Scalar multiplication as a special form of matrix multiplication Question
What do we gain or lose, conceptually, if we consider scalar multiplication as a special form of matrix multiplication?
Background
The question bothers me since I have been reading about dilations and scaling of geometrical objects in Paul Lockhart's book "Measurement". Geometrically, dilation is a transformation that stretches an object in one dimension by a certain factor. Analogously, the linear transformation
$$
\begin{pmatrix}
  1 & 0 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & \lambda
\end{pmatrix}
\cdot
\begin{pmatrix}
  x_1 \\
  x_2 \\
  x_3
\end{pmatrix}
$$
"stretches" the third component by the factor $\lambda$. Scaling is a geometric transformation that stretches an object in all dimensions by a certain factor. Analogously, the linear transformation
$$
\begin{pmatrix}
  \lambda & 0 & 0 \\
  0 & \lambda & 0 \\
  0 & 0 & \lambda
\end{pmatrix}
\cdot
\begin{pmatrix}
  x_1 \\
  x_2 \\
  x_3
\end{pmatrix}
$$
"stretches" all three components by the factor $\lambda$. This, however, can be written more succinctly using scalar multiplication:
$$
\lambda
\cdot
\begin{pmatrix}
  x_1 \\
  x_2 \\
  x_3
\end{pmatrix}.
$$
In fact, every scalar multiplication can be expressed as a multiplication with a special matrix, and it turns out to be a mere shortcut. On the face of it, this observation is not very spectacular; however, it raises interesting philosophical and conceptual questions as to the foundations of linear algebra.
For example, if scalar multiplication is only a nice-to-have shortcut, then isn't it in fact superfluous conceptually? Currently, scalar multiplication is taught as if it was a distinct concept, independent of matrix multiplication. What would change if we got rid of this shortcut? What could alternative axioms of vector spaces and moduls look like? What about linear transformations? What is easier, what is harder$-$not to write down, but conceptually?
I know that this topic is very broad, but I would like to collect opinions, ideas, examples.
 A: I disagree that "scalar multiplication is only a nice-to-have shortcut", or that is "superfluous conceptually". In fact the very definition of a vector space $V$ requires there to be a scalar multiplication.
After that comes the concept of a linear transformation, which again requires the scalar multiplication to be defined. Matrix multiplication is a convenient way to represent these, and that too only in case of finite dimensional vector spaces (or certain infinite dimensional vector spaces). 
So defining matrix multiplication and then saying that scalar multiplication is a special case is putting the cart in front of the horse in my opinion. 
A: Note: Some discussion with Siddharth has revealed a flaw in my earlier answer, which I've corrected now (with a complete reversal of conclusion).
The set of all endomorphisms of an Abelian group $(V, +)$ forms a ring $\operatorname{End}(G)$ with pointwise addition as the ring addition and composition of mappings as the ring multiplication. If there exists a subring $L \subseteq \operatorname{End}(G)$ with center $F = \operatorname{Z}(L)$, such that $F$ is a field and $F \cap \operatorname{End}(G) = (1)$, the ideal generated by the identity endomorphism, then $V$ is a vector space over the field $F$ with $L$ as its ring of linear operators.
Why does this work? [Edit: I'm not sure the following is entirely correct. I'll have to work on it further and update the answer]. In one direction (vector space $\to$ above characterization), it's obvious that the linear operators of a vector space $V$ do form a subring of the endomorphism ring of the Abelian group $(V, +)$, and that the subset of "scalar" operators form a subfield of this ring. As shown here (beautifully), the scalar operators are exactly the central elements of the ring of linear operators. Conversely, it's not hard to show that the Abelian group $(V, +)$ forms a (traditional) vector space over the field $F$ defined above, in a manner that makes $L$ (isomorphic to) the ring of all linear operators of this space.
