Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 2 / b

Apply part (a) to the compact positive operator $T^{\ast }T$. Its positive square root $B=|T|$ is defined by multiplying each eigenvector of eigenvalue $\lambda$ by $\sqrt{\lambda }$ and vanishing on the kernel. The eigenvalues tend to zero, so $B$ is compact. For every $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a partial isometry with initial space $(\ker T{)}^{\perp }$, final space $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot 0$ with many different unitary $U$.