Tensor as outer product This is a problem I am trying to solve and need help with. 

Given a $  \left( \begin{array}{} 0 \\ 2 \end{array} \right)$ tensor h such that h$(\quad ;A)=\alpha $h$(\quad ;B)$ for any two vectors $\vec A$ and $\vec B$ (where $\vec B \neq 0)$ where $\alpha$ is a number which may depend on $\vec A$ and $\vec B$. Show that there exits two one-forms such that h$=\tilde p\otimes \tilde q$

My attempt: I take vector $\vec V$ as input and I get $$h_{ij}V^iA^j=\alpha h_{ij}V^iB^j$$ for a general $\vec A $ and $\vec B$. Since  $\vec A $ and $\vec B$ are arbitrary if we take $\vec A = \vec B = \hat e_i$ then,
$$h_{ij}V^i=\alpha h_{ij}V^i$$ 
This seems to imply that $\alpha=1$ and I haven't arrived at the required result. Please help. I don't know if I have to use (Lorentz) metric anywhere here.
 A: 
I don't know if I have to use (Lorentz) metric anywhere here.

No you don't. This question is wholly about linearity and the definitions of tensors.
Choose a basis $\{\hat{e}_j\}_{j=0}^N$ where my index runs from nought to $N$ to be in keeping with standard notation in relativity, but that is the only link: this question is general.
Our tensor $h(\cdot,\,\cdot)$ will be wholly specified by the values $h_{j\,k}=h(\hat{e}_j,\,\hat{e}_k)$ since values for all other vectors follow by bilinearity of $h(\cdot,\,\cdot)$ and linear superposition of these values.
So now, by our basic assumption: 
$$\begin{array}{c}
h_{j\,1} = h(\hat{e}_j,\,\hat{e}_1) = \alpha_1\,h(\hat{e}_j,\,\hat{e}_0);\;j=0\cdots N\\
h_{j\,2} = h(\hat{e}_j,\,\hat{e}_2) = \alpha_2\,h(\hat{e}_j,\,\hat{e}_0);\;j=0\cdots N\\
\vdots\\
h_{j\,N} = h(\hat{e}_j,\,\hat{e}_N) = \alpha_N\,h(\hat{e}_j,\,\hat{e}_0);\;j=0\cdots N\\
\end{array}$$
for some constant numbers $\alpha_1,\,\alpha_2,\,\cdots,\,\alpha_N$. Writing $\alpha_0\stackrel{def}{=}1$ and $\beta_j = h(\hat{e}_j,\,\hat{e}_0)$ we now have:
$$h_{j\,k}=h(\hat{e}_j,\,\hat{e}_k)=\beta_j\,\alpha_k$$
and you should be able to go from here.
