If $T$ is a normal operator such that $T^2 = T^3$ then $T$ is idempotent I have tried to solve this problem for some days now, but I am stuck with a lot of calculations that lead nowhere. Any hints or suggestions will be the most appreciated.
Question
Let $V$ be a finite-dimensional space over $\mathbb{C}$ and $T \in L(V)$ be a normal operator such that $T^3 = T^2$. Show that $T$ is idempotent. 
 A: Hint: A normal operator is diagonalizable.  Thus its minimal polynomial has only simple roots.
A: Since $T$ is normal, we can find a basis where its matrix is diagonal. The possible eigenvalues are $0$ and $1$.
A diagonal matrix with only $0$ or $1$ on the diagonal is idempotent.
A: A normal operator such as $T$ which acts on a finite-dimensional vector space over $\Bbb C$ may always be diagonalized by a unitary $U$:
$U^\dagger TU = \text{diag}(\mu_1, \mu_2, \ldots \mu_n); \tag 1$
we note that the transformation
$T \to U^\dagger T U \tag 2$
preserves powers of $T$, that is,
$U^\dagger T^m U = (U^\dagger T U)^m, \; m \in \Bbb N; \tag 3$
it is easy to see (3) via induction with base case $m = 1$ which holds trivially:
$U^\dagger T U = U^\dagger T U; \tag 4$
if now
$U^\dagger T^k U = (U^\dagger T U)^k, \tag 5$
then
$U^\dagger T^{k + 1} U = U^\dagger T^k T U = U^\dagger T^k UU^\dagger T U = (U^\dagger T U)^k U^\dagger T U = (U^\dagger T U)^{k + 1}, \tag 6$
where we have used the definition of unitarity,
$U^\dagger U = UU^\dagger = I \tag 7$
in the derivation (6).
If we now diagonalize $T$ and use the given relationship
$T^3 = T^2, \tag 8$
then we find that
$\text{diag}(\mu_1^3, \mu_2^3, \ldots, \mu_n^3) = \text{diag}(\mu_1^2, \mu_2^2, \ldots, \mu_n^2), \tag 9$
that is,
$\mu_i^3 = \mu_i^2, \; 1 \le i \le n; \tag{10}$
now if $\mu_i \ne 0, \tag{11}$
then from (10),
$\mu_i = 1; \tag{12}$
since therefore each
$\mu_i \in \{0, 1\}, \tag{13}$
we have
$\mu_i^2 = \mu_i, \; 1 \le i \le n, \tag{14}$
or
$(\text{diag}(\mu_1, \mu_2, \ldots, \mu_n))^2 = \text{diag}(\mu_1^2, \mu_2^2, \ldots, \mu_n^2) = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n); \tag{15}$
now since, from (1) and (7),
$T = U \text{diag}(\mu_1, \mu_2, \ldots, \mu_n) U^\dagger, \tag{16}$
we see, using (3) with $T$ replaced by $\text{diag}(\mu_1, \mu_2, \ldots, \mu_n)$ (legitimate since the proof is the same in either case),
$T^2 =  (U \text{diag}(\mu_1, \mu_2, \ldots, \mu_n) U^\dagger)^2 = U(\text{diag}(\mu_1, \mu_2, \ldots, \mu_n)^2 U^\dagger$
$= U\text{diag}(\mu_1^2, \mu_2^2, \ldots, \mu_n^2) U^\dagger = U\text{diag}(\mu_1, \mu_2, \ldots, \mu_n)U^\dagger = T, \tag{17}$
showing $T$ is idempotent.  $OE\Delta$.
