I'm working on a question and am looking for some advice:
$$ T: M_{2}(\Bbb R) \to M_{2}(\Bbb R),\\ T\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = A\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}$$
where $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
Find the transformation matrix $T$ with respect to the basis:
$$ e_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, e_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, e_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, e_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} $$
I have transformed each of the basis vectors according to the given transformation. Then expressed each as a linear combination of the basis vectors. Organizing the coefficients into a matrix I get
$$ \begin{pmatrix} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \end{pmatrix} $$
The only way I can see this being correct is if a vector we wish to transform was written as a column vector instead of a $2\times 2$ matrix in order to satisfy matrix multiplication rules. This will then produce another column vector, rather than a $2\times 2$ matrix.
Does this seem correct? I don't seem to have any problems when we are working in $\Bbb R^{n}$, only with this $M_{2}(\Bbb R)$ space.
Thanks
This is fine, and indeed, it's to be expected. If you have a linear transformation $T : V \to W$, and find a matrix representation $M$ of $T$ with respect to bases $B$ and $C$ for $V$ and $W$ respectively, then for any $v \in V$, $$[Tv]_C = M[v]_B.$$ This is the way that matrices for linear transformations are supposed to work: if you transform a vector into a coordinate column vector, and multiply it by the matrix for the transformation, then the resulting column vector is a coordinate vector with respect to the other basis.
In this case, our spaces $V$ and $W$ are both the space of $2 \times 2$ matrices. These are $4$-dimensional spaces, so any matrix for $T$ must be $4 \times 4$. In order to use this matrix, we have to multiply it to $4 \times 1$ coordinate vectors with respect to the basis. Similarly, the result will be a $4 \times 1$ coordinate vector.