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If we assume that $T^2 = T$, and that $T$ is unitary, I want to prove that $T=T^*$.

I know that there is a solution here: $T^2 = T$ and $T$ is normal implies $T$ is hermitian for a normal matrix, but in my case I'm assuming $T$ is unitary, and I'm asking for proof verification since I couldn't find any reference and my solution seemed really simple, giving me doubts.

Since $T$ is unitary we know that $$T^*T=I$$. Let us multiply by $T$ from the right and we get: $$ T^*T^2=T$$

But from $T^2=T$ we get that: $$T^*T = T$$

Therefore $T=I$ and is hermitian (and even more than that, I have only one possible form for the matrix).

Is this correct? Thanks!

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That's fine. More generally if $T$ is invertible and $T^2=T$ then $T=I$.

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  • $\begingroup$ Yep, actually we don't need to assume that it is hermitian and the proof is still very simple. thanks! $\endgroup$ – Mickey Jul 5 '17 at 19:05

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