7.1 Proposition: Every eigenvalue of a self-adjoint operator is real.
Proof: Suppose $T$ is a self-adjoint operator on $V$. Let $\lambda$ be an eigenvalue of $T$, and $v$ a non-zero vector in $V$ such that $Tv = \lambda v$ Then \begin{align} \lambda \|v\|^2 &= \langle \lambda v, v\rangle\\ &= \langle Tv, v\rangle\\ &= \langle v, Tv\rangle \end{align}
I don't understand that last step. how come $\langle Tv, v\rangle = \langle v, Tv\rangle$ and not its conjugate?
For the sake of not leaving the question unanswered,
The definition of the adjoint $T^*: W \rightarrow V$ of an operator $T:V \rightarrow W$ is the relation
$$\forall v\in V, w \in W : \, \langle Tv,w \rangle_W = \langle v, T^*w\rangle_V$$
Here, $V$ and $W$ are inner-product spaces (real, complex, Hermitian or Hilbert - choose your favorite), while $\langle\cdot,\cdot\rangle_V$ and $\langle\cdot,\cdot\rangle_W$ denote the inner product in $V$ and $W$, respectively.
Next, the definition of self-adjoint operator $T : V \rightarrow V$ is just $T = T^*$. In this case, for any $v \in V$, we get
$$\langle Tv, v\rangle = \langle v, T^*v\rangle = \langle v, Tv \rangle$$