TITLE: Intuition behind definition of transpose map
QUESTION [8 upvotes]: The linear map $T^t:V' \to V'$ (where $V'$ is the set of all linear maps from $V$ to its scalar field $\Bbb F$) is defined by :
$$T^t(f)(v)=f(T(v))$$
This looks like some "commutative" definition which helps in many cases (for example, to prove the existence of a basis for which triangular matrix exists). Can someone elaborate on the motivation of this definition.

REPLY [9 votes]: A priori this has nothing to do with linearity. When you have a map $T$ that "beams" points $x\in X$ to a new location $y\in Y$, and if there is some temperature distribution $f:\ Y\to{\mathbb R}$ in $y$-space then you can ask for the temperature at the image point $y=T(x)$ without actually beaming the point $x$. This defines a "virtual" temperature $f^*$ in  $x$-space, where $f^*$ is given by the formula
$$f^*:\ X\to{\mathbb R},\quad x\mapsto f^*(x):=f\bigl(T(x)\bigr)\ .$$
Since this makes sense for any  any temperature distribution $f$ on $Y$ we have here a law that produces for any function $f:\ Y\to{\mathbb R}$ a function $f^*:\ X\to{\mathbb R}$. This law is completely determined by the "beaming map" $T$; therefore it is allowed to call it $T^*$.
Now if $X$ and $Y$ are vector spaces and $T$ as well as $f$ are linear then it turns out that $f^*$ and $T^*$ are linear as well. Therefore  the map $T^*$ restricted to $Y'$ can be considered as a linear map from $Y'$ to $X'$. It is then denoted by $T'$ and called the transpose of $T$.