Simplify Vector of Traces Let $A_0,\dots, A_k, B \in \mathbb{R}^{n\times n}$.
Can the following expression be simplified?
$$\begin{pmatrix}\mathsf{tr}(A_0B^t)\\
\mathsf{tr}(A_1B^t)\\
\vdots\\
\mathsf{tr}(A_kB^t)
\end{pmatrix}.$$
For example, it seems plausible one can pull out the matrix $B^t$ via appropriately appealing to the tensor product.
Can this be done?
 A: I'm not sure how you want to simplify this but if you index $A$ and $B$ like $A_{kij}$ and $B_{ij}$ then you are really looking to do this:
$$C_k=\sum_{i,j}A_{kij}B_{ij}$$
This is not a tensor product, unless $A$ can be written as a tensor but this would require that $k=n$, which I am not sure it is.
A: $A, B\in M_{m×n}(\Bbb{R})$ with $A=(a_{ij}),B=(b_{ij})$, Hadamard product $A\circ B=(c_{ij})$ where $c_{ij}=a_{ij}\cdot b_{ij}$
Now $\operatorname{tr}(AB^T):= $ Sum of all entries in the Hadamard product of $A\circ B$ $= \sum_{i, j} c_{ij}$

Let $A_{l}=(a_{ij}^{l})_{n×n}$ and $B=(b_{ij}) $ . Then $\operatorname{tr}({A_{l}B^T}) =\sum_{i, j}a_{ij}^{l}b_{ij}$ $,\space \forall 0\le l\le k$
$\begin{align}\begin{pmatrix}\operatorname{tr}(A_0B^t)\\
\operatorname{tr}(A_1B^t)\\
\vdots\\
\operatorname{tr}(A_kB^t)
\end{pmatrix}&=
\begin{pmatrix}\sum_{ij}a^{0}_{ij}b_{ij}\\\sum_{ij}a^{1}_{ij}b_{ij}\\\vdots\\\sum_{ij}a^{k}_{ij}b_{ij}
\end{pmatrix}_{(k+1)×1}\\&=\sum_{i=1}^{n}\sum_{j=1}^{n} b_{ij} \begin{pmatrix}a^{0}_{ij}\\a^1_{ij}\\\vdots\\a^k_{ij}\end{pmatrix}
\end{align}$
