Proposition 13

Let E be a complex Hilbert space and $u$ an endomorphism of E. The following conditions are equivalent:

(i) The endomorphism $u$ is normal;

(ii) There exists a unitary element $v$ of $\mathscr{L}(E)$, permutable with $|u|$, such that $u=v|u|$.

Let $(j,|u|)$ be the polar decomposition of $u$. Suppose that $u$ is normal. Then $|u^*|= (uu^*)^{1/2}= (u^*u)^{1/2}=|u|$. Prop. 11 then implies $|u|j=|u^*|j=ju^*j=j|u|$. Moreover, $j$ leaves stable the supplementary orthogonal subspaces Ker($|u|$) $=$ Ker($j$) and Im($|u|$) $=$ Im($j$) (prop. 10). Let $v$ be the element of $\mathscr{L}(E)$ which coincides with $j$ on Ker($u$)$^{\circ}$ and with the identity mapping on Ker($u$). Since $j$ induces an isometry of Ker($u$)$^{\circ}$ onto Im($u$) $=$ Ker($u^*$)$^{\circ}=$ Ker($u$)$^{\circ}$ (since $u$ is normal), the endomorphism $v$ is unitary; moreover, it is permutable with $|u|$, since $j|u|=|u|j$, and one has $u=v|u|$.

Conversely, let $v$ be a unitary element of $\mathscr{L}(E)$, permutable with $|u|$, such that $u=v|u|$. One has $uu^*=v|u|^2v^*=|u|^2vv^*=|u|^2=u^*u$, therefore $u$ is normal.

Let E be a complex Hilbert space and $u\in \mathscr{L}(E)$. Let $(j,|u|)$ be the polar decomposition of $u$. It may happen that $j$ commutes with $|u|$ without $u$ being normal (Exercise 11 of I, p. 189).