Left multiplications for each linear map Problem
During studying linear algebra, I've stumbled upon linear maps. So obviously, for any $(m \times m)$-matrix $A$, the map
$$
\lambda_A (B)=AB
$$
is a linear map from $\mathfrak{M}_{m,n}$ (the set of all $(m \times n)$-matrices) to itself. Now we are looking for the converse:

Let $L:\mathfrak M_{m,n}\to\mathfrak M_{m,n}$ be a linear map. Is $L=\lambda_A$ for some $A\in\mathfrak M_{m,m}$?

My Attempt #1
First, I tried using the natural projections $\pi_j:V_1\times\cdots\times V_n\to V_j$ and embeddings $\iota_j:V_j\to V_1\times\cdots\times V_n$ defined as
$$
\pi_j(v_1,\cdots,v_n)=v_j,\qquad\iota_j(v_j)=(0,\cdots,v_j,\cdots,0)
$$
(only the $j$-th coordinate of $\iota_j(v_j)$ equals $v_j$, everything else is zero). So first we could identify $\mathfrak M_{m,n}$ as $\mathbb R^m\times\cdots\times\mathbb R^m$ ($n$ times). Then I came up with this small example:

Let $L:\mathfrak M_{2,2}\to\mathfrak M_{2,2}$ be defined as
$$
L\begin{pmatrix} 
a & b \\
c & d
\end{pmatrix}=\begin{pmatrix}
3a + 2c & 3b + 2d \\
a & b\end{pmatrix}.
$$
Then
$$
(\pi_1 \circ L\circ\iota_1)\mathbf e_1=\begin{pmatrix}3 \\ 1\end{pmatrix},
\qquad
(\pi_2 \circ L\circ\iota_2)\mathbf e_2=\begin{pmatrix}2 \\ 0\end{pmatrix}
$$
(where $\{\mathbf e_1,\mathbf e_2\}$ is the standard basis for $\mathbb R^2$) and thus $L=\lambda_A$ where
$$
A=\begin{pmatrix}3 & 2 \\ 1 & 0\end{pmatrix}.
$$

Now seeing this example, I guessed that for a given linear map $L$, the matrix $A\in\mathfrak M_{m,m}$ defined as
$$
A=
\begin{pmatrix}
(\pi_1 \circ L \circ \iota_1)\mathbf e_1 & \cdots & (\pi_m \circ L \circ \iota_m)\mathbf e_m
\end{pmatrix}
$$
would satisfy the equation $L=\lambda_A$, but I don't really get how to prove this.
My Attempt #2
In the example listed above, every row of the resulting matrix is a linear combination of the rows of the original matrix. So I thought about finding the coefficients of the linear combination; for example,
$$
\pi_i(L(B^\mathbf t))=\sum_{j=1}^m a_{ij}\pi_j(B^\mathbf t) \qquad(i=1,\cdots,m).
$$
Then defining $A=(a_{ji})$ would do. But does the equation above always have a solution?
Overall
This post has became quite long for me just writing up everything I could think of. To sum up, I would like to know how to carry on my attempts, or come up with a completely new one. Also in my attempts, I didn't really use the fact that $L$ is linear, so I wonder how to utilize it in my answer.
 A: For a concrete example, in $\mathfrak M_{2,2}$ the map of matrix transposition (easily seen to be linear) is not given by left multiplication by any fixed matrix$~A$. If it were, applying the requirement to the identity matrix gives the necessary condition $AI=I$ or $A=I$, but this only remaining candidate evidently does not work for non-symmetric matrices.
A: If $n>1$, the statement is false. Just note that if $A=(a_{ij})_{1\leqslant i,j\leqslant n}$ and if $B$ has every entry equal to $0$, except that the $j$th of the first line is equal to $1$, then $\lambda_A(B)$ is a matrix such that every column is null, except that the entries of the $j$th column are $a_{1,j}, a_{2,j},\ldots,a_{m,j}$. So, take any linear map from $\mathfrak{M}_{m,n}$ for which this does not occur. As, for instance, the linear map$$L\bigl((b_{ij})_{1\leqslant i\leqslant m,1\leqslant j\leqslant n}\bigr)=\begin{bmatrix}b_{11}&0&0&\ldots&0\\0&0&0&\ldots&0\\&\vdots&&\ddots&\vdots\\0&0&0&\ldots&0\end{bmatrix}.$$
A: $\mathfrak M_{m,n}$ is an $mn$-dimensional space. Therefore the dimension of the vector space of all linear operators on $\mathfrak M_{m,n}$ is $(mn)^2$. However, the dimension of the subspace consisting of all linear operators of the form $\lambda_A$ is only $m^2$ (because $A$ is $m\times m$). Therefore, unless $n=1$, there is always some linear operator on $\mathfrak M_{m,n}$ that is not in the form of $\lambda_A$.
