I'm struggling to simplify the last step of a $(p,q)$ tensor and how its components change with a linear change of basis on the associated vector space. So far I have:
Given a vector space $V$ over some field $F$ with bases $\mathcal{B}=(e_{1}, \dots, e_{n})$ and $\mathcal{B}'=(\overline{e}_{1}, \dots, \overline{e}_{n})$ which are related linearly by a matrix $A=(a_{j}^{i})$
Now let $\alpha$ be a $(p,q)$ tensor on $V$
$$\begin{align} \alpha = \sum_{{i_{\beta}, j_{\gamma}}}{\alpha_{k_{1}, \dots, k_{p}}^{l_{1}, \dots, l_{q}} e^{i_{1}} \otimes \cdots \otimes e^{i_{p}} \otimes e_{j_{1}} \otimes \cdots \otimes e_{j_{q}}} \\ \alpha = \sum_{{i_{\beta}, j_{\gamma}}}{\overline{\alpha}_{k_{1}, \dots, k_{p}}^{l_{1}, \dots, l_{q}} \overline{e}^{i_{1}} \otimes \cdots \otimes \overline{e}^{i_{p}} \otimes \overline{e}_{j_{1}} \otimes \cdots \otimes \overline{e}_{j_{q}}} \end{align}$$
From here, I'm trying to find the components of $\alpha$ with respect to the new basis $\mathcal{B}'$:
$$ \overline{\alpha}_{i_{1}, \dots, i_{p}}^{j_{1}, \dots, j_{q}} = \alpha(\overline{e}_{1}, \dots, \overline{e}_{i_{p}}, \overline{e}^{j_{1}}, \dots, \overline{e}^{j_{q}}) $$
$$ =\sum_{l_{{\gamma}, k_{\beta}}}{\alpha^{l_{1}\dots l_{q}}_{k_{1} \dots k_{p}} e^{i_{1}} \otimes e^{i_{1}} \cdots \otimes e^{i_{p}} \otimes e_{j_{1}} \otimes \cdots \otimes e_{j_{q}}} \left(\sum_{\mu}{a_{i_{1}}^{\mu}e_{\mu}}, \dots, \sum_{\mu}{a_{i_{p}}^{\mu}e_{\mu}}, \sum_{\nu}{b^{j_{1}}_{\nu}e^{\nu}}, \dots, \sum_{\nu}{b^{j_{q}}_{\nu}e^{\nu}} \right) $$
I'm not sure how to simplify this, any help would be appreciated.
As recommended I consider a $(1,1)$-tensor with summation convention: $$ \boldsymbol{\alpha}={\alpha^i}_j\; e_i\otimes e^j={\overline{\alpha}^i}_j\; \overline{e}_i\otimes \overline{e}^j\,. $$ Using $$ \overline{e}_j={a^\mu}_j\;e_\mu\,,\quad \overline{e}^i={b_\nu}^i\;e^\nu $$ we get \begin{align} {\overline{\alpha}^i}_j\stackrel{!}=\boldsymbol{\alpha}(\overline{e}^i,\overline{e}_j)= \boldsymbol{\alpha}\Big({b_\nu}^i\;e^\nu,{a^\mu}_j\;e_\mu\Big)= {b_\nu}^i\;{a^\mu}_j\;\boldsymbol{\alpha}\Big(e^\nu,e_\mu\Big)\stackrel{!}= {b_\nu}^i\;{a^\mu}_j\;{\alpha^\nu}_\mu\,. \end{align} The Kronecker deltas $$ \overline{e}^i(\overline{e}_j)=\delta^i_j\,,\quad e^\nu(e_\mu)=\delta^\nu_\mu $$ and so on have been used in the equals signs labelled with $!\,.$ They are the mechanism that picks the $(i,j)$-th (resp. the $(\nu,\mu)$-th) component of the tensor when it eats $\overline{e}^i$ and $\overline{e}_j\,,$ resp. $e^\nu$ and $e_\mu\,.$
To show what can go wrong when you don't apply the principle of summation convention that a dummy index can appear only twice:
For a $(2,2)$-tensor you $\color{red}{cannot}$ write \begin{align} \boldsymbol{\alpha}(\overline{e}_j,\overline{e}_k,\overline{e}^i,\overline{e}^\ell) =\boldsymbol{\alpha}\Big({a^\mu}_j\;e_\mu\,,{a^{\color{red}{\mu}}}_k \;e_{\color{red}\mu}\,, {b_\nu}^i\;e^\nu\,,{b_{\color{red}{\nu}}}^\ell\;e^{\color{red}{\nu}}\Big)\,. \end{align} Replace the second pair of $\color{red}{\mu}$ and $\color{red}{\nu}$ by, say, $\sigma$ and $\rho$ and restart.