Cosine-Sine Block Decomposition

Apply [F1] to $A$ to obtain unitary $L_0,R_0$ and $C_0=\mathrm{diag}(c_1,\dots ,c_k)$ with $A=L_0{C}_0{R}_0^{\ast }$ and $c_i\ge 0$. Since $A$ is the compression of the unitary $U$ to the first block, $\mathrm{\parallel }Av\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }$ for every $v$, so each singular value satisfies $c_i\le 1$. Put $S_0=\mathrm{diag}(s_i)$ with $s_i=\sqrt{1-c_i^2}$.

The first block row of $U{U}^{\ast }=I$ gives $A{A}^{\ast }+B{B}^{\ast }=I$, hence $(L_0^{\ast }B)(L_0^{\ast }B{)}^{\ast }=S_0^2$. For every $i$ with $s_i>0$, the row $s_i^{-1}{e}_i^{\ast }{L}_0^{\ast }B$ has norm one, and these rows are mutually orthogonal by the displayed Gram identity. Extend them to an orthonormal basis using [F3], and choose the signs of the completed rows so that a unitary $W$ satisfies $L_0^{\ast }B=-S_0{W}^{\ast }$.

The first block column of $U^{\ast }U=I$ gives $A^{\ast }A+C^{\ast }C=I$, hence $(C{R}_0{)}^{\ast }(C{R}_0)=S_0^2$. For every $i$ with $s_i>0$, the column $s_i^{-1}C{R}_0{e}_i$ has norm one, and these columns are mutually orthogonal. Extend them to an orthonormal basis using [F3]; the resulting unitary $Z$ satisfies $C{R}_0=Z{S}_0$. Therefore $\tilde{U}:=\mathrm{diag}(L_0^{\ast },Z^{\ast })U\mathrm{diag}(R_0,W)=\left(\begin{array}{cc}{C}_0& -S_0\\ S_0& E\end{array}\right)$ for some $E$.

The upper-right block of $\tilde{U}{\tilde{U}}^{\ast }=I$ gives $S_0{E}^{\ast }=C_0{S}_0$. Thus, for each $i$ with $s_i>0$, column $i$ of $E$ is $c_i{e}_i$. The lower-right blocks of $\tilde{U}{\tilde{U}}^{\ast }=I$ and ${\tilde{U}}^{\ast }\tilde{U}=I$ give $E{E}^{\ast }=E^{\ast }E=C_0^2$. For such $i$, the prescribed entry $E_{ii}=c_i$ already contributes $c_i^2$ to row $i$'s squared norm, so $E{E}^{\ast }=C_0^2$ forces every other entry in that row to vanish. Thus both the rows and columns indexed by $P=\{i:s_i>0\}$ have no cross-block entries. On $K=\{i:s_i=0\}$ one has $c_i=1$; the $K\times K$ block of $E$ has orthonormal rows and columns and is therefore unitary. This also covers $c_i=0$, where the corresponding row and column are zero, and $c_i=1$, where the index belongs to $K$.

Let $Q$ be that unitary $K\times K$ block and replace $Z$ by $Z^{\mathrm{\prime }}=Z\mathrm{diag}(I,Q)$. Since $Q$ acts only where $S_0=0$ and $C_0=I$, this leaves $C{R}_0=Z^{\mathrm{\prime }}{S}_0$ unchanged and changes the lower-right block to $C_0$. Hence $U=\mathrm{diag}(L_0,Z^{\mathrm{\prime }})\left(\begin{array}{cc}{C}_0& -S_0\\ S_0& C_0\end{array}\right)\mathrm{diag}(R_0^{\ast },W^{\ast })\mathrm{.}$ Both outside factors are unitary, and $C_0^2+S_0^2=I$ by definition. The argument includes $c_i=0$, $c_i=1$, repeated values, and empty positive or zero supports.