Prove that $T$ is a normal operator if and only if $\|{Tv}\|^2 = \|{T^*v}\|^2$ for every vector $v \in V$

$(\Rightarrow)$ $T$ is normal, so we have $TT^* = T^*T$. Then $\langle TT^*v, v \rangle = \langle T^*v, T^*v \rangle = \|T^*v\|^2$ On ther other hand $\langle T^*Tv, v \rangle = \langle Tv, Tv \rangle = \|Tv\|^2$, so $\|T^*v\|^2 = \|Tv\|^2$

But I don't know how to show the other implication.


$\newcommand\ip[2]{\langle#1,#2\rangle}$For complex scalars this is very simple. If $||Tv||^2=||T^*v||^2$ it follows that $$\ip{v}{T^*Tv}=\ip{v}{TT^*v}$$for every $v$. So you only need to show that if $\ip v{Av}=0$ for all $v$ then $A=0$. This depends on the fact that we're talking about complex scalars (consider a rotation of the plane for a conterexample in $\Bbb R^2$).

The fact that $\ip{w+v}{A(w+v)}=0$ shows that $$\ip v{Aw}+\ip w{Av}=0.$$ Replacing $w$ by $iw$ shows that $$-i\ip v{Aw}+i\ip w{Av}=0.$$Hence $\ip w{Av}=0$ for all $v,w$, so $Av=0$.

Edit It's also easy in the real case. I have to confess I got a hint from those other posts; however one can get the idea from those other posts that it depends on the spectral theorem, and it's much easier than that. Say $A=T^*T-TT^*$ As above we see that $\ip v{Aw}+\ip w{Av}=0$. But now $\ip w{Av}=\ip{Av}w=\ip v{A^*w}$. So $\ip v{(A+A^*)w)}=0$, or $A+A^*=0$. Otoh it's clear that $A^*=A$. So $A=0$.

  • $\begingroup$ @B.Mehta You're aware that in general $(XY)^*=Y^*X^*$, right? And that $X^{**}=X$? What do you get when you apply those two facts to $(T^*T-TT^*)^*$??? $\endgroup$ May 28 '18 at 15:37
  • $\begingroup$ Ah my mistake, for some reason I read that as $A^* = A$ for all real operators which seemed very false, instead of just for this particular $A$. Thanks! $\endgroup$
    – B. Mehta
    May 28 '18 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.