Exercise 2.32

[archmageirvine]

The proof here for Exercise 2.32 appears to assume P^2=P implies P is a projector. I am not certain this is true (the converse is certainly true, a projector satisfies P^2=P as indicated by Exercise 2.16), but could there not be other non-projectors that also satisfy P^2=P ?

[Fadelis98]

P^2 = P is just the definition of a Projector

[Techmaster21]

Your concern is 100% valid, I had the same issue with the given proofs I found online. $P^2 = P \implies P \text{ is a projector}$ is only true if $P$ is also taken to be Hermitian. Otherwise we can have, for instance,

$$ \begin{aligned} P &= \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \\ P^2 &= \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} = P \end{aligned} $$

Or for a more complex example (pun intended)

$$ \begin{aligned} P &= \begin{bmatrix} -\frac{1}{2}\sqrt{5} + \frac{1}{2} & i \\ i & \frac{1}{2}\sqrt{5} + \frac{1}{2} \end{bmatrix} \\ P^2 &= \begin{bmatrix} -\frac{1}{2}\sqrt{5} + \frac{1}{2} & i \\ i & \frac{1}{2}\sqrt{5} + \frac{1}{2} \end{bmatrix} \begin{bmatrix} -\frac{1}{2}\sqrt{5} + \frac{1}{2} & i \\ i & \frac{1}{2}\sqrt{5} + \frac{1}{2} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2}\sqrt{5} + \frac{1}{2} & i \\ i & \frac{1}{2}\sqrt{5} + \frac{1}{2} \end{bmatrix} = P \end{aligned} $$

In both cases, P is not a projector. For one they are not Hermitian, which all projectors are.