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?

If $\alpha_{ij}A^iB^j = 0$ is zero for arbitrary $(A^i)$ and $(B^j)$, then take $A^i = \delta^i_k$ and $B^j = \delta^j_\ell$. So $0 = \alpha_{ij} \delta^i_k \delta^j_\ell = \alpha_{k\ell}$. Since $k$ and $\ell$ were arbitrary, we're done.
