Why is $T$ normal and idempotent?

Question

Let $V$ be an inner-product space, finite-dimensional over $\mathbb{C}$. Let the operator $T:V\to V$ satisfy $$T^2 = \frac{1}{2}(T+T^*)\,.$$

1. I'd like to prove that $T$ is normal, i.e., $T^*T = TT^*$,
2. and that $\,T^2 - T = 0\,$ holds.

So I've tried the obvious:

$$ \langle T(v), T^*(v) \rangle = \langle T^2(v), v \rangle = \frac{1}{2} \langle (T+T^*)(v), v \rangle = \frac{1}{2} \left(\langle T(v),v \rangle + \langle T^*(v), v\rangle \right) $$

I don't see we can infer something smart from those equalities.

Any ideas?

Answer 1

Solving for $T^*$, we see that $T^* = 2T^2 - T$. This should show easily why $T^*T = TT^*$. (Hint: Multiply $T$ to one side of $T^*$ you just solved for, and do it on the other side. What do you get?)

Edit: As for the second part, show that for all $x\in V$, $\langle (T^2-T)x,(T^2-T)x \rangle =0$. Use the identity for $T^2$ above, and use the fact that $T$ is normal. Things will cancel nicely.

Answer 2

Since $T^{\ast}$ is a polynomial in $T$, we have that $T$ is normal.

Now it follows that $T$ is diagonalizable, so consider any eigen-value $\lambda$ of $T$ : Since $$ 2T^2 - T = T^{\ast} \Rightarrow 2\lambda^2 - \lambda = \overline{\lambda} $$ So if $\lambda = x+iy$, then $$ x^2 - y^2 + 2ixy - 2x = 0 \Rightarrow xy=0 $$ Either way, $\lambda \in \mathbb{R}$. Hence $T = T^{\ast}$, whence $T^2 = T$.