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?
 A: $A^iB^j$ are the component of the tensor product of two vector $\mathbf{A}\otimes\mathbf{B}$. Among all these tensors there are also the tensors
$\mathbf{e}_i\otimes\mathbf{e}_j,$ where $B=\{\mathbf{e}_1,\ldots,\mathbf{e}_n\}$ is a base of the vector space $V,$ and $B'=\{\mathbf{e}_i\otimes\mathbf{e}_j,\ i,j=1,\ldots,n\}$ is a base of the space of rank $2$ tensors over $V.$
$\alpha_{i,j}A^iB^j$ is the inner product of the tensor $\boldsymbol\alpha$ and the tensor $\mathbf{A}\otimes\mathbf{B}$
$$
\alpha_{i,j}A^iB^j=\boldsymbol\alpha:(\mathbf{A}\otimes\mathbf{B})
$$
You know that if the inner product of a vector for each element of a base vanish, then the vector is the null vector. This is true also for the inner product vector space of tensors.
A: 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 $h,k.$
A: 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.
