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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.

janusz
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1 Answers1

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$\newcommand\ip[2]{\langle#1,#2\rangle}$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 $A=0$.

David C. Ullrich
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  • @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 $(T^T-TT^)^$??? – David C. Ullrich May 28 '18 at 15:37
  • 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! – B. Mehta May 28 '18 at 15:39