If $T$ is a normal operator such that $T^2 = T^3$ then $T$ is idempotent

Question

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.

Answer 1

Hint: A normal operator is diagonalizable. Thus its minimal polynomial has only simple roots.

Answer 2

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. $OE\Delta$.

Answer 3

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.