If $T^2=TT^*$ then can i conclude that $T=T^*$?

Question

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?

Answer 1

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.

Answer 2

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$.