(1). The singular value decomposition leads tot eh polar decomposition: Every operator $A$ can be written as $A=UP$, where $U$ is unitary and $P$ is positive. In this decomposition the positive part $P$ is unique, $P=|A|$. The unitary part $U$ is unique if $A$ is invertible.

(2). An operator $A$ is normal if and only if the factors $U$ and $P$ in the polar decomposition of $A$ commute.

(3). We have derived the polar decomposition from the singular value decomposition. Show that it is possible to derive the latter from the former.

Solution.  

(1). By the singular value decomposition, there exists unitaries $W$ and $Q$ such that $$\bex A=WSQ^*, \eex$$ and thus $$\bex A=WQ^*\cdot QSQ^*. \eex$$ Setting $$\bex U=WQ^*,\quad P=QSQ^*=|A|, \eex$$ we are completed.

(2). $\ra$: By density argument, we may assume $A$ is invertible. Suppose $A$ is normal and $A=UP$ is the polar decomposition, then by the spectral theorem, there exists a unitary $V$ such that $$\bex A=V\vLm V^*,\quad \vLa=\diag(\lm_1,\cdots,\lm_n). \eex$$ By the uniqueness part of (1), $$\bex U=V\sgn(\vLm)V^*,\quad P=V|\vLm|V^*, \eex$$ and thus $UP=PU=A$. $\la$: Suppose $A=UP$ is the polar decomposition with $UP=PU$, then $$\bex A^*A=PU^*UP=P^2, \eex$$ $$\bex AA^*=UP\cdot(UP)^*=PU\cdot (PU)^* =PUU^*P=P^2. \eex$$

(3). Suppose $A=UP$ is the polar decomposition, then by the spectral theorem, there exists a unitary $V$ such that $$\bex P=V\diag(s_1,\cdots,s_n)V^*,\quad s_i\geq 0. \eex$$ Hence, $$\bex A=UV\cdot \diag(s_1,\cdots,s_n)\cdot V^*. \eex$$

[Bhatia.Matrix Analysis.Solutions to Exercises and Problems]ExI.2.4