I have an expression:
$$ W = \sum_{k,k'}(\mathbf{q}\otimes\mathbf{q})(\gamma^k\gamma^{k'}) $$
And $\gamma$ is:
$$ \gamma^{k} = \mathbf{f^{k}} \cdot \mathbf{Ae} $$
Basically, expression W is a product of two dot products. $\mathbf{A}$ and $\mathbf{f^{k}}$ are constants with respect to $\mathbf{e}$. $\mathbf{q}$ is a 3x3 tensor, $\mathbf{f^{k}}$ is a 6x1 vector, $\mathbf{A}$ is a 6x6 tensor, and $\mathbf{e}$ is a 6x1 vector.
The goal is to extract the $\mathbf{e}$ terms into a form of:
$$ W = \sum_{k,k'}(\mathbf{q}\otimes\mathbf{q})...expression(\mathbf{f^{k,k'}},\mathbf{A})...(\mathbf{e}\otimes\mathbf{e}) $$
I looked up the tensor and dot product identities and I couldn't find a way to derive it. The only approach that made sense to me is using Einstein notation, and I would like to ask if someone can check my work:
The $\gamma$ terms can be written as:
$$ \gamma\gamma=\delta_{ki}f_{k}A_{ij}e_{j}\delta_{nl}f_{n}A_{lm}e_{n} $$
$$ \gamma\gamma=\delta_{ki}\delta_{nl}f_{k}A_{ij}f_{n}A_{lm}e_{j}e_{n} $$
$$ \gamma\gamma=\delta_{ki}\delta_{nl}f_{k}A_{ij}f_{n}A_{lm} (e \otimes e) $$
Then I am stuck at how to resolve the Kroneckers back into dot products.