Definition of adjoint operator (asking for intuition) Definition of the adjoint operator: A linear operator T on an inner product space V is said to have an adjoint operator $T^{*}$ on V if $\langle T(u),v \rangle= \langle u,T^{*}(v) \rangle$.
Question: Why people come up with that definition? It does not sound intuitive to me. $T^{*}$ is the transpose conjugate of T right, and does that definition follow from definition of inner product space?
 A: The point of the definition is to extend the notion of the "conjugate transpose" so that it makes sense on an arbitrary inner product space. I'm not sure what you mean by "does that definition follow from definition of inner product space".  However, I think it might be helpful to see why if $V = \Bbb C^n, W = \Bbb C^m$ with the usual inner product and $T:V \to W$ is the operator on $V$ defined by $T(x) = Ax$, then the adjoint operator $T^*: W \to V$ is $T^*(x) = A^*x$. In other words, taking the adjoint is "the same as" taking the conjugate transpose.
Let $A'$ denote the conjugate-transpose of $A$.  Recall that the usual inner product on $\Bbb C^n$ is given by 
$$
\langle x,y\rangle = y'x = \sum_{k=1}^n x_k \bar y_k.
$$
If we define $T(x) = Ax$ and $S(x) = A'x$, then we find that for $x \in V$ and $y \in W$, we have
$$
\langle T(x),y \rangle = y'(Ax) = (y'A)x = (A'y)'x = \langle x,S(y) \rangle.
$$
So, $S$ is indeed the adjoint operator to $T$.
A: The adjoint on inner product spaces comes from a more general construction. If $X$ and $Y$ are Banach spaces and $T : X \to Y$ is a bounded linear operator, then $T$ induces a map from the dual of $Y$ to the dual of $X$, that is a $T^*:Y^*\to X^*$ defined by 
$T^*y^*(x)=y^*(T(x))\tag 1$ 
So, if $\mathbb F$ is the scalar field of the spaces $X$ and $Y$, we have that $T^*$ sends an arbitrary  $y^*:Y\to \mathbb F$ to a $T^*y^*:X\to \mathbb F$, which acts on an arbitrary $x\in X$ as in $(1).$
The reason this definition is useful is that knowledge of the properties of  the dual space often provides answers to questions about the space itself. 
Of course, one has to check that $T^*y^*$ is a bounded linear operator. Linearity is immediate, and boundedness follows from the calculation
$|y^*(T(x))| \leq \| y^* \| \| T \| \| x \| \tag2$
To specialize this to your case, suppose $X=Y=V$ an inner product space and $T:V\to V$ is a bounded linear operator. By the Riesz theorem, there is a bijection 
$v\leftrightarrow \langle \cdot,v\rangle\  \text{between the elements of}\  V\  \text{and those of}\  V^*\tag 3$ 
Let $y,w\in V$ be the elements corresponding to $y^*$ and $T^*y^*$, respectively. Then, $\langle T(v),y\rangle=\langle v,w\rangle$. But, $T^*$ sends $y^*$ to $T^*y^*$ so applying the correspondence $(3)$, we have $T^*y=w$, from which it follows that 
$\langle T(v),y\rangle=\langle v,T^*y\rangle \tag4$
