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?

• Note that $x^\mu$ only behaves like a vector if the coordinate transformations are Lorentz transforms (or more generally linear transformations). – jacob1729 May 24 at 14:30
• If it eats covariant vectors (and spits out scalars),it's contravariant. If it eats contravariant vectors, it's covariant. – WillO May 24 at 14:30
• The covariant vector should be a row vector. – Jon May 24 at 14:56

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.