Example for linear transformations as vector spaces? I am currently studying a book on Functional Analysis by George Bachman and Lawrence Narici. In one of the first chapters, they describe how a (linear) mapping $A$ can map from a vector space $X$ to a different vector space $Y$:
$$A:X\rightarrow Y$$
which has the (linear) properties that for any scalars $\alpha$ and $\beta$ and any vectors $x$ and $y$, we have:
$$A(\alpha x + \beta y) = \alpha A(x)+\beta A(y)$$
Now the critical part: They state that "the collection of all linear transformations mapping $X$ into $Y$ can be viewed as a vector space by defining addition of the linear transformations $A$ and $B$ to be that transformation which takes $x$ inton $Ax+Bx$; symbollically, we have:
$$(A+B)x=Ax+Bx$$
and for scalar multiplication:"
$$(\alpha A)x=\alpha A x$$
where they omit the parantheses around $(x)$ for brevity. I see how these operations above imply linearity, but how can the collection of these transformations be interpreted as a vector space? As far as I have gathered, a vector space should have basis vectors. What would these basis vectors be in the case of $A$?
I have an unfinished example below, which you could use to explain your answer. Otherwise feel free to use your own examples, if this is easier.
Example: $x \in X \in \mathbb{R}^2$ and $y \in Y \in \mathbb{R}^3$. In this case $A$ in $y=Ax$ would be a $3 \times 2$ matrix. What would be the basis vectors of the vector space interpretation of $A$?
 A: Regarding your example: Any $3 \times 2$ matrix can be written as a linear combination of $\{E_{ij} \}$, where $E_{ij}$ is $0$ everywhere except in position $(i,j)$.
A better way of looking at vector spaces is to concentrate on their properties from an algebraic point of view: the essential property is linearity, the fact that all elements in the vector space can be multiplied by a scalar and added together to form new elements in the vector space. A function between vector spaces is linear if it preserves this essential property: that the image of a linear combination is again a linear combination is very simple way.
All vector spaces have a basis, and that is why linear functions can be expressed in terms of matrix multiplication.
A: Note that having basis vectors is not a defining property of vector spaces. Moreover, while in standard set theory you can prove that every vector space has a basis, you can't always explicitly give one.
In your example, it is easy to give a basis. A standard choice would be the set of matrices $E_{ij}$ that have an $1$ in row $i$/column $j$, and $0$ everywhere else:
$$\pmatrix{a&b\\c&d\\e&f}
 = a\pmatrix{1&0\\0&0\\0&0}
 + b\pmatrix{0&1\\0&0\\0&0}
 + \ldots
 + f\pmatrix{0&0\\0&0\\0&1}$$
An example of a vector space where you can't explicitly give a basis is $\mathbb R$ as a vector space over $\mathbb Q$ (clearly $\mathbb R$ is a group under addition, and you certainly can multiply a real number by a rational number, giving again a real number; the remaining vector space axioms all reduce to the corresponding number arithmetic laws).
