If $A$ is a contraction, show that $$\bex A^*(I-AA^*)^{1/2}=(I-A^*A)^{1/2}A^*. \eex$$ Use this to show that if $A$ is a contraction on $\scrH$, then the operators $$\bex U=\sex{\ba{cc} A&(I-AA^*)^{1/2}\\ (I-A^*A)^{1/2}&-A^* \ea}, \eex$$ $$\bex V=\sex{\ba{cc} A&-(I-AA^*)^{1/2}\\ (I-A^*A)^{1/2}&A^* \ea} \eex$$ are unitary operators on $\scrH\oplus \scrH$.
Solution.
(1). By the singular value decomposition, there exist unitaries $W,Q$ such that $$\bex A=WSQ^*,\quad S=\diag(s_1,\cdots,s_n),\quad s_i\geq 0, \eex$$ and hence $$\bex A^*=QSW^*. \eex$$ Consequently, $$\beex \ba{rlrl} AA^*&=WS^2W^*,&A^*A&=QS^2Q^*,\\ I-AA^*&=W(I-S^2)W^*,&I-A^*A&=Q(I-S^2)Q^*,\\ (I-AA^*)^{1/2}&=W\vLm W^*,& (I-A^*A^{1/2}&=Q\vLm Q^*, \ea \eeex$$ where $$\bex \vLm=\diag\sex{\sqrt{1-s_1^2},\cdots,\sqrt{1-s_n^2}}. \eex$$ Thus, $$\beex \bea A^*(I-AA^*)^{1/2}&=QS\vLm W^*\\ &=Q\diag\sex{s_1\sqrt{1-s_1^2},\cdots, s_n\sqrt{1-s_n^2}}W^*\\ &=Q\vLm S W^*\\ &=(I-A^*A)^{1/2} A^*. \eea \eeex$$
(2). As noticed in (1), $A$ is a contraction is equivalent to say that $A^*$ is a contraction. Direction computations with $$\bex A^*(I-AA^*)^{1/2}=(I-A^*A)^{1/2}A^*,\quad A(I-A^*A)^{1/2}=(I-AA^*)^{1/2}A \eex$$ yields the fact that $U,V$ are unitary.
