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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.

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    $\begingroup$ It is true: check the eigenvalues of $T$ and $T^*$. They are real $\endgroup$ – user251257 Aug 29 '16 at 14:58
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$T$ 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, $T=T^*$.

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  • $\begingroup$ Did it need normality of $T$ anywhere ? $\endgroup$ – user228169 Aug 29 '16 at 15:25
  • $\begingroup$ i did not use the normality of $T$. If $T$ is normal, and $Tf=T^*f =\lambda f$, then $\bar{\lambda}\lambda f = T^*Tf=T^2f=\lambda^2 f$, i.e., $\lambda = \bar{\lambda}$. Hence, $T$ is selfadjoint. $\endgroup$ – Ignat Domanov Aug 29 '16 at 15:33

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