If $T^2=TT^*$ then can i conclude that $T=T^*$? let $B(H)$ be all bounded operator on Hilbert space H. If $T^2=TT^*$ then can i conclude that $T=T^*$? I think this is true if T is one to one. Can i construct an example that shows it is not true for any T?
 A: For a bounded linear operator $T$, we have that $H = \mathrm{ker}(T) \oplus \overline{\mathrm{Im}(T^*)}$.
The two operators $T$ and $T^*$ coincide on $\mathrm{Im}(T^*)$ thanks to the hypothesis: we can rewrite it as $T^* T^* = T T^*$, so if $y = T^*x$, then $$Ty = TT^*x = T^*T^*x = T^*y$$
so they also coincide on its closure (being continuous).
It remains to show that they coincide on $\mathrm{Ker}(T)$, in other words that $T^*$ is zero on $\mathrm{Ker}(T)$. But if $Tx = 0$, then:
$$\langle T^*x, T^*x \rangle = \langle x, TT^*x \rangle = \langle x, T^2x \rangle = 0$$
Therefore $T^*x = 0$. QED.
A: The answer is YES, and the proof is nontrivial. (I hope that somebody else comes up with a simpler proof.)
If $T$ is a bounded operator in a Hilbert space the $T=S+A$, where $S=\frac{1}{2}(T+T^*)$ symmetric and $Α=\frac{1}{2}(T῏T^*)$ antisymmetric.
The target is to show that $A=0$.
So, $T^2=TT^*$ implies that $(S+A)(2A)=0$ or 
$$SA=-A^2.\tag{1}$$
Now 
$$
SA=-AA^2 \Longrightarrow S^2A=S(SA)=S(-A^2)=-(SA)A=A^3,
$$ 
and in general, for every $\lambda\in\mathbb C$, 
$$
S^nA=(-1)^nA A^{n}\Longrightarrow  \lambda^nS^nA=(-1)^nA \lambda^nA^{n}\Longrightarrow
\mathrm{e}^{\lambda S}A=A\,\mathrm{e}^{-\lambda A},
$$
where $e^{tB}=\sum_{n=0}^\infty \frac{B^n}{n!}$. Such operator (the exponential) always exists, and it is bounded. We also get that
$$
A=\mathrm{e}^{-\lambda S}A\,\mathrm{e}^{-\lambda A}.
$$
As $A$ is antisymmetric, and $S$ symmetric, then $\mathrm{e}^{(\lambda-\bar\lambda)A}$ and 
$\mathrm{e}^{(\lambda+\bar\lambda)S}$ are orthogonal matrices, which means that
$$
\|\mathrm{e}^{(\lambda-\bar\lambda)A}\|=\|\mathrm{e}^{(\lambda+\bar\lambda)S}\|=1,
$$
and hence is we set
$$
f(\bar\lambda)=\mathrm{e}^{(\lambda+\bar\lambda)S}A\mathrm{e}^{(\lambda-\bar\lambda)A}=\mathrm{e}^{(\lambda+\bar\lambda)S}\mathrm{e}^{-\lambda S}A\,\mathrm{e}^{-\lambda A}\mathrm{e}^{(\lambda-\bar\lambda)A}
$$
or
$$
f(\bar\lambda)=\mathrm{e}^{(\lambda+\bar\lambda)S}A\mathrm{e}^{(\lambda-\bar\lambda)A}=\mathrm{e}^{\bar\lambda S}A\,\mathrm{e}^{-\bar\lambda A},
$$
then
$$
\|f(\bar\lambda)\|=\|\mathrm{e}^{(\lambda+\bar\lambda)S}A\mathrm{e}^{(\lambda-\bar\lambda)A}\|\le \|A\|.
$$
Thus the entire analytic function $f(\bar\lambda)=\mathrm{e}^{\bar\lambda S}A\,\mathrm{e}^{-\bar\lambda A}$ is bounded, and hence constant. Thus
$$
A=f(0)=f(\lambda)=\mathrm{e}^{\lambda S}A\,\mathrm{e}^{-\lambda A},
$$
and hence
$$
A\mathrm{e}^{\lambda A}=\mathrm{e}^{\lambda S}A, \quad\text{for all}\,\,\lambda\in\mathbb C.
$$
Dffferentiating the above with respect to $\lambda$ and setting $\lambda=0$, we get
$$
A^2=SA \tag{2}
$$
Now $(1)$ and $(2)$ imply that $A^2=0$. This means that for all $x\in H$
$$
0=\langle A^2 x,x\rangle=\langle A x,-Ax\rangle=-\|Ax\|^2,
$$
and hence $Ax=0$, and thus A=0$.
