Proposition 10

Let $E_1$ and $E_2$ be complex Hilbert spaces and let $u\in \mathscr{L}(E_1; E_2)$.

a) The initial subspace and the final subspace of $|u|$ are both equal to the initial subspace of $u$ and one has $\||u|\|=\|u\|$;

b) There exists a unique partially isometric mapping $j$ of $E_1$ into $E_2$ such that Ker($j$) $=$ Ker($u$) and $u=j|u|$;

c) The initial (resp. final) subspace of $j$ is equal to that of $u$;

d) Let $u_1$ be a positive element of $\mathscr{L}(E_1)$ and $j_1$ a partially isometric element of $\mathscr{L}(E_1; E_2)$ such that $u=j_1u_1$ and Ker($j_1$) $=$ Ker($u_1$). Then $u_1=|u|$ and $j_1=j$.

For every $x\in E_1$, one has

This proves that Ker($u$) $=$ Ker($|u|$) and $\||u|\|=\|u\|$. Since $|u|$ is hermitian, the closure of the image of $|u|$ is the orthogonal complement of its kernel (EVT, V, p. 41, prop. 4), that is to say the initial space of $u$, whence a).

Formula (2) implies that there exists an isometric mapping $v$ of Im($|u|$) onto Im($u$) such that $v(|u|(x)) =u(x)$ for every $x\in E_1$. Let $j$ be the unique element of $\mathscr{L}(E_1; E_2)$ which extends $v$ and vanishes on Im($|u|$)$^{\circ}=$ Ker($|u|$) $=$ Ker($u$). Then $j$ has the properties of b). The uniqueness of $j$ follows from the decomposition E = Ker($u$)$\oplus$ Im($|u|$).

The initial subspace of $j$ is Ker($j$)$^{\circ}=$ Ker($u$)$^{\circ}$, the initial space of $u$. Its final subspace is $j$(Ker($u$)$^{\circ}$) $=j$(Im($|u|$)) $=$ Im($u$), the final subspace of $u$. This proves c).

Let now $u_1$ and $j_1$ be as in d). We have $u^*u=u_1j_1^*j_1u_1$. The mapping $j_1^*j_1$ is the orthogonal projector with kernel Ker($j_1$) $=$ Ker($u_1$) (EVT, V, p. 41, prop. 5 (ii)) and hence with image Im($u_1$). Therefore $u^*u=u^2_1$ and hence $u_1= (u^*u)^{1/2}=|u|($I, p. 118, prop. 16). The uniqueness assertion of b) finally implies that $j_1=j$.