TITLE: How to interpret the adjoint?
QUESTION [50 upvotes]: Let $V \neq \{\mathbf{0}\}$ be a inner product space, and let $f:V \to V$ be a linear transformation on $V$.
I understand the definition1 of the adjoint of $f$ (denoted by $f^*$), but I can't say I really grok this other linear transformation $f^*$.
For example, it is completely unexpected to me that to say that $f^* = f^{-1}$ is equivalent to saying that $f$ preserves all distances and angles (as defined by the inner product on $V$).
It is even more surprising to me to learn that to say that $f^* = f$ is equivalent to saying that there exists an orthonormal basis for $V$ that consists entirely of eigenvectors of $f$.
Now, I can follow the proofs of these theorems perfectly well, but the exercise gives me no insight into the nature of the adjoint.
For example, I can visualize a linear transformation $f:V\to V$ whose eigenvectors are orthogonal and span the space, but this visualization tells me nothing about what $f^*$ should be like when this is the case, largely because I'm completely in the dark about the adjoint in general.
Similarly, I can visualize a linear transformation $f:V\to V$ that preserves lengths and angles, but, again, and for the same reason, this visualization tells me nothing about what this implies for $f^*$.
Is there (coordinate-free, representation-agnostic) way to interpret the adjoint that will make theorems like the ones mentioned above less surprising?

1 The adjoint of $f:V\to V$ is the unique linear transformation $f^*:V\to V$ (guaranteed to exist for every such linear transformation $f$) such that, for all $u, v \in V$,
$$ \langle f(u), v\rangle  = \langle u, f^*(v)\rangle \,.$$

REPLY [6 votes]: It's difficult to give an intuitive description of the adjoint. Note that the adjoint is here even before we have scalar products. It's just that a scalar product allows to interpret the adjoint of a map $A:\ V\to V$ in one and the same space $V$.
A linear map $A:\ V\to W$ from one vector space $V$ to some other vector space $W$ (of any dimensions) produces for each vector $x \in V$  a vector $y:=Ax\in W$. 
Assume now that on $W$ a linear function $\phi:\ W\to{\mathbb R}$, i.e., an element of $W^*$, is given, which assigns, e.g., to each point $y\in W$ a temperature value $\phi(y)$, or computes for each $y\in W$ the first coordinate with respect to some basis of $W$. Then the function
$$\psi:\quad V\to{\mathbb R},\qquad x\mapsto\phi\bigl(Ax\bigr)$$
computes for each point $x\in V$ the temperature felt at $Ax\in W$, "even before $x$ is actually mapped to $W$". In this way we can regard $\psi$ as a "virtual" temperature distribution on $V$. It is obvious that $\psi$ is a linear function from $V$ to ${\mathbb R}$, i.e., an element of $V^*$.
What we have described here for one $\phi\in W^*$ can of course be done with every $\phi\in W^*$: For each such $\phi$ we shall get a corresponding $\psi\in V^*$. All in all the map $A:\ V\to W$ given at the beginning induces a certain map $$W^*\to V^*, \quad \phi\to\psi\ .$$
This map is called the transpose of $A$ and is denoted by $A^*$. By definition we have the identity
$$A^*\phi.x\ =\ \phi. Ax\qquad\forall x\in V,\ \forall\phi\in W^*\ .$$
Here the . means that the linear functional on the left of the dot is applied to the vector on the right of the dot.
An example: Assume that $(e_k)_{1\leq k\leq n}$ is a basis in $V$ and $(f_i)_{1\leq i\leq m}$ is a Basis in $W$. Then $A$ has a certain matrix $[a_{ik}]$ with respect to these bases. Now  let $$\phi:=f_1^*:\quad y\mapsto y_1\tag{1}$$
be the functional that computes the first coordinate of any given vector $y\in W$. Then $\phi.Ax$ is the first coordinate $y_1$ of the vector $y:=Ax$. We all know that $$y_1=\sum_{k=1}^n a_{1k} x_k\ .\tag{2}$$ Since we can write $x_k$ as $x_k=e_k^*.x$ we can interpret $(2)$ as $$A^*\phi.x=\phi.Ax=\sum_{k=1}^n a_{1k} e_k^*.x\qquad(x\in V)\ ,$$
or, given the definition $(1)$ of $\phi$:
$$A^* f_1^*=\sum_{k=1}^n a_{1k} e_k^*\ .$$