Tensorial direct product The direct product of two tensors is also a tensor. I would like to know if we can write a tensor as a product of only two other tensors. For example, how to find $A^{\mu}$ and $ B^{\nu}$ so that $\eta^{\mu\nu}= A^{\mu}B^{\nu}$?
 A: This is not always possible. For example, the tensor
$$
\delta^{ij} = \begin{cases}1\quad\text{when}\,i=j\\0\quad\text{otherwise}\end{cases}
$$
cannot be written as $A^iB^j$ which we can see from the fact that the matrix of $\delta^{ij}$ is the identity matrix which is full rank while the matrix of $A^iB^j$ is rank one.
Minkowski metric $\eta^{\mu\nu}$ is also full rank and hence there do not exist $A^\mu$ and $B^\nu$ such that $\eta^{\mu\nu} = A^\mu B^\nu$.
In general, you need more than one term of the form $A^iB^j$ to represent a given tensor $C^{ij}$
$$
C^{ij}=\sum_{k=1}^r A_k^iB_k^j
$$
where $r$ is the rank of $C^{ij}$ and $A_k$ and $B_k$ are tensors. This is a restatement of the singular value decomposition for matrices and has a generalization, called tensor rank decomposition, to tensors with arbitrary number of indices.
A: The product $A^\mu B^\nu$ is a tensor (assuming  $A^\mu$ and $B^\nu$ are tensors).
However, for a general tensor like $Q^{\mu\nu}$,
no, $Q^{\mu\nu}$ cannot be generally written as product like $A^\mu B^\nu$.

However, it may written as a sum of such products
$Q^{\mu\nu}=\sum_i A_i^\mu\ B_i^\nu$.
For example, if $Q^{\mu\nu}$ is anti-symmetric, then it cannot be written as $A^\mu B^\nu$... but would require something like
$A^\mu B^\nu - B^\mu A^\nu$.
