Suppose $A_{\mu \nu}$ is a (0,2)-tensor and $B^\mu$ is a vector. Show that $A_{\mu\nu}B^\mu$ is a covector 
Problem: Suppose $A_{\mu \nu}$ is a (0,2)-tensor and $B^\mu$ is a vector. Show that $A_{\mu\nu}B^\mu$ is a covector.

I am confused, the other day they taught me that $A_{\mu\nu}B^\mu$ is in fact NOT a covector, but the components of a covector. They also taught me that, if $A = A_{\alpha \beta} dx^\alpha dx^\beta$ is a (0,2) tensor and $B=B^\rho \partial_\rho$ is a vector, then the product is a covector. This is roughly how, here is my attempt:
$$
AB = A_{\alpha \beta} dx^\alpha dx^\beta B^\rho \partial_\rho = 
A_{\alpha \beta} B^\rho \frac{\partial x^\alpha}{\partial x^\rho}  dx^\beta
= A_{\alpha \beta} B^\rho \delta^\alpha_\rho  dx^\beta = \underbrace{A_{\alpha \beta} B^\alpha}_{\equiv C_\beta}  dx^\beta = C_\beta d x^\beta
$$
Is a covector by definition because it is a map from the tanget space to the real numbers.
What is wrong with my solution? What are they looking for when they ask me to show that $A_{\mu\nu}B^\mu$ is a covector?
 A: Copied from a comment: 
I believe your confusion comes from the fact that physicists like to talk about $V_\mu$ as being the covector $V$ as opposed to just one of its components. This is a practice that expedites calculations but one does sacrifice clarity for the sake of simplicity. So, what they are asking you to do is show that $B^{\mu}A_{\mu\nu}\equiv C_\nu$ are the components of a covector (which you've already done). 
You are finished. 
Bravo.
I hope this helped!
A: Let $e_{\lambda}$ be a basis of a vector space $V$ and $dx^{\nu}$ the corresponding dual basis in $V^*$, namely $dx^{\nu}(e_{\lambda})\colon V^*\times V\to \mathbb{R} =\delta_{\lambda}^{\mu}$.


*

*$A_{\mu\nu}$ is shorthand for $A_{\mu\nu}\,dx^{\mu}\otimes dx^{\nu}$.

*$B^{\lambda}$ is shorthand for $B^{\lambda}\,e_{\lambda}$.


$A_{\mu\nu}B^{\mu}$ is shorthand for 
$$
A_{\mu\nu}\,dx^{\mu}\otimes dx^{\nu}(B^{\lambda}\,e_{\lambda}) = A_{\mu\nu} B^{\lambda}\ dx^{\mu}\otimes dx^{\nu}(e_{\lambda}) = A_{\mu\nu} B^{\lambda}\ dx^{\mu}\delta^{\nu}_{\lambda} = A_{\mu\nu} B^{\nu}\ dx^{\mu}
$$
which is by definition a covector.
