If $\alpha_{ij} A^i B^j = 0$ and $A^i$ and $B^j$ are arbitrary vectors, then prove that $\alpha_{ij}= 0$. This problem appeared in my Differential Geometry class, the professor explained the problem by, first taking an arbitrary vector and demonstrating that $\alpha_{ii} = 0$. and then proceeded to demonstrate that, $A^l = B^m = 1, (1\leqq l \leq n, 1\leqq m \leqq n, l\neq m)$. I get the proof somewhat. Can any of you elucidate it or give an alternative proof?

Fix, $h,k$, then if you take
$$
\begin{aligned}
A^i=\begin{cases}
1, && i=h,\\
0, && i\neq h,
\end{cases}
\end{aligned}
\qquad
\begin{aligned}
B^j=\begin{cases}
1, && j=k,\\
0, && j\neq k
\end{cases}
\end{aligned}
$$
then
$$
\alpha_{ij}A^iB^j=0\implies\alpha_{hk}=0,
$$
for the arbitrariness of $h,k,$ this is true for all $h,k.$
