Transformation matrix of linear map on vector space whose elements are matrices I'm starting to understand how to compute and think about the transformation matrix of a linear transformation $L:\mathbb{R}^n \to \mathbb{R}^m$. However, I'm trying to understand how this translates to a vector spaces of matrices. For example, consider the linear map $L:\mathbb{R}^{n \times m} \to \mathbb{R}^{m \times p}$, and the standard basis $\beta=\{E_{ij}|1\leq i \leq n, 1\leq j \leq m\}$ where each $E_{ij}$ contains zeros except in the $i$-th row and $j$-th column, which contains a ones. From this it follows that any matrix $Q\in \mathbb{R}^{n \times m}$ can be written as a linear combination of these basis matrices. Applying the transformation $L$ to this matrix $Q$ we obtain another matrix $L(Q)\in \mathbb{R}^{m \times p}$. In the case of vectors the transformation matrix $A$ of $L$ has columns containing the coordinates of each basis vector $L(v_i)$ w.r.t. to basis $\beta_W$. However, I'm having difficulty translating this idea to the context in which the vectors are $m\times n$ matrices.
Any help would be much appreciated. Thanks,
 A: I think the key here is you are converting your matrices into vectors by writing them as linear combinations of the basis vectors.
For example given an ordered basis (order matters for transformation matrices, you always have to keep the basis vectors in the same order!) $\beta_{3\times 3}=\{ E_{11},E_{12},E_{13},E_{21},E_{22},E_{23},E_{31},E_{32},E_{33}\}$:
$\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{bmatrix} = a_{11}\cdot E_{11}+a_{12}\cdot E_{12}+\cdot \cdot \cdot +a_{33}\cdot E_{33} \equiv(a_{11},a_{12},...,a_{33})\in \mathbb{R}^{3\cdot 3}$
The beauty about vector spaces is you can treat whatever you are working with, in this case matrices, as a vector, so you are essentially converting your $n\times m$ matrix into a vector in $\mathbb{R}^{n\cdot m}$, you find the matrix of the linear transformation as you would for any linear transformation because you are back to familiar territory, and then that matrix allows you to input a vector (a matrix written as a linear combination of the ordered basis vectors of the domain) and you get as output another vector (another matrix written as linear combination of the ordered basis vectors of the codomain).
Hope this helps!
