A Faithful Module M. Reid ch. 2.2.
Reid defines an A-module M over a ring A and a mapping
$\mu_f: M \to M, m \mapsto fm, f \in A, m \in M$.
He goes on to define $A \to \text{End M}$, $f \mapsto \mu_f$ and notes that it is a ring homomorphism if and only if the multiplicative and distributive axioms for scalar multiplication are satisfied:
$\mu_{f+g}=\mu_f+\mu_g$
$\mu_{fg}=\mu_f\mu_g$
$\mu_{1_A}=1_{\text{End M}}$
He notes that $\mu_f$ is A-linear and says the following (I quote):
As far as M is concerned, $\mu(A) \subset \text{End M}$ is the only part of A that matters, and I sometimes write $A'=\mu(A)$. If $\mu$ is injective, then $A \cong A'$ and linear algebra in M reflects all the properties of A; in traditional algebraic terminology, M is a faithful A-module.
Question: In what sense does linear algebra in M "reflect" all the properties of A?
 A: I think this is an informal way of saying that the isomorphism $A \cong A'$ is not just any isomorphism; it is in particular the map $f \mapsto \mu_f$. Or maybe in other words, "the isomorphism $A \cong A'$ is an isomorphism of $A$-algebras".
In linear algebra over a field $k$, "scaling by an element of $k$" is an endomorphism of any vector space. In fact, scaling by $\alpha \in k$ is exactly the map $\mu_\alpha$ defined above. With respect to any basis, one of these endomorphisms always corresponds to a multiple of the identity matrix; namely $\mu_\alpha$ corresponds to $\alpha I_n$ (where $n$ is the dimension of whatever vector space we're looking at).
Scalar multiples of the identity matrix behave exactly like scalars do – if you have some statement about elements of $k$, you can replace each element of $k$ with the corresponding multiple of the identity matrix, and that resulting statement holds for those matrices if and only if the original statement holds for the elements of $k$. This is the "reflects" part. In other words, $f \mapsto \mu_f$ is an isomorphism onto its image, or yet other words $f \mapsto \mu_f$ is injective.
Actually, this is a little bit wrong – can you spot the problem?
If $V$ is a trivial (aka $0$-dimensional) vector space, then the matrices representing endomorphisms of $V$ have dimension $0 \times 0$ (these are "empty matrices"). There is exactly one empty matrix (in the same way that there is exactly one empty set), so in fact all scalar multiples of the identity matrix are exactly the same. The map $f \mapsto \mu_f$ is not injective in this case.
The finally correct conclusion is that over a field $k$, every module is faithful, except the zero module, which is not faithful.

Test your understanding!
Which of the following modules are faithful?

*

*$\mathbb{Z}/24\mathbb{Z}$ as a $\mathbb{Z}$-module

*$\mathbb{R}[x,y]$ as an $\mathbb{R}[x]$-module

*$\mathbb{R}[x,y]/(y^2)$ as an $\mathbb{R}[x]$-module

*$\mathbb{R}[x,y]/(y^2)$ as an $\mathbb{R}[y]$-module

*$\mathbb{R}^7$ as an $M_{7 \times 7}(\mathbb{R})$-module

