Motivation of Adjoint Transformations Linear Algebra Done Right introduces the concept of Adjoint transfomations as 

Suppose $T \in \mathcal{L}(V, W)$. The adjoint of T is the function
  $T^{*} : W \rightarrow V$ such that:
$\langle T v, w\rangle=\left\langle v, T^{*} w\right\rangle$
for every $v \in V$ and $w \in W$

But this is introduced without any motivation. We work through some examples and find $T^*$ and I think it is interesting it exists. But other than being neat what is the motivation to introduce such a concept? What does the above definition say about $V,W$ or $T$? Why do we need this concept? I can't find a good motivation else where as well.
 A: Without reference to any euclidean structure, for any vector space $V$ you have a "duality bracket" $\langle , \rangle : V\times V^*\to K$ defined by $\langle v, f\rangle = f(v)$. 
Of course when you have a linear form $f:W\to K$ and a linear map $T:V\to W$ you also get the composite $f\circ T : V\to K$ so you have $T^* : W^* \to V^*$, $T^*f := f\circ T$. 
This satisfies the following equation $\langle T(v), f\rangle = f(T(V)) = T^*(f)(v) = \langle v, T^*(f)\rangle$. 
This is very natural, all I've used is composition of maps and evaluation of maps on things : it feels motivated, and I don't know how your book does it, but there's a very interesting theory of duality between $V$ and $V^*$ so this duality bracket and this $T^*$ are natural, motivated objects.
Now when you have a scalar product $\langle, \rangle$ on each vector space $V,W$, the thing is that you have a canonical identification (under certain hypotheses : finite dimensionality or completeness) $V\to V^*$ that sends of course $w\mapsto \langle -, w\rangle$ and under this identification, the duality bracket mentioned above coincides with the scalar product (if $f=\langle -,w\rangle$, then $\langle v, f\rangle = f(v) = \langle v, w\rangle$). 
So with this in mind, a natural question is : ok I know what the duality bracket becomes, what does $T^*$ become ? Of course now $T^* : W^* \to V^*$ becomes $W\to V$ with our identifications and we just have to determine what $T^*$ does on vectors. But vectors are determined by their scalar product on other vectors so to find what $T^*w$ is we just have to test against various $v$'s, in other words we need to know what $\langle v, T^*w\rangle$ is . But we have identified what the duality bracket is, and we know $T^*$ on linear forms, so with all our identifications, we should have $\langle v, T^*w\rangle = \langle Tv,w\rangle$. 
Now it turns out that we can prove that there is such a $T^*w$ (under certain hypotheses like finite dimensionality or completeness), and so this desired equality becomes the definition of $T^*$. 
So $T^*$ is just what it needs to be in order to complete the identification "$V\approx V^*$" in this euclidean context, where we identify duality brackets with scalar products. 
