Let $T:V\to V$ be a linear map, where $V$ is an inner product space. Given $T^2=\frac{T+T^*}{2}$, prove that $T$ is normal and that $T=T^2$.

Question

Let $T:V\to V$ be a linear map, where $V$ is a finite-dimensional inner product space.

Given $T^2=\frac{T+T^{*}}{2}$, I would like to prove that $T$ is normal and that $T=T^2$.

The first part, to prove that $T$ is normal, is quite easy:
from $T^2=\frac{T+T^{*}}{2}$ we'll have $\,T^{*}=2T^2 - T$, and as $TT^{*} = 2T^3 -T^2 = T^{*}T$.

Now, I need to prove that $T=T^*$, then $T = T^2$ is immediate.
I have been thinking about this question for some days and I can't figure out in what technique to prove that $T = T^{*}$.

Answer 1

Hint: Each eigenvalue $\lambda$ of $T$ must satisfy $\lambda^2 = \frac{\lambda + \bar \lambda}{2}$. For $\lambda = a + bi$ ($a,b$ real), this means that $$ (a + bi)^2 = a \implies \begin{cases} a^2 - a - b^2 = 0\\ 2 ab = 0. \end{cases} $$ Consider the cases of $a = 0,b = 0$ separately.

Answer 2

We have $T^2 = \frac{T + T^*}{2}$. Therfore, $(T^2)^* = \frac{T^* + (T^*)^*}{2} = \frac{T + T^*}{2} = T^2$. Combining, $(T^2)^* = T^2$. Therefore $T^2$ is Hermitian. Diagonalize $T^2$ to write $T^2 = V D V^{-1}$, which indicates $T = V D^{\frac{1}{2}} V^{-1}$. It indicates $T$ is also Hermitian.

Answer 3

for a slightly different approach, write
$T^2=\frac{1}{2} \big(T+T^*\big) = (T^*)^2$
$T=\frac{1}{2}\big(T+T^*\big)+\frac{1}{2}\big(T-T^*\big)$
and estimate the skew hermitian part by considering
$\dim \ker\Big(\big(T-T^*\big)\Big) = \dim \ker\Big(T\big(T-T^*\big)T^*\Big)$
(due to normality of $T$)

$T\big(T-T^*\big)T^*$
$=T^2 T^* - T(T^*)^2$
$= \frac{1}{2}\Big(\big(T+T^*\big)T^* - T\big(T+T^*\big)\Big)$
$= \frac{1}{2}\Big(TT^*+(T^*)^2 - T^2-TT^*\Big)$
$=\mathbf 0$
$\longrightarrow \big(T-T^*\big) = \mathbf 0$
$\longrightarrow T=\frac{1}{2}\big(T+T^*\big) = T^2$

Answer 4

The following also works in an infinite-dimensional inner product space.

The initial condition $\,T^2=\frac12 (T^*+T)\,$ shows that $\,T^2$ is self-adjoint. Writing it equivalently as $$T^2-T \;=\; \frac{T^* -T}{2}\tag{1}$$ we get that $\,T$ is idempotent iff $\,T$ is self-adjoint.
Multiply the condition in its initial form with $(1)$ to obtain $$T^3\big(T-\mathbb 1\big) \;=\;T^2\big(T^2-T\big) \;=\; \frac14\big((T^*)^2 -T^2 -T^*T +TT^*\big) \;=\; 0\,,$$ the vanishing is due to the self-adjointness of $T^2$ and $T$ being normal.
Normal operators have the feature that their kernel does not grow upon iteration(${}^1$), i.e., $\ker T^r=\ker T\,$ for all $r=2,3,\dots\,$ Hence $\,\operatorname{im}(T-\mathbb 1)\subset\ker T^3=\ker T$, and $\,T\,(T-\mathbb 1)=0\,$ follows.

---

$\quad 1\;$ See for instance Lemma $1$ in this lecture .