This is Exercise 16.(c) from Conway's Functional Analysis book.

Suppose $H$ is a Hilbert space and $T$ is a compact operator on $H$. Assuming the result that $\exists A$ positive operator and $U$ a bounded operator such that $T=UA$

(NOTE: here $A= \sqrt {T^* T} $ and $U$ is the operator defined as in the construction of the polar decomposition i.e. $\|Uh\|=\|h\|$ when $h \perp \ker(T)$ and $Uh=0$ when $h \in\ker(T)$ i.e. $U(Ah)=Th$ when $h \perp \ker(A)=\bar{Ran(A)}$ and $Uh=0$ when $h\in\ker(A)$ ) ,

I have to prove :

$T=AU \iff T $ is normal

I tried to compute products and tried to show the one direction that ($T=AU \implies T $ is normal) , but since $U$ is just assumed to be bounded, not unitary, got stuck!


For $T$ Normal ,assuming $U$ to be unitary ( we can do that provided $T$ is normal), have been able to show that $AU=UA$ .

But for the other direction, ($T=AU=UA \implies T$ normal), got, $T^*T=AU^*UA$ and $TT^* = AU^*UA$ . So if U is at least normal, we are done! (as mentioned in an answer) . But , we have to be careful, because by construction of $U$ and $A$, it is not clear to me how is $U$ normal. ( since we are trying to show that $T$ is normal, we cannot assume beforehand that $U$ is unitary, like the other direction)

Please help!


I read through the exercise in Conway's book
(Second edition from 1990, Chapter II, §7, Exercise 16, Part c, it's on page 59),
and it deals with the polar decomposition of $\,T=U\!A\,$ for some bounded operator $T$ on $H$. The bounded operators $A$ and $U$ are uniquely determined by $T$ when asked to satisfy

  • $A=\sqrt{T^*T}=|T| $
  • $U$ is a partial isometry with $\,\ker U=\ker T\,$ and $\,\operatorname{im} U =\overline{\,\operatorname{im}T}\,$.

Note that $\,U^*\!\!\:U\,$ is the projector onto $\,\overline{\operatorname{im}|T|}\,$ which implies $\,A=U^*\!\!\:U\!A=U^*T$.
Multiplying with $U$ from the left yields $\,UU^*T=T$.

Consequences for the polar decomposition of $\,T^*$ $$=(U\!A)^*=AU^*= U^*\big(U\!AU^*\big)$$ Thus $\,U|T|U^*=\,|T^*|\,$ and $\,T^*=U^*|T^*|\,$ is the (unique) polar decomposition of $\,T^*$.

In the sequel $T$ is not required to be compact. I would guess that mentioning that property is due to the early placement of the exercise in the book in which Conway proceeds from the special to the more general.

It is shown that $\:T=AU\iff T$ is normal.

If $T=U\!A=AU\,$ then $T^*=AU^*=U^*\!A$, so the assumption tells us that $A$ also commutes with $U^*$. Hence $$\,T^*= U^*\big(U\!AU^*\big)= U^*UU^*\!A = U^*\!A$$ is the polar decomposition of $\,T^*$. We have $\,UT^*=A\,$ by the above "Note that ..." applied to $\,T^*$, and conclude that $T\,T^*=A\,UT^*=A^2=T^*T$.

$\,T=U\!A\,$ yields $\,T^*=AU^*$. Hence $\,UA^2U^* =T\,T^*=T^*T= A^2\,$ by normality. Taking the positive square-root (which is unique) gives $\,U\!AU^* = A$, being equivalent to $\,UA = A\,U$ by the properties of $U$.

  • $\begingroup$ Please explain how you get "Note that one always has $𝑈^*𝑇=𝐴$" and also from that how you get $UT^*=A$ in the first part. $\endgroup$ – reflexive Mar 17 '19 at 8:06
  • $\begingroup$ @Coherent Good action from yours to ask for details, it improved the answer! And thanks for the bounty award. $\endgroup$ – Hanno Mar 17 '19 at 14:14

With $T=AU$ you get $TT^* = AUU^*A$, and with $T=UA$ you get $T^*T=AU^*UA$. Now it is important that $U$ is unitary (or at least normal). In that case both expressions are the same.

If $U$ is not normal then this does not have to be true, take for example $A=\Bbb 1$ and $U=T$, then $T=UA=AU$ but $T$ is not necessarily normal.

  • $\begingroup$ Is it true that in $T=UA$, U is unitary iff T is normal. Then I think you can use that U is unitary in the first part. Otherwise you only know that T is normal implies U is unitary, so you can't use it to prove that T is normal. $\endgroup$ – reflexive Mar 9 '19 at 1:31
  • 2
    $\begingroup$ No, take $U=T$ and $A=\Bbb1$. You can have normal $T$ without having unitary $U$. You have likely missed at some point that $T=UA$ is supposed to be the polar decomposition, which is unique and has $U$ unitary. $\endgroup$ – s.harp Mar 9 '19 at 11:31

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