How numerical radius help us to conclude an operator is normal and partial isometry? In Furuta's book, "Invitation to Linear Operators" there is a theorem, theorem 2 in 3.7.3, that says:
If $T^k=T$ for some integer $k\ge 2$ and if $w(T)\le 1$, then $T$ is the direct sum of a unitary operator and zero, that is $T$ is normal and partial isometry.
I have tried several ways to solve this but unfortunately I couldn't do that.
Since I need to proof of this theorem, I will be glad if someone could help me for it. 
 A: The polynomial $p(\lambda)=\lambda(\lambda^{k-1}-1)$ has distinct roots
$$
   0,\alpha,\alpha^{2},\cdots,\alpha^{k-1},\;\;\;\; \alpha=e^{2\pi i/(k-1)}.
$$
Therefore, $X$ decomposes into the direct (not necessarily orthogonal) sum
$$
      X = M_0\oplus M_1\oplus M_2\oplus \cdots \oplus M_{k-1},\\
      M_0=\mathcal{N}(T),\;\;\;M_{j}=\mathcal{N}(T-\alpha^{j}I).
$$
In order to show that $T$ is normal on $M_1\oplus M_2\oplus\cdots\oplus M_{k-1}$, it is necessary and sufficient to show that $M_l\perp M_m$ for $l \ne m$. Let $x_l \in M_l$ and $x_m \in M_m$ be unit vectors with $l \ne m$. Then
\begin{align}
    &(T(x_l+\rho e^{i\theta}x_m),x_l+\rho e^{i\theta}x_m) \\
  & =\alpha^{l}+\rho e^{-i\theta}\alpha^{l}(x_l,x_m)+\rho e^{i\theta}\alpha^{m}(x_m,x_l)+\rho^{2}\alpha^{m} \\
  & = \alpha^{l}+\rho^{2}\alpha^{m}+\rho\alpha^{l}e^{-i\theta}(x_l,x_m)+\rho\alpha^{m}\overline{e^{-i\theta}(x_l,x_m)} \\
  & = \alpha^{l}+\rho^{2}\alpha^{m}+\alpha^{(l+m)/2}2\rho\Re(\alpha^{(l-m)/2}e^{-i\theta}(x_l,x_m)) \\
  & = \alpha^{l}\{1+2\rho\alpha^{-(l-m)/2}Re(\alpha^{(l-m)/2}e^{-i\theta}(x_l,x_m))\}+\rho^{2}\alpha^{m}.
\end{align}
Similarly,
$$
       \|x_l+\rho e^{i\theta}x_m\|^{2}=1+2\rho\Re(e^{-i\theta}(x_l,x_m))+\rho^{2}
$$
Choose $\theta$ so that $\Re(e^{-i\theta}(x_l,x_m))=0$, which is the same as choosing $\theta$ so that
$$
    e^{-i\theta}(x_l,x_m)=\pm i|(x_l,x_m)|.
$$
Then, ignoring $\rho^{2}$ terms (thinking of $\rho$ as small):
$$
     \frac{(T(x_l+\rho e^{i\theta}x_m),x_l+\rho e^{i\theta}x_m)}{\|x_l+\rho e^{i\theta}x_m\|^{2}}\approx \alpha^{l}\left(1\pm 2\rho \alpha^{-(l-m)/2}\Im(\alpha^{(l-m)/2})|(x_l,x_m)|\right).
$$
The right side can be shown to be greater than 1 in magnitude by choosing $\pm$, if necessary, unless $(x_l,x_m)=0$ or unless $\alpha^{(l-m)/2}=\pm 1$. So $(x_l,x_m)=0$.
