Let $V$ be a finite dimensional complex inner product space , $T$ be a normal operator on $V$ such that $T^*T=T^2$ , then is it true that $T^*=T$ i.e. $T$ is self-adjoint ?
I only know that $T$ is self-adjoint when $T$ is normal and idempotent . Please help.
$T$ is self-adjoint even if we drop the assumption of normality.
Proof.
To prove that $T=T^*$ it is sufficient to show that all eigenvalues of the selfadjoint matrix $\frac{T-T^*}{2i}$ are zero.
Assume that there exists a nonzero eigenvalue $\lambda$ of $\frac{T-T^*}{2i}$, i.e., (we drop $2i$) $$ (T-T^*) f = \lambda f\ne 0.\qquad\qquad (1) $$ Since $T^2 =T^*T=(T^2)^* = (T^*)^2$, it follows that
$$ 0=T^*(T-T^*).\qquad\qquad\qquad (2) $$ By (1) and (2), $$ 0\cdot f=T^*(T-T^*)f = T^*\lambda f\quad (3). $$ By (1) and (3), $T^*f=0$ and $Tf=\lambda f$. Hence, $$ 0=T^*\lambda f=T^*(\lambda f) =T^*(Tf)=T^2f=\lambda^2f, $$ a contradiction to $\lambda f\ne 0$ (see (1)). Thus, $T=T^*$.