Proposition 11

Let $E_1$ and $E_2$ be complex Hilbert spaces and $u\in \mathscr{L}(E_1; E_2)$. Let $(j,|u|)$ be the polar decomposition of $u$.

a) We have $|u|=j^*u=u^*j$;

b) We have $|u^*|=ju^*=uj^*$;

c) The polar decomposition of $u^*$ is $(j^*,|u^*|)$.

Let us put I = Ker($u$)$^{\circ}$ and F = Im($u$) for the initial subspace and the final subspace of $u$; moreover, we have I = Ker($|u|$)$^{\circ}=$ Im($|u|$) (prop. 10, a)). The mapping $j^*j$ is the orthogonal projector of $E_1$ onto I (loc. cit. and EVT, V, p. 41, prop. 5 (ii)). Therefore $j^*u=j^*j|u|=|u|$, then $u^*j= (j^*u)^*=|u|^*=|u|$, whence a).

Similarly, one calculates $u^*$ = $|u|j^*= (j^*j|u|)j^*$ = $j^*(j|u|j^*)$. The endomorphism $j|u|j^*$ of $E_2$ is positive. The linear mapping $j^*$ is partially isometric, with initial space F and final space I (EVT, V, p. 41, Prop. 5), and the linear mappings of I into F (resp. of F into I) deduced from $j$ and $j^*$ by passage to the subspaces are isomorphisms inverse to one another (loc. cit.). Hence Ker($j|u|j^*$) $=$ Ker($|u|j^*$) $=$ Ker($j^*$), since the image of $j^*$ is contained in Ker($u$)$^{\circ}=$ Ker($|u|$)$^{\circ}$. According to Prop. 10, d), the pair $(j^*, j|u|j^*)$ is the polar decomposition of $u^*$. This proves c), and assertion b) is then deduced from assertion a) applied to $u^*$.