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If $T$ is normal operator and idempotent i.e. $T=T^2$, then show that $T$ is self adjoint

Robert Lewis
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user794607
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    Please solve this – user794607 May 31 '20 at 04:10
  • Welcome to MSE! Please check out our site guidelines. In particular, it is a requirement for you to include some of your own thoughts and efforts towards solving a problem (they don't have to have gone anywhere useful). This helps us gauge your level, and write our answers appropriately. If you don't include your efforts, your post may be voted down and closed. – user793679 May 31 '20 at 04:15
  • @user793679 I tried it many times and I failed since T is normal so TT=TT AND T is idempotent so T =T^2 and TT T=TTT what will be the next step to solve the problem – user794607 May 31 '20 at 04:38
  • Since we have adjoints, we must be working in a Hilbert space. Are we in $\Bbb{R}^n$ or $\Bbb{C}^n$? Maybe an arbitrary finite-dimensional inner product space? Or are we in a (possibly infinite-dimensional) Hilbert space? – user793679 May 31 '20 at 04:41
  • The finite dimensional case is here: https://math.stackexchange.com/questions/319197/normal-idempotent-operator-implies-self-adjointness?rq=1 – Kavi Rama Murthy May 31 '20 at 04:47

2 Answers2

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Note that $T^*$ is also normal and idempotent. Then, \begin{align*} \|T^* Tx - Tx\|^2 &= \langle T^*Tx - Tx, T^* Tx - Tx\rangle \\ &= \langle TT^*x - TTx, T^*Tx - Tx \rangle \\ &= \langle T^*x - Tx, T^*T^*Tx - T^*Tx\rangle \\ &= \langle T^*x - Tx, T^*Tx - T^*Tx\rangle = 0, \end{align*} hence $T^*T = T$. Symmetrically, $TT^* = T^*$, and hence $T = T^*$.

user793679
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If $V$ is an inner product space of finite dimension over $\mathbb{R}$

Hint $T$ is normal i.e., $T=UDU^*$, where $U^{-1}=U^*$ and $D$ is diagonal.

Unknown
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