Prove that the projection operator $\mathbb P_+\equiv|+z\rangle\!\langle +z|$ is Hermitian 
Use Dirac notation (the properties of kets, bras and inner products) directly to establish that the projection operator $\mathbb{\hat P}_+$ is Hermitian. Use the fact that $\mathbb{\hat P}^2_+=\mathbb{\hat P}_+$ to establish that the eigenvalues of the projection operator are $1$ and $0$. 

I know how to prove this using mathematical notation, i.e. for any $x,y\in V$ we must show that $\langle x, \ \mathbb{\hat P}_+y\rangle = \langle \mathbb{\hat P}_+x, \ y\rangle$ but how can I prove the way the book suggested, i.e. using Dirac notation and the properties of kets and bras?
 A: Any projection operator can be written in the form
$$
P = \sum_{j = 1}^r |\psi_j \rangle \langle \psi_j |
$$
Where $\psi_1,\dots,\psi_n$ is an orthonormal basis of our Hilbert space.  Given $\psi = c_1\psi_1 + \cdots + c_n \psi_n$, we calculate
$$
\langle\psi| P = 
\langle \psi | \left(\sum_{j = 1}^r |\psi_j \rangle \langle \psi_j | \right) =
\sum_{j = 1}^r \langle \psi \mid \psi_j \rangle \langle \psi_j | =\\
\sum_{j = 1}^r \langle \psi_j \mid \psi \rangle^* \langle \psi_j | = 
\sum_{j = 1}^r c_j^* \langle \psi_j |
$$
This is the bra corresponding to the ket $P |\psi \rangle = \sum_{j=1}^r c_j | \psi_j \rangle$.  So, $P$ is self-adjoint.
A: I assume from the comments that $P$ is a rank-1 projection, and so it is of the form
$$P=|\psi\rangle\langle\psi|$$
for some vector $|\psi\rangle$ in the Hilbert space. Observe that, in a sense, $|\psi\rangle^*=\langle\psi|$ in Dirac notation, whence
$$P^* = (\langle\psi|)^*(|\psi\rangle)^*,$$
and since the $*$ is involutive one has
$$P^*=|\psi\rangle\langle\psi| = P.$$
The "idempotency" $P^2 = P$ comes from a direct computation
$$P^2 = |\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi|=P$$
since $|\psi\rangle$ is assumed to be a vector of norm one, so that $\Vert\psi\Vert^2 = \langle\psi|\psi\rangle=1$.
A: I figured out the solution to my question. It is similar to the answers given but I will write it anyways for future reference.
If $\mathbb{\hat P}_+$ is Hermitian then $\langle \psi|\mathbb{\hat P}_+|\phi\rangle=\langle\phi|\mathbb{\hat P}_+|\psi\rangle^*.$ 
$$\langle\psi|\mathbb{\hat P}_+|\phi\rangle=\langle\psi|+z\rangle\langle+z|\phi\rangle=\langle+z|\psi\rangle^*\langle\phi|+z\rangle^*=\left(\langle\phi|+z\rangle\langle+z|\psi\rangle\right)^*=\left(\langle\phi|\mathbb{\hat P}_+|\psi\rangle\right)^*.$$
Therefore $\mathbb{\hat P}_+$ is Hermitian.
For an eigenstate $\mathbb{\hat P}_+|\lambda\rangle=\lambda|\lambda\rangle$ so $$\mathbb{\hat P}_+^2|\lambda\rangle=\lambda\mathbb{\hat P}_+|\lambda=\lambda^2|\lambda\rangle$$ but since $\mathbb{\hat P}_+^2=\mathbb{\hat P}_+$ we have $\lambda^2=\lambda$ therefore $\lambda=0,1.$
A: Two basic facts are that

*

*if $|w\rangle$ and $|v\rangle$ are two vectors, then

$$
(|w\rangle\langle v|)^\dagger=|v\rangle\langle w|
$$
(*)


*in the general case $P=\sum\limits_{i}|i\rangle\langle i|$, where $|i\rangle$ are vectors of an othonormal basis. In your case $P=|\psi\rangle\langle \psi|$
(**)
Therefore, in order to prove that $P^\dagger=P$, we have simply that $P=|\psi\rangle\langle \psi|=(|\psi\rangle\langle \psi|)^\dagger=P^\dagger$.
To see this (*) we use the uniqueness of the Hermitian operator defined as
$$
(|v\rangle, A|w\rangle)=(A^\dagger|v\rangle, |w\rangle)
$$
with $A=|v\rangle\langle w|$, by substitution is easy to see that the unique way to obatin equality is setting $A^\dagger=|w\rangle\langle v|$.
For the (**) that enable us to use the same proof scheme for the general case of $P=\sum\limits_{i}|i\rangle\langle i|$, we can use the linearity of the inner product in its second argument to obtain $P^\dagger=\sum\limits_{i}(|i\rangle\langle i|)^\dagger$:
$$
(|v\rangle, (A+B)|w\rangle)=(|v\rangle, A|w\rangle)+(|v\rangle, B|w\rangle)=(A^\dagger|v\rangle, |w\rangle)+(B^\dagger|v\rangle, |w\rangle)=(|v\rangle, (A+B)^\dagger|w\rangle)
$$
