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Let $T\in \mathcal{L}(V)$ be an operator such that $T^{2}=T$. Prove the following are equivalent:

a) $T$ is self-adjoint

b) $T$ is normal

c) $v-Tv$ is orthogonal to $Tv$ for all $v \in V$

Attempt:

$\bullet a) \Rightarrow b)$ $T^{2}=T$ is the definition of idempotent operator. Since $T$ is idempotent and self-adjoint, $TT^{*}=T^{2}=T^{*}T$.

$\bullet b) \Rightarrow a)$ A normal, idempotent linear operator must be self-adjoint

So I have $a) \Leftrightarrow b)$ but I don't know how to get c) in there.

Shambhala
  • 991
  • $a\implies c$ should be easy. As for $c\implies a$ why not come up with a basis for V via T's eigenvectors and run Gram Schmidt? – user8675309 Oct 30 '21 at 00:05
  • It seems, @user8675309, that the op is struggling getting c. Telling them $a\to c$ is should be easy, is not helpful. – amWhy Oct 31 '21 at 20:25
  • @amWhy That was not helpful. I gave a hint for $c\implies a$ which leaves the ball in OP's court. If the hint doesn't resonate or $a\implies c$ remains a sticking point then they can say so. A lot of the time if someone is stuck on an $\iff$ and they manage to solve one direction (perhaps with a hint), this can unlock ideas for the other half. – user8675309 Nov 01 '21 at 04:43

1 Answers1

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$b) \implies c)$

Assume $T$ is normal. This implies

$$\text{Im}(T)^\perp=\ker(T^*)=\ker(T)=\text{Im}(I-T).$$

Therefore, $\left \langle (I-T)v|Tw \right \rangle=0,\forall v,w \in V$. If $v=w$, $v-Tv$ is orthogonal to $Tv$.

$c) \implies a)$

Assume $v-Tv$ is orthogonal to $Tv$ for all $v\in V$. This means

$$0=\left \langle Tv|v-Tv \right \rangle=\left \langle Tv|(I-T)v \right \rangle=\left \langle ((I-T)^* T)v|v \right \rangle.$$

If $V$ is a finite $\mathbb{C}$-space and $P\in \mathcal{L}(V)$ is such that $\left \langle Pv|v \right \rangle=0,\forall v\in V$, then $P=0$. This implies

$$0=(I-T)^* T=T-T^*T.$$

That is, $T=T^*T$.

$$T^*=(T^*T)^*=T^*(T^*)^*=T^*T=T,$$

so $T$ is self-adjoint.

Shambhala
  • 991