# Intuition behind definition of transpose map

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.

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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 $T$.

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Given any linear map between vector spaces $T: V → W$ you can define $T^t: W^* → V^*$ between the algebraic dual spaces of $V$ and $W$ in this way. Then that map $\cdot^t: L(V, W) → L(W^*, V^*)$ is a contravariant functor in the category of vector spaces and linear maps. In particular $(ST)^t = T^tS^t$. (At least for $𝔽 = ℝ,ℂ$) it also restricts to the category of normed spaces and bounded linear maps. So if $V$ and $W$ are normed spaces and $V', W'$ are the continuous dual spaces, $\cdot^t$ maps the space of bounded maps $\mathscr{L}(V, W)$ into $\mathscr{L}(W', V')$.

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There really are a few ways of thinking about transpose/adjoint maps, not all of which apply to the situation you described, but let me include them all the same.

The first is if you have a bilinear scalar product $\langle \cdot, \cdot \rangle: V \otimes V \to \mathbb F$ on your vector space $V$ (if it is non-degenerate, this is called an inner product). In this case, we usually define the transpose of $T^t$ of $T$ to be the operator which satisfies $$\langle v_1,Tv_2 \rangle = \langle T^t v_1, v_2 \rangle.$$

Another definition comes from the following: Every $\mathbb F$ vector space has a dual, $V^* = \{ f: V \to \mathbb F\}$. The better way of seeing this is that in the category of $\mathbb F$-vector spaces, the map $\text{Hom}(\cdot, \mathbb F)$ is a contravariant functor which takes $V$ to $V^* = \text{Hom}(V,\mathbb F)$. But to properly be a functor, $\text{Hom}(\cdot, \mathbb F)$ must also do something to the morphisms! In particular, if $T \in \text{Hom}(V,V)$ then $$T^t = \text{Hom}(T,\mathbb F) : \text{Hom}(V,\mathbb F) \to \text{Hom}(V,\mathbb F),$$ and this operates as $f \mapsto f\circ T$ (this is really the only thing it could be).

Now we can consolidate the pictures by slightly abusing how we think about scalar products. Every vector space comes with a 'canonical pairing' $\langle \cdot,\cdot\rangle : V^* \otimes V \to \mathbb F$ given by $\langle f, v \rangle = f(v)$, and so our definition of $T^t(f) = f\circ T$ can be written in this notation as $$\langle f,Tv\rangle = f(Tv) = (f\circ T)(v) = \langle T^tf,v\rangle$$ which is the same relation as before! In fact, if $V$ is additionally a finite dimensional vector space (or a Hilbert space) then any non-degenerate bilinear pairing defines an isomorphism $^\sharp : V \to V^*$ via $v^\sharp = \langle\cdot, v \rangle$, allowing us to properly translate between the $V^*\otimes V$ and $V \otimes V$ pictures since $V \cong V^*$.

Finally, if one views $V$ as a manifold (since all vector spaces are smooth manifolds) then at any point $v \in V$ we have $T_v V \cong V$. In this case, $T: V \to V$ is a smooth map on manifolds, and $T^t = T^*$ is pullback map. This means that on $0$-forms (i.e. functions $V \to \mathbb F$) that $T^t f = f \circ T$.

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The transpose of an operator plays a "dual" role to the operator itself - in much the same way that the dual of a vector space helps understand the vector space itself.

Perhaps the best illustration of this phenomenon is when the vector space is an infinite dimensional complex Hilbert space $H$. Continuous (a.k.a. bounded) linear operators on $H$ are very interesting in themselves (See, for instance, http://en.wikipedia.org/wiki/Fredholm_integral_equation). There, the relation between $T$ and $T^{\ast}$ is particularly relevant.

For instance, if $TT^{\ast}=T^{\ast}T$, then you have the very beautiful spectral theorem.

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