How to determine if a tensor is covariant or contravariant? In special relativity, the coordenates of a event are in general written using a 4-vector: $$x^{\mu} = \binom{ct}{\textbf{x}}$$ where $\textbf{x} = (x,y,z)$ are the spacial coordenates. This is a contravariant vector, and its covariant representation is written as $$x_{\mu} = \binom{ct}{-\textbf{x}}.$$ However, I know that the contravariant and covariant vectors are defined by the way that their components transforms. If $A$ is contravariant and $B$ is covariant:  $$A'^{\mu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}} A^{\nu}, \ \ \ \ B' _{\mu} = \frac{\partial x^{\mu}}{\partial x'^{\nu}} B_{\nu}.$$ Using this definition, how can I show that $x_{\mu}$ is in fact covariant? 
 A: As a side issue, coordinates don't really act like vectors. Infinitesimal changes in the coordinates do. It makes a difference when spacetime isn't flat or when you use coordinates that aren't Minkowski. But this is not really relevant to your question.
Consider the coordinate transformation $(t,x)\mapsto(t',x')=(\alpha t,\alpha x)$, where $\alpha$ is a positive constant. This is a change of units. By comparing this with the rules for transforming vectors and covectors, you'll see that the pair of coordinates acts like a vector, not a covector.
If the metric was given by the line element $dt^2-dx^2$ in the original coordinates, then it's given by $\alpha^{-2}dt'^2-\alpha^{-2}dx'^2$ in the new coordinates. Then lowering an index gives the components $(\alpha^{-1}t',\alpha^{-1}x')$ for the covector in the primed coordinate system. You can now check that this behaves according to the rule for transforming covectors.
A: Simply by looking what it's a function of. If it's a multilinear function of vectors than it's covariant. If it's a function of Co vectors(linear functionals) it's contravariant. 
