I'm trying to prove to myself why the following claim is true. I think there is some "trivial" explanation, but I'm missing something. The claim is: Let $ E_1, E_2 \subset F^n, W \subset M_n(F)$ and $ Q_1 ,Q_2 \in M_n(F)$ non singular. so $W=E_1\otimes E_2 \iff Q_1WQ_2=(Q_1E_1)\otimes(E_2Q_2)$
The notation $\otimes$ is $A\otimes B=span\left\{ x\otimes y\,|\,x\in A,\,y\in B\right\}$ for $A,B\subset F^n $.
I see that a vector in $QiEi$ is of the form $\left[\begin{array}{c} \sum_{j=1}^{n}Q_{1,j}^{i}\overrightarrow{x}_{j}\\ .\\ .\\ .\\ \sum_{j=1}^{n}Q_{n,j}^{i}\overrightarrow{x}_{j} \end{array}\right]$
so an element in $(Q1E1)⊗(E2Q2)$ would be $\left[\begin{array}{ccc} & \sum_{j=1}^{n}Q_{1,j}^{1}\overrightarrow{x}_{j}\cdot\left[\begin{array}{ccccc} \sum_{j=1}^{n}Q_{1,j}^{2}\overrightarrow{y}_{j} & . & . & . & \sum_{j=1}^{n}Q_{n,j}^{2}\overrightarrow{y}_{j}\end{array}\right]\\ & .\\ & .\\ & .\\ & \sum_{j=1}^{n}Q_{n,j}^{1}\overrightarrow{x}_{j}\cdot\left[\begin{array}{ccccc} \sum_{j=1}^{n}Q_{1,j}^{2}\overrightarrow{y}_{j} & . & . & . & \sum_{j=1}^{n}Q_{n,j}^{2}\overrightarrow{y}_{j}\end{array}\right] \end{array}\right] $
But I think there is a much more "elementary" explanation that I'm missing, And one that wont include a lot of Sigma notations etc.
I'll be happy to hear what I'm missing.
Thank you!