Polar decomposition

For a bounded operator $A$ on a Hilbert space, set $|A|=(A^{\ast }A{)}^{1/2}$. There is a unique partial isometry $U$, zero on $\ker A$, satisfying $A=U|A|$; on the range of $|A|$, define $U(|A|x)=Ax$ and extend by continuity. This is well-defined and isometric because $\Vert Ax{\Vert }^2=⟨A^{\ast }Ax,x⟩=\Vert |A|x{\Vert }^2$. For an invertible square matrix, $U$ is unitary.