If any given linear transformation $T$ over a complex inner product vector space satisfies this equation: $T^2 = T + T^*$, prove or disprove that $T$ is hermitian.
I've tried finding disproving example firsts but couldn't find the, So I went on proving the statement.
From $T^2=T+T^*$ I can take the adjoint and thus getting $(T^2)^* = (T+T^*)^* = T+T^* = (T^*)^2$ and therefore getting that ${T^*}^2 = T^2$. But that doesn't yet prove me that $T$ is hermitian. I also tried to show that $\langle Tv,u\rangle = \langle v,Tu\rangle$ for every $u,v$ but didn't work that out either. How can I take it from here?
Thanks.
You have $T^*=T^2-T$. So, $T^*$ commutes with $T$, so $T$ is normal, and so diagonalizable by a Hermitian matrix. If $\alpha$ is an eigenvalue of $T$ then $\overline\alpha=\alpha^2-\alpha$. Then $\alpha^2=\alpha+\overline\alpha$ is real, and $\alpha=a$ or $ai$ for some real $a$. But in the latter case $-ia=-a^2-ia$, so $a=0$. So $T$ is normal, with real eigenvalues, and so is Hermitian.