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Let $X$ be a Hilbert space and $\lVert Tx \rVert =\lVert T^*x\rVert$, where $T^*$ is the adjoint of $T$. Prove that $TT^*=T^*T$.

I proved in this way. $\lVert Tx \rVert^2 -\lVert T^*x\rVert^2=\langle Tx,Tx \rangle - \langle T^*x, T^*x \rangle = \langle T^*Tx,x\rangle - \langle TT^*x,x\rangle =\langle(TT^*-T^*T)x,x \rangle=0$, then $TT^*=T^*T$.

However, I don't think $\langle Tv,v \rangle =0$ implies $T=0$ in general.

I can't find other way to prove the statement though.

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    $\begingroup$ The implication $\langle Tv,v\rangle=0$ for all $v$ implies $T=0$ holds for self-adjoint operators, and $TT^*-T^*T$ is certainly self-adjoint. $\endgroup$ – Aweygan May 8 '17 at 22:42
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If $b(x,y)$ is a sesquilinear form on a complex vector space $X$, then $b(x,y)$ can be recovered from its quadratic form $q(x)=b(x,x)$ through the identity $$ b(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}q(x+i^n y). $$ So $(T^*Tx,x)=(TT^*x,x)$ holds for all $x$ iff $(T^*Tx,y)=(TT^*x,y)$ for all $x,y$, which is equivalent to $T^*T=TT^*$.

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