Definition of non-degenerate metric tensor We know that a metric has a property which is called non-degeneracy. I was searching for what does that mean and saw it associated with the fact that $det(g_{\mu\nu})\neq0$. How does this relate to that?
 A: If $\text{det } g = 0$, then $\text{ker }  g \neq \{\vec{0}\}$, ie there is some vector $X \in \text{ker } g$,  such that $g(X,\ast) $ gives zero 1-form, so $g(X,Y)=0 $ for any $Y$.
A: 1- A degenerate matrix is a matrix whose rank is smaller than its dimension.
2- A singular (non-invertible) matrix is one that has a vanishing determinant.
Equivalence of the two : A matrix whose rank is smaller than it's dimension when diagonalized will have at least one zero eigenvalue, and consequently a vanishing determinant.
A: related: https://math.stackexchange.com/q/160882/224026
see also: http://en.wikipedia.org/wiki/Metric_tensor:
" From the coordinate-independent point of view, a metric tensor is defined to be a nondegenerate symmetric bilinear form on each tangent space that varies smoothly from point to point. "
A: That $\det g \neq 0$ can be used as the definition of what it means for a metric $g$ to be non-degenerate. As noticed in another answer, an alternative definition is that $g$ has full rank. These two definitions are equivalent by the rank-nullity theorem.
I don't agree with the "when diagonalized" part of the accepted answer (I can't comment unfortunately). The matrix $$\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$$ has rank smaller than its dimension but cannot be diagonalized. It does have a zero eigenvalue so it is not invertible. I believe the relevant result to quote here is the rank-nullity theorem: the rank plus the dimension of the kernel equals the size of the matrix. So when a matrix is degenerate, the kernel is not empty, so there is a zero eigenvalue, so the matrix is not invertible, and vice versa.
