'Standard matrix' for linear transformation on matrix spaces? We know for linear transformation $T: R^n \rightarrow R^m$, there is a standard matrix A such that $T(x) = Ax$. But for linear transformation $T: R^{n_1 \times n_2}\rightarrow R^{m_1 \times m_2}$, does it have something similar?
 A: If you're thinking that there might be a matrix $A$ so that $T : \Bbb{R}^{a \times b} \to \Bbb{R}^{c \times d}$ satisfies $T(M) = AM$, or something like this, then no, this doesn't work.
If there were such a matrix, you could think about what kind of dimension it'd be. In order to multiply to $M \in \Bbb{R}^{a \times b}$ on the left, it would have to have $a$ columns, i.e. be a $\underline{\hspace{15pt}} \times a$ matrix, and the product $AM$ would be a $\underline{\hspace{15pt}} \times b$ matrix. So, in order for this product to belong to $\Bbb{R}^{c \times d}$, we would require $d = b$ right off the bat. Without this, finding such a matrix is doomed to fail.
Even if we restrict to square matrices of the same size, you can look at $T$ as the transpose operator, e.g.
$$T : \Bbb{R}^{2 \times 2} \to \Bbb{R}^{2 \times 2} : M \mapsto M^\top.$$
This map is linear. If there were some $A \in \Bbb{R}^{2 \times 2}$ such that $T(M) = AM$, then we would have
$$A = AI = T(I) = I^\top = I,$$
and so $M^\top = T(M) = IM = M$. But this is not true in general!
All that said, we can represent such a map as a matrix. Like all maps between finite-dimensional vector spaces, a linear transformation $T : \Bbb{R}^{a \times b} \to \Bbb{R}^{c \times d}$ does have a matrix representation. You must choose a basis each for $\Bbb{R}^{a \times b}$ and $\Bbb{R}^{c \times d}$, compute the image of each of the former basis, and express them as coordinates in the latter basis. These coordinate vectors form columns of the matrix. To use this matrix, you multiply coordinate column vectors with respect to the former basis, and you get back the image of the vector, as a coordinate column vector with respect to the latter basis.
To call one of these matrices "standard", we need "standard" bases. We have a "standard" basis for $\Bbb{R}^n$, and computing the matrix for a transformation with respect to this basis produces the standard matrix.
Matrices, on the other hand, have a set of vectors that we could almost call a standard basis. For example, in $\Bbb{R}^2$, we have the four matrices:
$$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.$$
But, bases must be ordered (otherwise this changes the matrix for the transformation), and unlike the standard basis for $\Bbb{R}^n$, it's not so obvious which order is more "natural". Should it be written as I wrote, or should the middle two matrices be switched?
This is enough to say that there isn't really a "standard" matrix, even though there definitely will be other (not-so-standard) matrices.
