The following is a question from a course on quantum mechanics:
Prove that $\hat P_i = \lvert i \rangle \langle i \rvert$ is a Projection operator as long as $\lvert i \rangle$ is normalized. Physically, the Projection operator projects out the part of the state that is linear in $\lvert i \rangle$.
I just don't understand the solution to this, which is as follows:
I wasn't being lazy in not typesetting the solution above, I just didn't see the point in doing so since my question is, how do we know $\lvert i \rangle$ is normalised?
When I see the word 'normalised', I normally think of some numerical factor required in a function which when integrated over all space/real-line comes out to unity. For example, the factor $\dfrac{1}{\sigma \sqrt{2\pi}}$ in a Gaussian function: $$G(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{\left(x-\mu\right)/2{\sigma}^2}$$
$$\int_{-\infty}^\infty G(x)dx=\int_{-\infty}^\infty \frac{1}{\sigma \sqrt{2\pi}}e^{\left(x-\mu\right)/2{\sigma}^2}dx=1\tag{1}$$
Since $(1)$ is equal to unity, I interpret this as $G(x)$ being normalized.
In the solution, they start with $\hat P_i = \lvert i \rangle \langle i \rvert$ and then square it to get ${\hat P_i}^2 = \lvert i \rangle \langle i \rvert i \rangle \langle i \rvert$ and then they take the conjugate transpose ${\hat P_i}^\dagger=\left(\lvert i \rangle \langle i \rvert\right)^\dagger$. But, how does any of this show that $\lvert i \rangle$ is normalized?
As far as I can tell the solution doesn't seem to prove $\lvert i \rangle$ is normalized, it just assumes it is.
An alternate way of asking this question is, how does the so-called 'proof' embedded as an image prove that $\lvert i \rangle$ is normalized?
$|i\rangle$ is called normalized when $\langle i|i\rangle=1$, that is, it has unit norm. Clearly, in the proof, $\hat P_i^2 = \hat P_i$ if and only if $\langle i|i\rangle=1$.