Ambiguity and motivation for the definition of a linear transformation. A linear operator is defined as a transformation $T:V \to W$ such that $V=W$. I was just given the zero transformation and the transformation $(x,y) \to (x,0)$ as examples of linear operators. Wouldn't this suggest that $V=0$ for any vector space $V$ and that $R^2 = R$ or am I misunderstanding the definition? Does, for example, $T:V \to W$ mean that $T$ maps from $V$ to any subspace of $W$? If the answer is yes, why is this the accepted definition?
Edit: It seems that the range of a function $f:V \to W$ is always a subset of the co-domain $W$. Does this mean that any transformation from a higher dimension to a lower dimension is an operator? For example, the statement $T:R^3 \to R^3$ is equivalent to the statement $T: R^3 \to R$ if the range of $T$ is $R$.
 A: It looks like you are misunderstanding the definition of a function. The notation $T:V\to W$ means that $T$ is a function whose domain is $V$ and whose codomain is $W$. Formally, a function $f:A \to B$ is a binary relation such that for each $a \in A$, there is a unique $b \in B$ such that $(a,b) \in f$, or $f(a) = b$. These functions can be surjective, in which case the range is equal to the codomain. In general, functions are not surjective, in which case the range of the function is a proper subset of the codomain.
A linear transformation $T$ is a function between two vector spaces $V$ and $W$, where $V$ and $W$ are not necessarily the same set, such that $T$ satisfies some additional properties. Namely, it is linear in the sense that for $x,y \in V$, $T(x + y) = T(x) + T(y)$ and for $c$ in the underlying field, $T(cx) = cT(x)$.
The special term linear operator is used to distinguish a linear transformation that maps elements from a vector space $V$ to the same space $V$. Symbolically, $T:V\to V$. The function $0: V \to V$ that sends each $v \in V$ to $0 \in V$ is a (linear) operator because the domain and the codomain are the same set; it is not surjective. In general, neither linear transformations nor linear operators are the zero transformation, as the linear operator $(x,y) \mapsto (x, 0)$ shows us. In general, linear transformations are not surjective. Just consider $(x,y) \mapsto (x, 0)$ as an example of a linear transformation that isn't surjective.
A: When we say $T : V \to W$, we don't necessarily mean that $T$ is surjective, only that for every $v$, we have $T(v) \in W$. In other words, it is fine if there are things in $W$ that $T$ does not map anything to; we can still say $T : V \to W$. This is true for functions in general.
