3{13}, \frac4{13}, \frac{12}{13}\rangle$. It is a unit vector in a three dimensional Euclidian space. The given metric can be written as
$$g_{ij}=\delta_{ij}-n_in_j$$
where $\delta_{ij}$ is the usual Euclidan metric.
This shows that the given metric is the induced metric on the orthogonal to $n^i$ subspace.
A: Having a zero column in a diagonalization is bad (since the metric would be degenerate), but also bad would be if somehow it looked like $$dl^2=dx^2+dy^2+dz^2$$ or $$dl^2=-dx^2-dy^2-dz^2.$$ S you also want to avoid the metric being positive definite or negative definite.  For more dimensions you'd also want to worry about having two spatial directions and two time directions!
So you want your metric to have a signature +--- or -+++ for a real spacetime, and for a lower number of dimensions, +-- or -++.
A: Expand your line element and obtain the metric $g_{ij}$.
It is of the form $$g_{ij}=\delta_{ij}-n_in_j$$ where $n=\langle \frac3{13}, \frac4{13}, \frac{12}{13}\rangle$ and so $n_in^i=1$ What you have now is a projection operator (because $g_{ij}g_{jk} = \delta_{ik}$, check it symbolically) which does this:
It takes any 3D vector $v$ and gives you its vector component along the plane perpendicular to the unit vector $n$, so it spans the 2D plane orthogonal to $n$
Proof:$$g_{ij}v^j=(\delta_{ij}-n_in_j)v^j=v_i-n_i(n_jv^j)$$ and the RHS is simply the vector $v$ minus its component along $n$, so you get the component orthogonal to $n$.
So this metric projects a 3D vector onto a plane.
In GR, these kinds of "degenerate metrics" are generally used when splitting space-time as a 3+1 foliation for solving initial value problems.
