# Showing that $U$ is unitary iff this kernel contains only zero vector

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?

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