In our introductory functional analysis class, we proved the following:

Theorem: Let $T \in B(H)$. Denote by $P$ the orthogonal projection onto $(\ker(T))^{\perp}$. There exists a unique operator $U \in B(H)$ satisfying $$T = U |T|$$ and $$U^{*} U = P. $$ This is the polar decomposition of $T$.

I also know the formula $U^{*} U = P$ is equivalent to the following statement:$$ || Ux || = ||x||, \qquad \text{for all} \ x \in (\ker T)^{\perp}$$ and $$ Ux = 0, \qquad \text{for all} \ x \in \ker T. $$

I now have to solve the following:

Problem: Let $T = U | T|$ be the polar decomposition of a bounded operator $T$. Prove that $U$ is unitary if and only if $\ker(T) = \ker(T^{*}) = \left\{0 \right\}. $

Attempt: $\Rightarrow $ This part is straightforward I think. Assume $U$ is unitary. So that means that $\langle U x, Uy \rangle = \langle x, y \rangle$ for all $x, y \in H$. Now let $x \in \ker(T)$. We wish to show that $x = 0$. I have $$ ||x||^2 = \langle x, x \rangle = \langle Ux, Ux \rangle = 0 $$ by the equivalent statement of the above theorem. Hence $x = 0$. However, I'm not sure how to show that $\ker(T^{*}) = \left\{0 \right\}$.

I wanted to use the relation $$\ker(T^{*}) = (\text{Im} (T))^{\perp} $$ but I don't know how.

$\Leftarrow$. Also this part is not clear to me. Let $x, y \in H$. I wish to show $\langle x, y \rangle = \langle U x, Uy \rangle$. How to use my assumptions to prove this?

  • $\begingroup$ Doesn't the polar decomposition tell you something about $|T|$ as well? That should help. $\endgroup$ – астон вілла олоф мэллбэрг Nov 16 at 9:10
  • $\begingroup$ For $"\Leftarrow" $ just use $\{0\}^\perp = H$. $\endgroup$ – Severin Schraven Nov 16 at 10:33
  • $\begingroup$ So $P$ is orthogonal projection onto $H$. Does that mean $P = I$ and then $U^{*} U = I$ so that $U $ is unitary? $\endgroup$ – Kamil Nov 16 at 16:00

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