linear map vs operator: raised to power In linear algebra, a linear map $T: V \rightarrow W$, so $T \in \mathcal{L}(V,W)$. When $T$ maps from $V$ to $V$ itself, then $T$ is an operator on $V$. 
Here is my problem, I read in a book saying that operator can be raised to powers, like $T^2$ for $T \in \mathcal{L}(V,V)$, while ordinary linear map can't, why?
 A: Suppose, for instance, that $T$ is a linear map from $\Bbb R^3$ to $\Bbb R^5$. What could $T^2(v)$ possibly mean? On the face of it it’s $T\big(T(v)\big)$. This clearly implies that $v\in\Bbb R^3$, since otherwise $T(v)$ makes no sense. But then $T(v)$ is in $\Bbb R^5$, so $T(v)$ is not in $\Bbb R^3$. But this means that $T(v)$ is not in the domain of $T$, so it makes no sense to try to find $T\big(T(v)\big)$: it’s like trying to do the matrix multiplication
$$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}\\
a_{41}&a_{42}&a_{43}\\
a_{51}&a_{52}&a_{53}
\end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix}\;.$$
$T^2$ makes sense only when the range of $T$ is a subset of the domain of $T$; if that’s not the case, there will always be some $v\in\operatorname{dom}\,T$ such that $T(v)\notin\operatorname{dom}\,T$, and for such a $v$ the expression $T^2(v)$ doesn’t make sense. On the other hand, if $T\in\mathscr{L}(V,V)$ for some $V$, then 
$$\operatorname{ran}\,T\subseteq V=\operatorname{dom}\,T$$
is definitely true.
Added: Lest I give the wrong impression, please note that dimension is not really important here. You’ve probably seen $P_2$, the space of all polynomials of degree at most $2$ with real coefficients and may know that it’s isomorphic to $\Bbb R^3$. Consider the map $$T:\Bbb R^3\to P_2:\langle a,b,c\rangle\mapsto ax^2+bx+c\;.$$ This is a linear map — an isomorphism, even! — between two vector spaces of dimension $3$, and here again $T^2$ makes no sense. If $v=\langle a,b,c\rangle\in\Bbb R^3$, we can compute $T(v)=ax^2+bx+c$, but this polynomial $ax^2+bx+c$ isn’t in the domain of $T$, so $T^2(v)=T(ax^2+bx+c)$ makes no sense.
